# Sum to 100

Sum to 100
You are encouraged to solve this task according to the task description, using any language you may know.
Task

Find solutions to the   sum to one hundred   puzzle.

Add (insert) the mathematical operators     +   or   -     (plus or minus)   before any of the digits in the
decimal numeric string   123456789   such that the resulting mathematical expression adds up to a
particular sum   (in this iconic case,   100).

Example:

```           123 + 4 - 5 + 67 - 89   =   100
```

Show all output here.

•   Show all solutions that sum to   100
•   Show the sum that has the maximum   number   of solutions   (from zero to infinity*)
•   Show the lowest positive sum that   can't   be expressed   (has no solutions), using the rules for this task
•   Show the ten highest numbers that can be expressed using the rules for this task   (extra credit)

An example of a sum that can't be expressed (within the rules of this task) is:   5074
(which, of course, isn't the lowest positive sum that can't be expressed).

*   (where   infinity   would be a relatively small   123,456,789)

## Ada

### The Package Sum_To

Between any two consecutive digits, there can be a "+", a "-", or no operator. E.g., the digits "4" and "5" occur in the string as either of the following three substrings: "4+5", "4-5", or "45". For the first digit, we only have two choices: "+1" (written as "1"), and "-1". This makes 2*3^8 (two times (three to the power of eight)) different strings. Essential is the generic function Eval in the package Sum_To calls the procedure Callback for each such string Str, with the number Int holding the sum corresponding to the evaluation of Str. The second generic procedure Print is for convenience. If the Sum fits the condition, i.e., if Print_If(Sum, Number), then Print writes Sum = Str to the output.

`package Sum_To is    generic      with procedure Callback(Str: String; Int: Integer);   procedure Eval;    generic      Number: Integer;      with function Print_If(Sum, Number: Integer) return Boolean;    procedure Print(S: String; Sum: Integer); end Sum_To;`

The implementation of Eval follows the observation above: Eval calls Rec_Eval with the initial string "1" and "-1". For each call, Rec_Eval recursively evaluates a ternary tree with 3^8 leafs. At each leaf, Rec_Eval calls Callback. The implementation of Print is straightforward.

`with Ada.Text_IO, Ada.Containers.Ordered_Maps; package body Sum_To is    procedure Eval is       procedure Rec_Eval(Str: String; Previous, Current, Next: Integer) is	 Next_Image: String := Integer'Image(Next); 	 -- Next_Image(1) holds a blank, Next_Image(2) a digit 	 function Sign(N: Integer) return Integer is 	    (if N<0 then -1 elsif N>0 then 1 else 0);       begin	 if Next = 10 then -- end of recursion	    Callback(Str, Previous+Current);	 else -- Next < 10	    Rec_Eval(Str & Next_Image(2), -- concatenate current and Next 		 Previous, Sign(Current)*(10*abs(Current)+Next), Next+1);	    Rec_Eval(Str & "+" & Next_Image(2), -- add Next 		 Previous+Current, Next, Next+1);	    Rec_Eval(Str & "-" & Next_Image(2), -- subtract Next 		 Previous+Current, -Next, Next+1);	 end if;      end Rec_Eval;    begin -- Eval      Rec_Eval("1", 0, 1, 2);  -- unary "+", followed by "1"      Rec_Eval("-1", 0, -1, 2); -- unary "-", followed by "1"   end Eval;    procedure Print(S: String; Sum: Integer) is      -- print solution (S,N), if N=Number   begin      if Print_If(Sum, Number) then 	 Ada.Text_IO.Put_Line(Integer'Image(Sum) & " = " & S & ";");      end if;   end Print; end Sum_To;`

### The First Subtask

Given the package Sum_To, the solution to the first subtask (print all solution for the sum 100) is trivial: Eval_100 calls Print_100 for all 2*3^8 strings, and Print_100 writes the output if the sum is equal to 100.

`with Sum_To;  procedure Sum_To_100 is    procedure Print_100 is new Sum_To.Print(100, "=");   procedure Eval_100 is new Sum_To.Eval(Print_100); begin   Eval_100;end Sum_To_100;`
Output:
``` 100 = 123+45-67+8-9;
100 = 123+4-5+67-89;
100 = 123-45-67+89;
100 = 123-4-5-6-7+8-9;
100 = 12+3+4+5-6-7+89;
100 = 12+3-4+5+67+8+9;
100 = 12-3-4+5-6+7+89;
100 = 1+23-4+56+7+8+9;
100 = 1+23-4+5+6+78-9;
100 = 1+2+34-5+67-8+9;
100 = 1+2+3-4+5+6+78+9;
100 = -1+2-3+4+5+6+78+9;```

### The other subtasks (including the extra credit)

For the three other subtasks, we maintain an ordered map of sums (as the keys) and counters for the number of solutions (as the elements). The procedure Generate_Map generates the Map by calling the procedure Insert_Solution for all 2*3^8 solutions. Finding (1) the sum with the maximal number of solutions, (2) the first sum>=0 without a solution and (3) the ten largest sums with a solution (extra credit) are done by iterating this map.

`with Sum_To, Ada.Containers.Ordered_Maps, Ada.Text_IO;use Ada.Text_IO; procedure Three_Others is    package Num_Maps is new Ada.Containers.Ordered_Maps     (Key_Type => Integer, Element_Type => Positive);   use Num_Maps;    Map: Num_Maps.Map;   -- global Map stores how often a sum did occur    procedure Insert_Solution(S: String; Sum: Integer) is      -- inserts a solution into global Map      use Num_Maps;      -- use type Num_Maps.Cursor;      Position: Cursor := Map.Find(Sum);   begin      if Position = No_Element then -- first solutions for Sum	 Map.Insert(Key => Sum, New_Item => 1); -- counter is 1      else -- increase counter for Sum	 Map.Replace_Element(Position => Position,			     New_Item => (Element(Position))+1);      end if;   end Insert_Solution;    procedure Generate_Map is new Sum_To.Eval(Insert_Solution);     Current: Cursor; -- Points into Map   Sum: Integer;    -- current Sum of interest   Max: Natural; begin   Generate_Map;    -- find Sum >= 0  with maximum number of solutions   Max := 0; -- number of solutions for Sum (so far, none)   Current := Map.Ceiling(0); -- first element in Map with Sum >= 0   while Has_Element(Current) loop      if Element(Current) > Max then	 Max := Element(Current); -- the maximum of solutions, so far	 Sum := Key(Current);     -- the Sum with Max solutions      end if;      Next(Current);   end loop;   Put_Line("Most frequent result:" & Integer'Image(Sum));   Put_Line("Frequency of" & Integer'Image(Sum) & ":" & 	      Integer'Image(Max));   New_Line;    -- find smallest Sum >= 0 with no solution   Sum := 0;   while Map.Find(Sum) /= No_Element loop      Sum := Sum + 1;   end loop;   Put_Line("Smallest nonnegative impossible sum:" & Integer'Image(Sum));   New_Line;    -- find ten highest numbers with a solution    Current := Map.Last; -- highest element in Map with a solution   Put_Line("Highest sum:" & Integer'Image(Key(Current)));   Put("Next nine:");   for I in 1 .. 9 loop -- 9 steps backward      Previous(Current);      Put(Integer'Image(Key(Current)));   end loop;    New_Line;end Three_others;`
Output:
```Most frequent result: 9
Frequency of 9: 46

Smallest nonnegative impossible sum: 211

Highest sum: 123456789
Next nine: 23456790 23456788 12345687 12345669 3456801 3456792 3456790 3456788 3456786```

## Aime

`integer b, i, j, k, l, p, s, z;index r, w; i = 0;while (i < 512) {    b = i.bcount;    j = 0;    while (j < 1 << b) {        data e;         j += 1;         k = s = p = 0;        l = j;        z = 1;        while (k < 9) {            if (i & 1 << k) {                e.append("-+"[l & 1]);                s += p * z;                z = (l & 1) * 2 - 1;                l >>= 1;                p = 0;            }            e.append('1' + k);            p = p * 10 + 1 + k;             k += 1;        }         s += p * z;         if (e != '+') {            if (s == 100) {                o_(e, "\n");            }             w[s] += 1;        }    }     i += 1;} w.wcall(i_fix, 1, 1, r); o_(r.back, "\n"); k = 0;for (+k in w) {    if (!w.key(k + 1)) {        o_(k + 1, "\n");        break;    }} i = 10;for (k of w) {    o_(k, "\n");    if (!(i -= 1)) {        break;    }}`
Output:
```123-45-67+89
123+4-5+67-89
12+3+4+5-6-7+89
12-3-4+5-6+7+89
1+23-4+5+6+78-9
1+2+3-4+5+6+78+9
-1+2-3+4+5+6+78+9
123+45-67+8-9
1+2+34-5+67-8+9
12+3-4+5+67+8+9
1+23-4+56+7+8+9
123-4-5-6-7+8-9
9
211
123456789
23456790
23456788
12345687
12345669
3456801
3456792
3456790
3456788
3456786```

## ALGOL 68

`BEGIN    # find the numbers the string 123456789 ( with "+/-" optionally inserted  #    # before each digit ) can generate                                        #     # experimentation shows that the largest hundred numbers that can be      #    # generated are are greater than or equal to 56795                        #    # as we can't declare an array with bounds -123456789 : 123456789 in      #    # Algol 68G, we use -60000 : 60000 and keep counts for the top hundred    #     INT max number = 60 000;    [ - max number : max number ]STRING solutions;    [ - max number : max number ]INT    count;    FOR i FROM LWB solutions TO UPB solutions DO solutions[ i ] := ""; count[ i ] := 0 OD;     # calculate the numbers ( up to max number ) we can generate and the strings leading to them  #    # also determine the largest numbers we can generate #    [ 100 ]INT largest;    [ 100 ]INT largest count;    INT impossible number = - 999 999 999;    FOR i FROM LWB largest TO UPB largest DO        largest      [ i ] := impossible number;        largest count[ i ] := 0    OD;    [ 1 : 18 ]CHAR sum string := ".1.2.3.4.5.6.7.8.9";    []CHAR sign char = []CHAR( "-", " ", "+" )[ AT -1 ];    # we don't distinguish between strings starting "+1" and starting " 1" #    FOR s1 FROM -1 TO 0 DO        sum string[  1 ] := sign char[ s1 ];        FOR s2 FROM -1 TO 1 DO            sum string[  3 ] := sign char[ s2 ];            FOR s3 FROM -1 TO 1 DO                sum string[  5 ] := sign char[ s3 ];                FOR s4 FROM -1 TO 1 DO                    sum string[  7 ] := sign char[ s4 ];                    FOR s5 FROM -1 TO 1 DO                        sum string[  9 ] := sign char[ s5 ];                        FOR s6 FROM -1 TO 1 DO                            sum string[ 11 ] := sign char[ s6 ];                            FOR s7 FROM -1 TO 1 DO                                sum string[ 13 ] := sign char[ s7 ];                                FOR s8 FROM -1 TO 1 DO                                    sum string[ 15 ] := sign char[ s8 ];                                    FOR s9 FROM -1 TO 1 DO                                        sum string[ 17 ] := sign char[ s9 ];                                        INT number := 0;                                        INT part   := IF s1 < 0 THEN -1 ELSE 1 FI;                                        IF s2 = 0 THEN part *:= 10 +:= 2 * SIGN part ELSE number +:= part; part := 2 * s2 FI;                                        IF s3 = 0 THEN part *:= 10 +:= 3 * SIGN part ELSE number +:= part; part := 3 * s3 FI;                                        IF s4 = 0 THEN part *:= 10 +:= 4 * SIGN part ELSE number +:= part; part := 4 * s4 FI;                                        IF s5 = 0 THEN part *:= 10 +:= 5 * SIGN part ELSE number +:= part; part := 5 * s5 FI;                                        IF s6 = 0 THEN part *:= 10 +:= 6 * SIGN part ELSE number +:= part; part := 6 * s6 FI;                                        IF s7 = 0 THEN part *:= 10 +:= 7 * SIGN part ELSE number +:= part; part := 7 * s7 FI;                                        IF s8 = 0 THEN part *:= 10 +:= 8 * SIGN part ELSE number +:= part; part := 8 * s8 FI;                                        IF s9 = 0 THEN part *:= 10 +:= 9 * SIGN part ELSE number +:= part; part := 9 * s9 FI;                                        number +:= part;                                        IF  number >= LWB solutions                                        AND number <= UPB solutions                                        THEN                                            solutions[ number ] +:= ";" + sum string;                                            count    [ number ] +:= 1                                        FI;                                        BOOL inserted := FALSE;                                        FOR l pos FROM LWB largest TO UPB largest WHILE NOT inserted DO                                            IF number > largest[ l pos ] THEN                                                # found a new larger number #                                                FOR m pos FROM UPB largest BY -1 TO l pos + 1 DO                                                    largest      [ m pos ] := largest      [ m pos - 1 ];                                                    largest count[ m pos ] := largest count[ m pos - 1 ]                                                OD;                                                largest      [ l pos ] := number;                                                largest count[ l pos ] := 1;                                                inserted := TRUE                                            ELIF number = largest[ l pos ] THEN                                                # have another way of generating this number #                                                largest count[ l pos ] +:= 1;                                                inserted := TRUE                                            FI                                        OD                                    OD                                OD                            OD                        OD                    OD                OD            OD        OD    OD;     # show the solutions for 100 #    print( ( "100 has ", whole( count[ 100 ], 0 ), " solutions:" ) );    STRING s := solutions[ 100 ];    FOR s pos FROM LWB s TO UPB s DO        IF   s[ s pos ] = ";" THEN print( ( newline, "        " ) )        ELIF s[ s pos ] /= " " THEN print( ( s[ s pos ] ) )        FI    OD;    print( ( newline ) );    # find the number with the most solutions #    INT max solutions := 0;    INT number with max := LWB count - 1;    FOR n FROM 0 TO max number DO        IF count[ n ] > max solutions THEN            max solutions := count[ n ];            number with max := n        FI    OD;    FOR n FROM LWB largest count TO UPB largest count DO        IF largest count[ n ] > max solutions THEN            max solutions := largest count[ n ];            number with max := largest[ n ]        FI    OD;    print( ( whole( number with max, 0 ), " has the maximum number of solutions: ", whole( max solutions, 0 ), newline ) );    # find the smallest positive number that has no solutions #    BOOL have solutions := TRUE;    FOR n FROM 0 TO max number    WHILE IF NOT ( have solutions := count[ n ] > 0 )          THEN print( ( whole( n, 0 ), " is the lowest positive number with no solutions", newline ) )          FI;          have solutions    DO SKIP OD;    IF have solutions    THEN print( ( "All positive numbers up to ", whole( max number, 0 ), " have solutions", newline ) )    FI;    print( ( "The 10 largest numbers that can be generated are:", newline ) );    FOR t pos FROM 1 TO 10 DO        print( ( " ", whole( largest[ t pos ], 0 ) ) )    OD;    print( ( newline ) ) END`
Output:
```100 has 12 solutions:
-1+2-3+4+5+6+78+9
12-3-4+5-6+7+89
123-4-5-6-7+8-9
123-45-67+89
123+4-5+67-89
123+45-67+8-9
12+3-4+5+67+8+9
12+3+4+5-6-7+89
1+23-4+56+7+8+9
1+23-4+5+6+78-9
1+2+3-4+5+6+78+9
1+2+34-5+67-8+9
9 has the maximum number of solutions: 46
211 is the lowest positive number with no solutions
The 10 largest numbers that can be generated are:
123456789 23456790 23456788 12345687 12345669 3456801 3456792 3456790 3456788 3456786
```

## AppleScript

Translation of: JavaScript

AppleScript is essentially out of its depth at this scale. The first task (number of distinct paths to 100) is accessible within a few seconds. Subsequent tasks, however, terminate only (if at all) after impractical amounts of time. Note the contrast with the lighter and more optimised JavaScript interpreter, which takes less than half a second to return full results for all the listed tasks.

`use framework "Foundation" -- for basic NSArray sort property pSigns : {1, 0, -1} --> ( + | unsigned | - )property plst100 : {"Sums to 100:", ""}property plstSums : {}property plstSumsSorted : missing valueproperty plstSumGroups : missing value -- data Sign :: [ 1 | 0 | -1 ] = ( Plus | Unsigned | Minus )-- asSum :: [Sign] -> Inton asSum(xs)    script        on |λ|(a, sign, i)            if sign ≠ 0 then                {digits:{}, n:(n of a) + (sign * ((i & digits of a) as string as integer))}            else                {digits:{i} & (digits of a), n:n of a}            end if        end |λ|    end script     set rec to foldr(result, {digits:{}, n:0}, xs)    set ds to digits of rec    if length of ds > 0 then        (n of rec) + (ds as string as integer)    else        n of rec    end ifend asSum -- data Sign :: [ 1 | 0 | -1 ] = ( Plus | Unisigned | Minus )-- asString :: [Sign] -> Stringon asString(xs)    script        on |λ|(a, sign, i)            set d to i as string            if sign ≠ 0 then                if sign > 0 then                    a & " +" & d                else                    a & " -" & d                end if            else                a & d            end if        end |λ|    end script     foldl(result, "", xs)end asString -- sumsTo100 :: () -> Stringon sumsTo100()    -- From first permutation without leading '+' (3 ^ 8) to end of universe (3 ^ 9)    repeat with i from 6561 to 19683        set xs to nthPermutationWithRepn(pSigns, 9, i)        if asSum(xs) = 100 then set end of plst100 to asString(xs)    end repeat    intercalate(linefeed, plst100)end sumsTo100  -- mostCommonSum :: () -> Stringon mostCommonSum()    -- From first permutation without leading '+' (3 ^ 8) to end of universe (3 ^ 9)    repeat with i from 6561 to 19683        set intSum to asSum(nthPermutationWithRepn(pSigns, 9, i))        if intSum ≥ 0 then set end of plstSums to intSum    end repeat     set plstSumsSorted to sort(plstSums)    set plstSumGroups to group(plstSumsSorted)     script groupLength        on |λ|(a, b)            set intA to length of a            set intB to length of b            if intA < intB then                -1            else if intA > intB then                1            else                0            end if        end |λ|    end script     set lstMaxSum to maximumBy(groupLength, plstSumGroups)    intercalate(linefeed, ¬        {"Most common sum: " & item 1 of lstMaxSum, ¬            "Number of instances: " & length of lstMaxSum})end mostCommonSum  -- TEST ----------------------------------------------------------------------on run    return sumsTo100()     -- Also returns a value, but slow:    -- mostCommonSum()end run  -- GENERIC FUNCTIONS --------------------------------------------------------- -- nthPermutationWithRepn :: [a] -> Int -> Int -> [a]on nthPermutationWithRepn(xs, groupSize, iIndex)    set intBase to length of xs    set intSetSize to intBase ^ groupSize     if intBase < 1 or iIndex > intSetSize then        {}    else        set baseElems to inBaseElements(xs, iIndex)        set intZeros to groupSize - (length of baseElems)         if intZeros > 0 then            replicate(intZeros, item 1 of xs) & baseElems        else            baseElems        end if    end ifend nthPermutationWithRepn -- inBaseElements :: [a] -> Int -> [String]on inBaseElements(xs, n)    set intBase to length of xs     script nextDigit        on |λ|(residue)            set {divided, remainder} to quotRem(residue, intBase)             {valid:divided > 0, value:(item (remainder + 1) of xs), new:divided}        end |λ|    end script     reverse of unfoldr(nextDigit, n)end inBaseElements -- sort :: [a] -> [a]on sort(lst)    ((current application's NSArray's arrayWithArray:lst)'s ¬        sortedArrayUsingSelector:"compare:") as listend sort -- maximumBy :: (a -> a -> Ordering) -> [a] -> a on maximumBy(f, xs)    set cmp to mReturn(f)    script max        on |λ|(a, b)            if a is missing value or cmp's |λ|(a, b) < 0 then                b            else                a            end if        end |λ|    end script     foldl(max, missing value, xs)end maximumBy -- group :: Eq a => [a] -> [[a]]on group(xs)    script eq        on |λ|(a, b)            a = b        end |λ|    end script     groupBy(eq, xs)end group -- groupBy :: (a -> a -> Bool) -> [a] -> [[a]]on groupBy(f, xs)    set mf to mReturn(f)     script enGroup        on |λ|(a, x)            if length of (active of a) > 0 then                set h to item 1 of active of a            else                set h to missing value            end if             if h is not missing value and mf's |λ|(h, x) then                {active:(active of a) & x, sofar:sofar of a}            else                {active:{x}, sofar:(sofar of a) & {active of a}}            end if        end |λ|    end script     if length of xs > 0 then        set dct to foldl(enGroup, {active:{item 1 of xs}, sofar:{}}, tail(xs))        if length of (active of dct) > 0 then            sofar of dct & {active of dct}        else            sofar of dct        end if    else        {}    end ifend groupBy -- tail :: [a] -> [a]on tail(xs)    if length of xs > 1 then        items 2 thru -1 of xs    else        {}    end ifend tail  -- intercalate :: Text -> [Text] -> Texton intercalate(strText, lstText)    set {dlm, my text item delimiters} to {my text item delimiters, strText}    set strJoined to lstText as text    set my text item delimiters to dlm    return strJoinedend intercalate --  quotRem :: Integral a => a -> a -> (a, a)on quotRem(m, n)    {m div n, m mod n}end quotRem -- replicate :: Int -> a -> [a]on replicate(n, a)    set out to {}    if n < 1 then return out    set dbl to {a}     repeat while (n > 1)        if (n mod 2) > 0 then set out to out & dbl        set n to (n div 2)        set dbl to (dbl & dbl)    end repeat    return out & dblend replicate -- foldr :: (a -> b -> a) -> a -> [b] -> aon foldr(f, startValue, xs)    tell mReturn(f)        set v to startValue        set lng to length of xs        repeat with i from lng to 1 by -1            set v to |λ|(v, item i of xs, i, xs)        end repeat        return v    end tellend foldr -- foldl :: (a -> b -> a) -> a -> [b] -> aon foldl(f, startValue, xs)    tell mReturn(f)        set v to startValue        set lng to length of xs        repeat with i from 1 to lng            set v to |λ|(v, item i of xs, i, xs)        end repeat        return v    end tellend foldl -- unfoldr :: (b -> Maybe (a, b)) -> b -> [a]on unfoldr(f, v)    set mf to mReturn(f)    set lst to {}    set recM to mf's |λ|(v)    repeat while (valid of recM) is true        set end of lst to value of recM        set recM to mf's |λ|(new of recM)    end repeat    lst & value of recMend unfoldr -- until :: (a -> Bool) -> (a -> a) -> a -> aon |until|(p, f, x)    set mp to mReturn(p)    set v to x     tell mReturn(f)        repeat until mp's |λ|(v)            set v to |λ|(v)        end repeat    end tell    return vend |until| -- map :: (a -> b) -> [a] -> [b]on map(f, xs)    tell mReturn(f)        set lng to length of xs        set lst to {}        repeat with i from 1 to lng            set end of lst to |λ|(item i of xs, i, xs)        end repeat        return lst    end tellend map  -- Lift 2nd class handler function into 1st class script wrapper -- mReturn :: Handler -> Scripton mReturn(f)    if class of f is script then        f    else        script            property |λ| : f        end script    end ifend mReturn`
Output:
```Sums to 100:

1 +2 +34 -5 +67 -8 +9
1 +2 +3 -4 +5 +6 +78 +9
1 +23 -4 +5 +6 +78 -9
1 +23 -4 +56 +7 +8 +9
12 +3 +4 +5 -6 -7 +89
12 +3 -4 +5 +67 +8 +9
123 +45 -67 +8 -9
123 +4 -5 +67 -89
123 -45 -67 +89
123 -4 -5 -6 -7 +8 -9
12 -3 -4 +5 -6 +7 +89
-1 +2 -3 +4 +5 +6 +78 +9```

## AutoHotkey

 This example is incomplete. The output is incomplete, please address the 2nd and 3rd task requirements. Please ensure that it meets all task requirements and remove this message.
`global Matches:=[]AllPossibilities100()for eq, val in matches	res .= eq "`n"MsgBox % resreturn AllPossibilities100(n:=0,  S:="") {	if (n = 0)							; First call		AllPossibilities100(n+1, n)				; Recurse	else if (n < 10){		AllPossibilities100(n+1, 	S ",-" n)		; Recurse. Concatenate S, ",-" and n		AllPossibilities100(n+1, 	S ",+" n)		; Recurse. Concatenate S, ",+" and n 		AllPossibilities100(n+1, 	S n)			; Recurse. Concatenate S and n	} else 	{							; 10th level recursion		Loop, Parse, S, CSV					; Total the values of S and check if equal to 100		{			SubS := SubStr(A_LoopField, 2)			; The number portion of A_LoopField			if (A_Index = 1)				Total := A_LoopField			else if (SubStr(A_LoopField, 1, 1) = "+")	; If the first character is + add				Total += SubS			else						; else subtract				Total -= SubS		}		if (Total = 100)			matches[LTrim(LTrim(StrReplace(S, ","), "0"),"+")] := true ; remove leading 0's, +'s and all commas	}}`
Outputs:
```-1+2-3+4+5+6+78+9
1+2+3-4+5+6+78+9
1+2+34-5+67-8+9
1+23-4+5+6+78-9
1+23-4+56+7+8+9
12+3+4+5-6-7+89
12+3-4+5+67+8+9
12-3-4+5-6+7+89
123+4-5+67-89
123+45-67+8-9
123-4-5-6-7+8-9
123-45-67+89```

## AWK

Translation of: Fortran IV

Awk is a weird language: there are no integers, no switch-case (in the standard language version), programs are controlled by data flow, the interpreter speed is moderate. The advantage of Awk are associative arrays, used here for counting how many times we get the same sum as the result of calculations.

`# # RossetaCode: Sum to 100, AWK.## Find solutions to the "sum to one hundred" puzzle. function evaluate(code){    value  = 0    number = 0    power  = 1    for ( k = 9; k >= 1; k-- )    {        number = power*k + number        op = code % 3        if ( op == 0 ) {            value = value + number            number = 0            power = 1        } else if (op == 1 ) {            value = value - number            number = 0            power = 1         } else if ( op == 2) {            power = power * 10        } else {        }        code = int(code / 3);    }    return value;    } function show(code){    s = ""    a = 19683    b = 6561     for ( k = 1; k <= 9; k++ )    {           op = int( (code % a) / b )        if ( op == 0 && k > 1 )             s = s "+"        else if ( op == 1 )            s = s "-"        else {        }              a = b        b = int(b / 3)        s = s  k    }    printf "%9d = %s\n", evaluate(code), s;}  BEGIN {    nexpr = 13122     print    print "Show all solutions that sum to 100"    print    for ( i = 0; i < nexpr; i++ ) if ( evaluate(i) == 100 ) show(i);        print    print "Show the sum that has the maximum number of solutions"    print    for ( i = 0; i < nexpr; i++ ) {        sum = evaluate(i);        if ( sum >= 0 )            stat[sum]++;    }    best = (-1);      for ( sum in stat )         if ( best < stat[sum] ) {             best = stat[sum]             bestSum = sum         }    delete stat    printf "%d has %d solutions\n", bestSum, best     print    print "Show the lowest positive number that can't be expressed"    print        for ( i = 0; i <= 123456789; i++ ){        for ( j = 0; j < nexpr; j++ )             if ( i == evaluate(j) )                 break;         if ( i != evaluate(j) )             break;    }    printf "%d\n",i     print    print "Show the ten highest numbers that can be expressed"    print    limit = 123456789 + 1;    for ( i = 1; i <= 10; i++ )     {        best = 0;        for ( j = 0; j < nexpr; j++ )        {            test = evaluate(j);            if ( test < limit && test > best ) best = test;        }        for ( j = 0; j < nexpr; j++ ) if ( evaluate(j) == best ) show(j)        limit = best    }}`
Output:
```Show all solutions that sum to 100

100 = 1+2+3-4+5+6+78+9
100 = 1+2+34-5+67-8+9
100 = 1+23-4+5+6+78-9
100 = 1+23-4+56+7+8+9
100 = 12+3+4+5-6-7+89
100 = 12+3-4+5+67+8+9
100 = 12-3-4+5-6+7+89
100 = 123+4-5+67-89
100 = 123+45-67+8-9
100 = 123-4-5-6-7+8-9
100 = 123-45-67+89
100 = -1+2-3+4+5+6+78+9

Show the sum that has the maximum number of solutions

9 has 46 solutions

Show the lowest positive number that can't be expressed

211

Show the ten highest numbers that can be expressed

123456789 = 123456789
23456790 = 1+23456789
23456788 = -1+23456789
12345687 = 12345678+9
12345669 = 12345678-9
3456801 = 12+3456789
3456792 = 1+2+3456789
3456790 = -1+2+3456789
3456788 = 1-2+3456789
3456786 = -1-2+3456789```

## C

### Optimized for speed

Works with: GCC version 5.1
Works with: Microsoft Visual Studio version 2015

Warning: this version requires at least four byte integers.

`/*  * RossetaCode: Sum to 100, C99, an algorithm using ternary numbers. * * Find solutions to the "sum to one hundred" puzzle. */ #include <stdio.h> #include <stdlib.h> /* * There are only 13122 (i.e. 2*3**8) different possible expressions, * thus we can encode them as positive integer numbers from 0 to 13121. */#define NUMBER_OF_EXPRESSIONS (2 * 3*3*3*3 * 3*3*3*3 )enum OP { ADD, SUB, JOIN };typedef int (*cmp)(const void*, const void*); // Replacing struct Expression and struct CountSum by a tuple like // struct Pair { int first; int last; } is possible but would make the source// code less readable. struct Expression{     int sum;    int code;}expressions[NUMBER_OF_EXPRESSIONS];int expressionsLength = 0;int compareExpressionBySum(const struct Expression* a, const struct Expression* b){    return a->sum - b->sum;} struct CountSum{     int counts;     int sum; }countSums[NUMBER_OF_EXPRESSIONS];int countSumsLength = 0;int compareCountSumsByCount(const struct CountSum* a, const struct CountSum* b){    return a->counts - b->counts;} int evaluate(int code){    int value  = 0, number = 0, power  = 1;    for ( int k = 9; k >= 1; k-- ){        number = power*k + number;        switch( code % 3 ){            case ADD:  value = value + number; number = 0; power = 1; break;            case SUB:  value = value - number; number = 0; power = 1; break;            case JOIN: power = power * 10                ; break;        }        code /= 3;    }    return value;    } void print(int code){    static char s; char* p = s;    int a = 19683, b = 6561;            for ( int k = 1; k <= 9; k++ ){        switch((code % a) / b){            case ADD: if ( k > 1 ) *p++ = '+'; break;            case SUB:              *p++ = '-'; break;        }        a = b;        b = b / 3;        *p++ = '0' + k;    }    *p = 0;    printf("%9d = %s\n", evaluate(code), s);} void comment(char* string){    printf("\n\n%s\n\n", string);} void init(void){       for ( int i = 0; i < NUMBER_OF_EXPRESSIONS; i++ ){        expressions[i].sum = evaluate(i);        expressions[i].code = i;    }    expressionsLength = NUMBER_OF_EXPRESSIONS;    qsort(expressions,expressionsLength,sizeof(struct Expression),(cmp)compareExpressionBySum);     int j = 0;    countSums.counts = 1;    countSums.sum = expressions.sum;    for ( int i = 0; i < expressionsLength; i++ ){        if ( countSums[j].sum != expressions[i].sum ){            j++;            countSums[j].counts = 1;            countSums[j].sum = expressions[i].sum;        }               else            countSums[j].counts++;    }    countSumsLength = j + 1;    qsort(countSums,countSumsLength,sizeof(struct CountSum),(cmp)compareCountSumsByCount);} int main(void){     init();     comment("Show all solutions that sum to 100");                const int givenSum = 100;    struct Expression ex = { givenSum, 0 };    struct Expression* found;    if ( found = bsearch(&ex,expressions,expressionsLength,        sizeof(struct Expression),(cmp)compareExpressionBySum) ){        while ( found != expressions && (found-1)->sum == givenSum )            found--;        while ( found != &expressions[expressionsLength] && found->sum == givenSum )            print(found++->code);    }     comment("Show the positve sum that has the maximum number of solutions");    int maxSumIndex = countSumsLength - 1;    while( countSums[maxSumIndex].sum < 0 )        maxSumIndex--;    printf("%d has %d solutions\n",         countSums[maxSumIndex].sum, countSums[maxSumIndex].counts);     comment("Show the lowest positive number that can't be expressed");    for ( int value = 0; ; value++ ){        struct Expression ex = { value, 0 };        if (!bsearch(&ex,expressions,expressionsLength,                sizeof(struct Expression),(cmp)compareExpressionBySum)){            printf("%d\n", value);            break;        }    }     comment("Show the ten highest numbers that can be expressed");    for ( int i = expressionsLength-1; i >= expressionsLength-10; i-- )        print(expressions[i].code);     return 0;}`
Output:
```Show all solutions that sum to 100

100 = 123+4-5+67-89
100 = 123-4-5-6-7+8-9
100 = 123-45-67+89
100 = 1+2+34-5+67-8+9
100 = 123+45-67+8-9
100 = 1+2+3-4+5+6+78+9
100 = 1+23-4+5+6+78-9
100 = 12-3-4+5-6+7+89
100 = 12+3+4+5-6-7+89
100 = -1+2-3+4+5+6+78+9
100 = 12+3-4+5+67+8+9
100 = 1+23-4+56+7+8+9

Show the positve sum that has the maximum number of solutions

9 has 46 solutions

Show the lowest positive number that can't be expressed

211

Show the ten highest numbers that can be expressed

123456789 = 123456789
23456790 = 1+23456789
23456788 = -1+23456789
12345687 = 12345678+9
12345669 = 12345678-9
3456801 = 12+3456789
3456792 = 1+2+3456789
3456790 = -1+2+3456789
3456788 = 1-2+3456789
3456786 = -1-2+3456789
```

### Optimized for memory consumption

Translation of: Fortran 95
Works with: GCC version 5.1

Warning: this program needs at least four byte integers.

`/*  * RossetaCode: Sum to 100, C11, MCU friendly. * * Find solutions to the "sum to one hundred" puzzle. * * We optimize algorithms for size. Therefore we don't use arrays, but recompute * all values again and again. It is a little surprise that the time efficiency  * is quite acceptable. */ #include <stdio.h> enum OP { ADD, SUB, JOIN }; int evaluate(int code){    int value  = 0, number = 0, power  = 1;    for ( int k = 9; k >= 1; k-- ){        number = power*k + number;        switch( code % 3 ){            case ADD:  value = value + number; number = 0; power = 1; break;            case SUB:  value = value - number; number = 0; power = 1; break;            case JOIN: power = power * 10                           ; break;        }        code /= 3;    }    return value;    } void print(int code){    static char s; char* p = s;    int a = 19683, b = 6561;            for ( int k = 1; k <= 9; k++ ){        switch((code % a) / b){            case ADD: if ( k > 1 ) *p++ = '+'; break;            case SUB:              *p++ = '-'; break;        }        a = b;        b = b / 3;        *p++ = '0' + k;    }    *p = 0;    printf("%9d = %s\n", evaluate(code), s);} int main(void){     int i,j;    const int nexpr = 13122;#define LOOP(K) for (K = 0; K < nexpr; K++)         puts("\nShow all solutions that sum to 100\n");    LOOP(i) if ( evaluate(i) == 100 ) print(i);     puts("\nShow the sum that has the maximum number of solutions\n");    int best, nbest = (-1);    LOOP(i){        int test = evaluate(i);        if ( test > 0 ){            int ntest = 0;            LOOP(j) if ( evaluate(j) == test ) ntest++;            if ( ntest > nbest ){ best = test; nbest = ntest; }        }    }    printf("%d has %d solutions\n", best,nbest);     puts("\nShow the lowest positive number that can't be expressed\n");    for ( i = 0; i <= 123456789; i++ ){        LOOP(j) if ( i == evaluate(j) ) break;         if ( i != evaluate(j) ) break;    }    printf("%d\n",i);     puts("\nShow the ten highest numbers that can be expressed\n");    int limit = 123456789 + 1;    for ( i = 1; i <= 10; i++ ) {        int best = 0;        LOOP(j){            int test = evaluate(j);            if ( test < limit && test > best ) best = test;        }        LOOP(j) if ( evaluate(j) == best ) print(j);        limit = best;    }     return 0;}`
Output:
```Show all solutions that sum to 100

100 = 1+2+3-4+5+6+78+9
100 = 1+2+34-5+67-8+9
100 = 1+23-4+5+6+78-9
100 = 1+23-4+56+7+8+9
100 = 12+3+4+5-6-7+89
100 = 12+3-4+5+67+8+9
100 = 12-3-4+5-6+7+89
100 = 123+4-5+67-89
100 = 123+45-67+8-9
100 = 123-4-5-6-7+8-9
100 = 123-45-67+89
100 = -1+2-3+4+5+6+78+9

Show the sum that has the maximum number of solutions

9 has 46 solutions

Show the lowest positive number that can't be expressed

211

Show the ten highest numbers that can be expressed

123456789 = 123456789
23456790 = 1+23456789
23456788 = -1+23456789
12345687 = 12345678+9
12345669 = 12345678-9
3456801 = 12+3456789
3456792 = 1+2+3456789
3456790 = -1+2+3456789
3456788 = 1-2+3456789
3456786 = -1-2+3456789
```

## C++

Works with: GCC version 5.1
Works with: Microsoft Visual Studio version 2010

For each expression of sum s, there is at least one expression whose sum is -s. If the sum s can be represented by n expressions, the sum -s can also be represented by n expressions. The change of all signs in an expression change the sign of the sum of this expression. For example, -1+23-456+789 has the opposite sign than +1-23+456-789. Therefore only the positive sum with the maximum number of solutions is shown. The program does not check uniqueness of this sum. We can easily check (modifying the program) that: sum 9 has 46 solutions; sum -9 has 46 solutions; any other sum has less than 46 solutions.

`/*  * RossetaCode: Sum to 100, C++, STL, OOP.  * Works with: MSC 16.0 (MSVS2010); GCC 5.1 (use -std=c++11 or -std=c++14 etc.). * * Find solutions to the "sum to one hundred" puzzle. */#include <iostream>#include <iomanip>#include <algorithm>#include <string>#include <set>#include <map> using namespace std; class Expression{    private:        enum { NUMBER_OF_DIGITS = 9 }; // hack for C++98, use const int in C++11        enum Op { ADD, SUB, JOIN };        int code[NUMBER_OF_DIGITS];    public:        static const int NUMBER_OF_EXPRESSIONS;        Expression(){            for ( int i = 0; i < NUMBER_OF_DIGITS; i++ )                code[i] = ADD;        }        Expression& operator++(int){ // post incrementation            for ( int i = 0; i < NUMBER_OF_DIGITS; i++ )                if ( ++code[i] > JOIN ) code[i] = ADD;                 else break;            return *this;                }        operator int() const{            int value = 0, number = 0, sign = (+1);            for ( int digit = 1; digit <= 9; digit++ )                switch ( code[NUMBER_OF_DIGITS - digit] ){                case ADD: value += sign*number; number = digit; sign = (+1); break;                case SUB: value += sign*number; number = digit; sign = (-1); break;                case JOIN:                      number = 10*number + digit;  break;            }            return value + sign*number;        }        operator string() const{            string s;            for ( int digit = 1; digit <= NUMBER_OF_DIGITS; digit++ ){                switch( code[NUMBER_OF_DIGITS - digit] ){                    case ADD: if ( digit > 1 ) s.push_back('+'); break;                    case SUB:                  s.push_back('-'); break;                }                s.push_back('0' + digit);            }            return s;        }};const int Expression::NUMBER_OF_EXPRESSIONS = 2 * 3*3*3*3 * 3*3*3*3; ostream& operator<< (ostream& os, Expression& ex){    ios::fmtflags oldFlags(os.flags());    os << setw(9) << right << static_cast<int>(ex)    << " = "        << setw(0) << left  << static_cast<string>(ex) << endl;    os.flags(oldFlags);    return os;} struct Stat{    map<int,int> countSum;    map<int, set<int> > sumCount;    Stat(){        Expression expression;        for ( int i = 0; i < Expression::NUMBER_OF_EXPRESSIONS; i++, expression++ )            countSum[expression]++;        for ( auto it = countSum.begin(); it != countSum.end(); it++ )            sumCount[it->second].insert(it->first);    }}; void print(int givenSum){    Expression expression;    for ( int i = 0; i < Expression::NUMBER_OF_EXPRESSIONS; i++, expression++ )        if ( expression == givenSum )             cout << expression;} void comment(string commentString){    cout << endl << commentString << endl << endl;} int main(){    Stat stat;     comment( "Show all solutions that sum to 100" );    const int givenSum = 100;    print(givenSum);     comment( "Show the sum that has the maximum number of solutions" );    auto maxi = max_element(stat.sumCount.begin(),stat.sumCount.end());    auto it = maxi->second.begin();     while ( *it < 0 ) it++;    cout << static_cast<int>(*it) << " has " << maxi->first << " solutions" << endl;     comment( "Show the lowest positive number that can't be expressed" );    int value = 0;     while(stat.countSum.count(value) != 0) value++;    cout << value << endl;     comment( "Show the ten highest numbers that can be expressed" );    auto rit = stat.countSum.rbegin();    for ( int i = 0; i < 10; i++, rit++ ) print(rit->first);     return 0;}`
Output:
```Show all solutions that sum to 100

100 = 1+2+3-4+5+6+78+9
100 = 1+2+34-5+67-8+9
100 = 1+23-4+5+6+78-9
100 = 1+23-4+56+7+8+9
100 = 12+3+4+5-6-7+89
100 = 12+3-4+5+67+8+9
100 = 12-3-4+5-6+7+89
100 = 123+4-5+67-89
100 = 123+45-67+8-9
100 = 123-4-5-6-7+8-9
100 = 123-45-67+89
100 = -1+2-3+4+5+6+78+9

Show the sum that has the maximum number of solutions

9 has 46 solutions

Show the lowest positive number that can't be expressed

211

Show the ten highest numbers that can be expressed

123456789 = 123456789
23456790 = 1+23456789
23456788 = -1+23456789
12345687 = 12345678+9
12345669 = 12345678-9
3456801 = 12+3456789
3456792 = 1+2+3456789
3456790 = -1+2+3456789
3456788 = 1-2+3456789
3456786 = -1-2+3456789
```

## C#

`using System;using System.Collections.Generic;using System.Linq; class Program{    static void Main(string[] args)    {        // All unique expressions that have a plus sign in front of the 1; calculated in parallel        var expressionsPlus = Enumerable.Range(0, (int)Math.Pow(3, 8)).AsParallel().Select(i => new Expression(i, 1));        // All unique expressions that have a minus sign in front of the 1; calculated in parallel        var expressionsMinus = Enumerable.Range(0, (int)Math.Pow(3, 8)).AsParallel().Select(i => new Expression(i, -1));        var expressions = expressionsPlus.Concat(expressionsMinus);        var results = new Dictionary<int, List<Expression>>();        foreach (var e in expressions)        {            if (results.Keys.Contains(e.Value))                results[e.Value].Add(e);            else                results[e.Value] = new List<Expression>() { e };        }        Console.WriteLine("Show all solutions that sum to 100");        foreach (Expression e in results)            Console.WriteLine("  " + e);        Console.WriteLine("Show the sum that has the maximum number of solutions (from zero to infinity)");        var summary = results.Keys.Select(k => new Tuple<int, int>(k, results[k].Count));        var maxSols = summary.Aggregate((a, b) => a.Item2 > b.Item2 ? a : b);        Console.WriteLine("  The sum " + maxSols.Item1 + " has " + maxSols.Item2 + " solutions.");        Console.WriteLine("Show the lowest positive sum that can't be expressed (has no solutions), using the rules for this task");        var lowestPositive = Enumerable.Range(1, int.MaxValue).First(x => !results.Keys.Contains(x));        Console.WriteLine("  " + lowestPositive);        Console.WriteLine("Show the ten highest numbers that can be expressed using the rules for this task (extra credit)");        var highest = from k in results.Keys                      orderby k descending                      select k;        foreach (var x in highest.Take(10))            Console.WriteLine("  " + x);    }}public enum Operations { Plus, Minus, Join };public class Expression{    protected Operations[] Gaps;    // 123456789 => there are 8 "gaps" between each number    ///             with 3 possibilities for each gap: plus, minus, or join    public int Value; // What this expression sums up to    protected int _one;     public Expression(int serial, int one)    {        _one = one;        Gaps = new Operations;        // This represents "serial" as a base 3 number, each Gap expression being a base-three digit        int divisor = 2187; // == Math.Pow(3,7)        int times;        for (int i = 0; i < 8; i++)        {            times = Math.DivRem(serial, divisor, out serial);            divisor /= 3;            if (times == 0)                Gaps[i] = Operations.Join;            else if (times == 1)                Gaps[i] = Operations.Minus;            else                Gaps[i] = Operations.Plus;        }        // go ahead and calculate the value of this expression        // because this is going to be done in a parallel thread (save time)        Value = Evaluate();    }    public override string ToString()    {        string ret = _one.ToString();        for (int i = 0; i < 8; i++)        {            switch (Gaps[i])            {                case Operations.Plus:                    ret += "+";                    break;                case Operations.Minus:                    ret += "-";                    break;            }            ret += (i + 2);        }        return ret;    }    private int Evaluate()        /* Calculate what this expression equals */    {        var numbers = new int;        int nc = 0;        var operations = new List<Operations>();        int a = 1;        for (int i = 0; i < 8; i++)        {            if (Gaps[i] == Operations.Join)                a = a * 10 + (i + 2);            else            {                if (a > 0)                {                    if (nc == 0)                        a *= _one;                    numbers[nc++] = a;                    a = i + 2;                }                operations.Add(Gaps[i]);            }        }        if (nc == 0)            a *= _one;        numbers[nc++] = a;        int ni = 0;        int left = numbers[ni++];        foreach (var operation in operations)        {            int right = numbers[ni++];            if (operation == Operations.Plus)                left = left + right;            else                left = left - right;        }        return left;    }}`
Output:
```Show all solutions that sum to 100
123-45-67+89
123-4-5-6-7+8-9
123+45-67+8-9
123+4-5+67-89
12-3-4+5-6+7+89
12+3-4+5+67+8+9
12+3+4+5-6-7+89
1+23-4+5+6+78-9
1+23-4+56+7+8+9
1+2+34-5+67-8+9
1+2+3-4+5+6+78+9
-1+2-3+4+5+6+78+9
Show the sum that has the maximum number of solutions (from zero to infinity)
The sum 9 has 46 solutions.
Show the lowest positive sum that can't be expressed (has no solutions), using the rules for this task
211
Show the ten highest numbers that can be expressed using the rules for this task (extra credit)
123456789
23456790
23456788
12345687
12345669
3456801
3456792
3456790
3456788
3456786
```

## Common Lisp

`(defun f (lst &optional (sum 100) (so-far nil)) "Takes a list of digits as argument"  (if (null lst)    (cond ((= sum 0) (format t "~d = ~{[email protected]~}~%" (apply #'+ so-far) (reverse so-far)) 1)          (t 0) )    (let ((total 0)          (len (length lst)) )      (dotimes (i len total)        (let* ((str1 (butlast lst i))               (num1 (or (numlist-to-string str1) 0))               (rem (nthcdr (- len i) lst)) )          (incf total            (+ (f rem (- sum num1) (cons num1 so-far))               (f rem (+ sum num1) (cons (- num1) so-far)) )))))))  (defun numlist-to-string (lst) "Convert a list of digits into an integer"  (when lst    (parse-integer (format nil "~{~d~}" lst)) ))  `
Output:
```>(f '(1 2 3 4 5 6 7 8 9))
100 = +123+45-67+8-9
100 = +123-45-67+89
100 = +123+4-5+67-89
100 = +123-4-5-6-7+8-9
100 = +12+3+4+5-6-7+89
100 = +12+3-4+5+67+8+9
100 = +12-3-4+5-6+7+89
100 = +1+23-4+56+7+8+9
100 = +1+23-4+5+6+78-9
100 = +1+2+34-5+67-8+9
100 = +1+2+3-4+5+6+78+9
100 = -1+2-3+4+5+6+78+9
12
```

The other subtasks are not yet implemented.

## D

Translation of: Java
`import std.stdio; void main() {    import std.algorithm : each, max, reduce, sort;    import std.range : take;     Stat stat = new Stat();     comment("Show all solutions that sum to 100");    immutable givenSum = 100;    print(givenSum);     comment("Show the sum that has the maximum number of solutions");    const int maxCount = reduce!max(stat.sumCount.keys);    int maxSum;    foreach(key, entry; stat.sumCount[maxCount]) {        if (key >= 0) {            maxSum = key;            break;        }    }    writeln(maxSum, " has ", maxCount, " solutions");     comment("Show the lowest positive number that can't be expressed");    int value = 0;    while (value in stat.countSum) {        value++;    }    writeln(value);     comment("Show the ten highest numbers that can be expressed");    const int n = stat.countSum.keys.length;    auto sums = stat.countSum.keys;    sums.sort!"a>b"        .take(10)        .each!print;} void comment(string commentString) {    writeln();    writeln(commentString);    writeln();} void print(int givenSum) {    Expression expression = new Expression();    for (int i=0; i<Expression.NUMBER_OF_EXPRESSIONS; i++, expression.next()) {        if (expression.toInt() == givenSum) {            expression.print();        }    }} class Expression {    private enum NUMBER_OF_DIGITS = 9;    private enum ADD = 0;    private enum SUB = 1;    private enum JOIN = 2;     enum NUMBER_OF_EXPRESSIONS = 2 * 3 * 3 * 3 * 3 * 3 * 3 * 3 * 3;    byte[NUMBER_OF_DIGITS] code;     Expression next() {        for (int i=0; i<NUMBER_OF_DIGITS; i++) {            if (++code[i] > JOIN) {                code[i] = ADD;            } else {                break;            }        }        return this;    }     int toInt() {        int value = 0;        int number = 0;        int sign = (+1);        for (int digit=1; digit<=9; digit++) {            switch (code[NUMBER_OF_DIGITS - digit]) {                case ADD:                    value += sign * number;                    number = digit;                    sign = (+1);                    break;                case SUB:                    value += sign * number;                    number = digit;                    sign = (-1);                    break;                case JOIN:                    number = 10 * number + digit;                    break;                default:                    assert(false);            }        }        return value + sign * number;    }     void toString(scope void delegate(const(char)[]) sink) const {        import std.conv : to;        import std.format : FormatSpec, formatValue;        import std.range : put;         auto fmt = FormatSpec!char("s");        for (int digit=1; digit<=NUMBER_OF_DIGITS; digit++) {            switch (code[NUMBER_OF_DIGITS - digit]) {                case ADD:                    if (digit > 1) {                        put(sink, '+');                    }                    break;                case SUB:                    put(sink, '-');                    break;                default:                    break;            }            formatValue(sink, digit, fmt);        }    }     void print() {        print(stdout);    }     void print(File printStream) {        printStream.writefln("%9d = %s", toInt(), this);    }} class Stat {    int[int] countSum;    bool[int][int] sumCount;     this() {        Expression expression = new Expression();        for (int i=0; i<Expression.NUMBER_OF_EXPRESSIONS; i++, expression.next()) {            int sum = expression.toInt();            countSum[sum]++;        }        foreach (key, entry; countSum) {            bool[int] set;            if (entry in sumCount) {                set = sumCount[entry];            } else {                set.clear();            }            set[key] = true;            sumCount[entry] = set;        }    }}`
Output:
```Show all solutions that sum to 100

100 = 1+2+3-4+5+6+78+9
100 = 1+2+34-5+67-8+9
100 = 1+23-4+5+6+78-9
100 = 1+23-4+56+7+8+9
100 = 12+3+4+5-6-7+89
100 = 12+3-4+5+67+8+9
100 = 12-3-4+5-6+7+89
100 = 123+4-5+67-89
100 = 123+45-67+8-9
100 = 123-4-5-6-7+8-9
100 = 123-45-67+89
100 = -1+2-3+4+5+6+78+9

Show the sum that has the maximum number of solutions

9 has 46 solutions

Show the lowest positive number that can't be expressed

211

Show the ten highest numbers that can be expressed

123456789 = 123456789
23456790 = 1+23456789
23456788 = -1+23456789
12345687 = 12345678+9
12345669 = 12345678-9
3456801 = 12+3456789
3456792 = 1+2+3456789
3456790 = -1+2+3456789
3456788 = 1-2+3456789
3456786 = -1-2+3456789```

## Elixir

`defmodule Sum do  def to(val) do    generate    |> Enum.map(&{eval(&1), &1})    |> Enum.filter(fn {v, _s} -> v==val end)    |> Enum.each(&IO.inspect &1)  end   def max_solve do    generate    |> Enum.group_by(&eval &1)    |> Enum.filter_map(fn {k,_} -> k>=0 end, fn {k,v} -> {length(v),k} end)    |> Enum.max    |> fn {len,sum} -> IO.puts "sum of #{sum} has the maximum number of solutions : #{len}" end.()  end   def min_solve do    solve = generate |> Enum.group_by(&eval &1)    Stream.iterate(1, &(&1+1))    |> Enum.find(fn n -> solve[n]==nil end)    |> fn sum -> IO.puts "lowest positive sum that can't be expressed : #{sum}" end.()  end   def  highest_sums(n\\10) do    IO.puts "highest sums :"    generate    |> Enum.map(&eval &1)    |> Enum.uniq    |> Enum.sort_by(fn sum -> -sum end)    |> Enum.take(n)    |> IO.inspect  end   defp generate do    x = ["+", "-", ""]    for a <- ["-", ""], b <- x, c <- x, d <- x, e <- x, f <- x, g <- x, h <- x, i <- x,        do: "#{a}1#{b}2#{c}3#{d}4#{e}5#{f}6#{g}7#{h}8#{i}9"  end   defp eval(str), do: Code.eval_string(str) |> elem(0)end Sum.to(100)Sum.max_solveSum.min_solveSum.highest_sums`
Output:
```{100, "-1+2-3+4+5+6+78+9"}
{100, "1+2+3-4+5+6+78+9"}
{100, "1+2+34-5+67-8+9"}
{100, "1+23-4+5+6+78-9"}
{100, "1+23-4+56+7+8+9"}
{100, "12+3+4+5-6-7+89"}
{100, "12+3-4+5+67+8+9"}
{100, "12-3-4+5-6+7+89"}
{100, "123+4-5+67-89"}
{100, "123+45-67+8-9"}
{100, "123-4-5-6-7+8-9"}
{100, "123-45-67+89"}
sum of 9 has the maximum number of solutions : 46
lowest positive sum that can't be expressed : 211
highest sums :
[123456789, 23456790, 23456788, 12345687, 12345669, 3456801, 3456792, 3456790,
3456788, 3456786]
```

## Forth

Works with: Gforth version 0.7.3

This solution uses `EVALUATE` on a string buffer to compute the sum. Given the copious string manipulations, `EVALUATE`, and the large byte-array used to keep sum counts, this implementation is optimized neither for speed nor for memory. On my machine it runs in about 3.8 seconds, compared to the speed-optimized C solution which runs in about 0.005 seconds.

`CREATE *OPS CHAR + C, CHAR - C, CHAR # C,CREATE 0OPS CHAR - C, CHAR # C,CREATE BUFF 43 C, 43 CHARS ALLOTCREATE PTR CELL ALLOTCREATE LIMITS 2 C, 3 C, 3 C, 3 C, 3 C, 3 C, 3 C, 3 C, 3 C,CREATE INDX   0 C, 0 C, 0 C, 0 C, 0 C, 0 C, 0 C, 0 C, 0 C,CREATE OPS 0OPS , *OPS , *OPS , *OPS , *OPS , *OPS , *OPS , *OPS , *OPS ,: B0   BUFF 1+ dup  PTR !  43 blank ;: B, ( c --)  PTR @ C!  1 PTR +! ;CREATE STATS 123456790 ALLOT  STATS 123456790 ERASE : inc ( c-addr c-lim u -- t|f)   1- tuck + >r swap dup rot + ( addr a-addr) ( R: l-addr)   BEGIN dup [email protected] 1+ dup [email protected] [email protected] =      IF drop 2dup =        IF 2drop FALSE rdrop EXIT   \ no inc, contents invalid       ELSE 0 over C! 1-  r> 1- >r  \ reset and carry       THEN     ELSE swap C! drop TRUE rdrop EXIT      THEN    AGAIN ;: INDX+   INDX LIMITS 9 inc 0= ;: SYNTH   B0  [CHAR] 0 B,  9 0 DO     INDX I + [email protected]  OPS I CELLS + @ + [email protected]     dup  [CHAR] # <> IF BL B, B, BL B, ELSE drop THEN     I [CHAR] 1 + B,   LOOP  BUFF COUNT ;: .MOST   cr ." Sum that has the maximum number of solutions" cr 4 spaces   STATS 0  STATS 1+ 123456789 bounds DO     dup I [email protected] <  IF drop drop I I [email protected] THEN   LOOP  swap STATS - . ." has " . ." solutions" ;: .CANT   cr ." Lowest positive sum that can't be expressed" cr 4 spaces   STATS 1+ ( 0 not positive)  BEGIN dup [email protected] WHILE 1+ REPEAT  STATS - . ;: .BEST   cr ." Ten highest numbers that can be expressed" cr 4 spaces   0 >r  [ STATS 123456789 + ]L   BEGIN  [email protected] 10 <  over STATS >= and   WHILE  dup [email protected] IF dup STATS - .  r> 1+ >r THEN  1-   REPEAT  r> drop ;: .   0 <# #S #> TYPE ;: .INFX   cr 4 spaces  9 0 DO      INDX I + [email protected]  OPS I cells + @ + [email protected]     dup  [char] # <> IF emit ELSE drop THEN  I 1+ .   LOOP ;: REPORT ( n)   dup 100 =  IF .INFX THEN   dup 0> IF STATS + dup  [email protected] 1+  swap c! ELSE drop THEN ;: >NUM   0. bl word count >number 2drop d>s ;: #   10 * + ;   \ numeric concatenation: +    >NUM + ;  \ infix +: -    >NUM - ;  \ infix -: .SOLUTIONS   cr ." Solutions that sum to 100:"   BEGIN SYNTH EVALUATE REPORT INDX+ UNTIL ;: SUM100   .SOLUTIONS .MOST .CANT .BEST cr ;`
Output:

Note: must start Gforth with a larger-than-default dictionary size:

`gforth -m 124M sum100.fs -e SUM100`
```Solutions that sum to 100:
-1+2-3+4+5+6+78+9
1+2+3-4+5+6+78+9
1+2+34-5+67-8+9
1+23-4+5+6+78-9
1+23-4+56+7+8+9
12+3+4+5-6-7+89
12+3-4+5+67+8+9
12-3-4+5-6+7+89
123+4-5+67-89
123+45-67+8-9
123-4-5-6-7+8-9
123-45-67+89
Sum that has the maximum number of solutions
9 has 46 solutions
Lowest positive sum that can't be expressed
211
Ten highest numbers that can be expressed
123456789 23456790 23456788 12345687 12345669 3456801 3456792 3456790 3456788 3456786```

## Fortran

### Fortran IV

Translation of: Fortran 95
Works with: gfortran version 5.1

The program below is written in Fortran IV. Nevertheless, Fortran IV had a variety of dialects. It did not work the same on every type of computer. The source code below is compiled without problems using today's compilers. It have not been checked on any old mainframe. Sorry, I have no access to CDC6000. The algorithm used is not very fast, but uses little memory. In practice, this program took about 15 seconds to complete the task on PC (in 2017). For comparison: the program written in C ++ (using maps and STL collections) took about 1 second on the same machine.

```C ROSSETACODE: SUM TO 100, FORTRAN IV
C FIND SOLUTIONS TO THE "SUM TO ONE HUNDRED" PUZZLE
C =================================================

PROGRAM SUMTO100
DATA NEXPRM1/13121/
WRITE(6,110)
110  FORMAT(1X/1X,34HSHOW ALL SOLUTIONS THAT SUM TO 100/)
DO 10 I = 0,NEXPRM1
10  IF ( IEVAL(I) .EQ. 100 ) CALL PREXPR(I)

WRITE(6,120)
120  FORMAT(1X/1X,
153HSHOW THE SUM THAT HAS THE MAXIMUM NUMBER OF SOLUTIONS/)
NBEST = -1
DO 30 I = 0, NEXPRM1
ITEST = IEVAL(I)
IF ( ITEST .LT. 0 ) GOTO 30
NTEST = 0
DO 20 J = 0, NEXPRM1
20  IF ( IEVAL(J) .EQ. ITEST ) NTEST = NTEST + 1
IF ( NTEST .LE. NBEST ) GOTO 30
IBEST = ITEST
NBEST = NTEST
30  CONTINUE
WRITE(6,121) IBEST, NBEST
121  FORMAT(1X,I8,5H HAS ,I8,10H SOLUTIONS/)

WRITE(6,130)
130  FORMAT(1X/1X,
155HSHOW THE LOWEST POSITIVE NUMBER THAT CAN'T BE EXPRESSED/)
DO 50 I = 0,123456789
DO 40 J = 0,NEXPRM1
40  IF ( I .EQ. IEVAL(J) ) GOTO 50
GOTO 60
50  CONTINUE
60  WRITE(6,131) I
131  FORMAT(1X,I8)

WRITE(6,140)
140  FORMAT(1X/1X,
150HSHOW THE TEN HIGHEST NUMBERS THAT CAN BE EXPRESSED/)
ILIMIT = 123456789
DO 90 I = 1,10
IBEST = 0
DO 70 J = 0, NEXPRM1
ITEST = IEVAL(J)
70  IF( (ITEST .LE. ILIMIT) .AND. (ITEST .GT. IBEST)) IBEST = ITEST
DO 80 J = 0, NEXPRM1
80  IF ( IEVAL(J) .EQ. IBEST ) CALL PREXPR(J)
90  ILIMIT = IBEST - 1
END

C     EVALUATE THE VALUE OF THE GIVEN ENCODED EXPRESSION
C     --------------------------------------------------
FUNCTION IEVAL(ICODE)
IC = ICODE
IEVAL = 0
N = 0
IP = 1
DO 50 K = 9,1,-1
N = IP*K + N
GOTO (10,20,40,30) MOD(IC,3)+1
10  IEVAL = IEVAL + N
GOTO 30
20  IEVAL = IEVAL - N
30  N = 0
IP = 1
GOTO 50
40  IP = IP * 10
50  IC = IC / 3
END

C     PRINT THE ENCODED EXPRESSION IN THE READABLE FORMAT
C     ---------------------------------------------------
SUBROUTINE PREXPR(ICODE)
DIMENSION IH(9),IHPMJ(4)
DATA IHPMJ/1H+,1H-,1H ,1H?/
IA = 19683
IB =  6561
DO 10 K = 1,9
IH(K) = IHPMJ(MOD(ICODE,IA) / IB+1)
IA = IB
10  IB = IB / 3
IVALUE = IEVAL(ICODE)
WRITE(6,110) IVALUE, IH
110  FORMAT(I9,3H = 1A1,1H1,1A1,1H2,1A1,1H3,1A1,1H4,1A1,1H5,1A1,1H6,1A1
1,1H7,1A1,1H8,1A1,1H9)
END```
Output:
```SHOW ALL SOLUTIONS THAT SUM TO 100

100 = +1+2+3-4+5+6+7 8+9
100 = +1+2+3 4-5+6 7-8+9
100 = +1+2 3-4+5+6+7 8-9
100 = +1+2 3-4+5 6+7+8+9
100 = +1 2+3+4+5-6-7+8 9
100 = +1 2+3-4+5+6 7+8+9
100 = +1 2-3-4+5-6+7+8 9
100 = +1 2 3+4-5+6 7-8 9
100 = +1 2 3+4 5-6 7+8-9
100 = +1 2 3-4-5-6-7+8-9
100 = +1 2 3-4 5-6 7+8 9
100 = -1+2-3+4+5+6+7 8+9

SHOW THE SUM THAT HAS THE MAXIMUM NUMBER OF SOLUTIONS

9 HAS       46 SOLUTIONS

SHOW THE LOWEST POSITIVE NUMBER THAT CAN'T BE EXPRESSED

211

SHOW THE TEN HIGHEST NUMBERS THAT CAN BE EXPRESSED

123456789 = +1 2 3 4 5 6 7 8 9
23456790 = +1+2 3 4 5 6 7 8 9
23456788 = -1+2 3 4 5 6 7 8 9
12345687 = +1 2 3 4 5 6 7 8+9
12345669 = +1 2 3 4 5 6 7 8-9
3456801 = +1 2+3 4 5 6 7 8 9
3456792 = +1+2+3 4 5 6 7 8 9
3456790 = -1+2+3 4 5 6 7 8 9
3456788 = +1-2+3 4 5 6 7 8 9
3456786 = -1-2+3 4 5 6 7 8 9```

### Fortran 95

Works with: gfortran version 5.1
`! RossetaCode: Sum to 100, Fortran 95, an algorithm using ternary numbers.!! Find solutions to the 'sum to one hundred' puzzle.!! We optimize algorithms for size. Therefore we don't use arrays, but recompute! all values again and again. It is a little surprise that the time efficiency ! is quite acceptable. Actually the code is more compact than the implementation! in C++ (STL maps and sets). We purposely break DRY and use magic values.! Nevertheless, it is Fortran 95, free form lines, do-endo etc. program sumto100     parameter (nexpr = 13122)     print *    print *, 'Show all solutions that sum to 100'    print *    do i = 0, nexpr-1        if ( ievaluate(i) .eq. 100 ) then            call printexpr(i)        endif        enddo     print *    print *, 'Show the sum that has the maximum number of solutions'    print *        ibest = -1    nbest = -1    do i = 0, nexpr-1        itest = ievaluate(i)        if ( itest .ge. 0 ) then            ntest = 0            do j = 0, nexpr-1                if ( ievaluate(j) .eq. itest ) then                    ntest = ntest + 1                endif            enddo            if ( (ntest .gt. nbest) ) then                ibest = itest                nbest = ntest            endif        endif    enddo    print *, ibest, ' has ', nbest, ' solutions'    print *!   do i = 0, nexpr-1    !       if ( ievaluate(i) .eq. ibest ) then!           call printexpr(i)!       endif    !   enddo     print *    print *, 'Show the lowest positive number that can''t be expressed'    print *    loop: do i = 0,123456789        do j = 0,nexpr-1            if ( i .eq. ievaluate(j) ) then                cycle loop            endif        enddo        exit    enddo loop    print *, i     print *    print *, 'Show the ten highest numbers that can be expressed'    print *    ilimit = 123456789    do i = 1,10        ibest = 0        do j = 0, nexpr-1            itest = ievaluate(j)            if ( (itest .le. ilimit) .and. (itest .gt. ibest ) ) then                ibest = itest            endif        enddo        do j = 0, nexpr-1                if ( ievaluate(j) .eq. ibest ) then                call printexpr(j)            endif            enddo        ilimit = ibest - 1;    enddo end  function ievaluate(icode)    ic = icode    ievaluate = 0    n = 0    ip = 1    do k = 9,1,-1        n = ip*k + n        select case(mod(ic,3))            case ( 0 )                ievaluate = ievaluate + n                n = 0                ip = 1            case ( 1 )                ievaluate = ievaluate - n                n = 0                ip = 1            case ( 2 )                ip = ip * 10        end select        ic = ic / 3                    enddoend  subroutine printexpr(icode)    character(len=32) s    ia = 19683    ib =  6561    s = ""    do k = 1,9        ic = mod(icode,ia) / ib        ia = ib        ib = ib / 3        select case(mod(ic,3))            case ( 0 )                if ( k .gt. 1 ) then                    s = trim(s) // '+'                endif            case ( 1 )                s = trim(s) // '-'        end select        s = trim(s) // char(ichar('0')+k)    end do    ivalue = ievaluate(icode)    print *, ivalue, ' = ', send`
Output:
``` Show all solutions that sum to 100

100  = 1+2+3-4+5+6+78+9
100  = 1+2+34-5+67-8+9
100  = 1+23-4+5+6+78-9
100  = 1+23-4+56+7+8+9
100  = 12+3+4+5-6-7+89
100  = 12+3-4+5+67+8+9
100  = 12-3-4+5-6+7+89
100  = 123+4-5+67-89
100  = 123+45-67+8-9
100  = 123-4-5-6-7+8-9
100  = 123-45-67+89
100  = -1+2-3+4+5+6+78+9

Show the sum that has the maximum number of solutions

9  has           46  solutions

Show the lowest positive number that can't be expressed

211

Show the ten highest numbers that can be expressed

123456789  = 123456789
23456790  = 1+23456789
23456788  = -1+23456789
12345687  = 12345678+9
12345669  = 12345678-9
3456801  = 12+3456789
3456792  = 1+2+3456789
3456790  = -1+2+3456789
3456788  = 1-2+3456789
3456786  = -1-2+3456789
```

### Batch processing

By the simple expedient of storing all evaluations in an array (which is not so large) and then sorting the array, the required results appear in a blink. The source is essentially F77 except for the usage of an array assignment of OP = -1, writing out the highest ten results via an array expression instead of a DO-loop, array OPNAME extending from -1 to +1, a CYCLE statement rather than a GO TO, and the use of the I0 format code. Subroutine DEBLANK is straightforward, and omitted. It was only to remove spaces from the text of the expression. Reading the expression from right to left is about as picky as left-to-right.
`      INTEGER NDIGITS,TARGET	!Document the shape.      PARAMETER (NDIGITS = 9, TARGET = 100)      INTEGER*1 OP(NDIGITS)	!A set of operation codes, associated with each digit.      INTEGER N,D,P		!Number, digit, power.      CHARACTER*1 OPNAME(-1:+1)	!Encodement of the operations.      PARAMETER (OPNAME = (/"-"," ","+"/))	!These will look nice.      CHARACTER*20 TEXT		!A scratchpad for the expression. Single digits only.      INTEGER I,L,H,ME		!Assistants.      LOGICAL CURSE		!Needed for a Comb Sort.      INTEGER LOOP,NHIT		!Some counters.      INTEGER ENUFF		!Collect the results.      PARAMETER (ENUFF = 20000)	!Surely big enough...      INTEGER VALUE(ENUFF)	!A table.      INTEGER V,VV,PV,VE	!For scanning the table.      INTEGER MSG		!I/O unit number.       MSG = 6		!Standard output.      WRITE(MSG,1) NDIGITS,TARGET	!Announce.    1 FORMAT ("To find expressions of ",I0," digits in order, "     1 "interspersed with + or -, adding to ",I0,/)      NHIT = 0		!No matches to TARGET.      LOOP = 0		!Because none have been made.      OP = -1		!Start the expression sequence. Calculate the value of the expression given by OP(i) i pairs.  100 LOOP = LOOP + 1		!Here we go again.      N = 0			!Clear the number.      D = 0			!No previous digits have been seen.      P = 1			!The power for the first digit.      DO I = NDIGITS,1,-1	!Going backwards sees the digits before the sign.        D = D + I*P			!Assimilate the digit string backwards.        IF (OP(I).EQ.0) THEN		!A no-operation?          P = P*10				!Yes. Prepare the power for the next digit leftwards.         ELSE     			!Otherwise, add or subtract the digit string's value.          N = N + SIGN(D,OP(I))			!By transferring the sign to D..          D = 0					!Clear, ready for the next digit string.          P = 1					!The starting power, again.        END IF				!So much for that step.      END DO			!On to the next.      IF (OP(1).EQ.0) N = N + D	!Provide an implicit add for an unsigned start.      VALUE(LOOP) = N		!Save the value for later...      IF (N.EQ.TARGET) THEN	!Well then?        NHIT = NHIT + 1			!Yay!        WRITE (TEXT,101) (OPNAME(OP(I)),I, I = 1,NDIGITS)	!Translate the expression.  101   FORMAT (10(A1,I1))		!Single-character operation codes, single-digit number parts.        CALL DEBLANK(TEXT,L)		!Squeeze out the no-operations, so numbers are together.        WRITE (MSG,102) N,TEXT(1:L)	!Result!  102   FORMAT (I5,": ",A)		!This should do.      END IF			!So much for that. Concoct the next expression, working as if with a bignumber in base three, though offset.  200 P = NDIGITS		!Start with the low-order digit.  201 OP(P) = OP(P) + 1		!Add one to it.      IF (OP(P).GT.1) THEN	!Is a carry needed?        OP(P) = -1			!Yes. Set the digit back to the start.        P = P - 1			!Go up a power.        IF (P.GT.0) GO TO 201		!And augment the next digit up.      END IF			!Once the carry fizzles, the increment is complete.      IF (OP(1).LE.0) GO TO 100	!A leading + is equivalent to a leading no-op. Contemplate the collection.  300 WRITE (6,301) LOOP,NHIT  301 FORMAT (/,I0," evaluations, ",I0," hit the target.")Crank up a comb sort.      H = LOOP - 1		!Last - First, and not +1.      IF (H.LE.0) STOP "Huh?"	!Ha ha.  310 H = MAX(1,H*10/13)	!The special feature.      IF (H.EQ.9 .OR. H.EQ.10) H = 11	!A twiddle.      CURSE = .FALSE.		!So far, so good.      DO I = LOOP - H,1,-1	!If H = 1, this is a BubbleSort.        IF (VALUE(I) .GT. VALUE(I + H)) THEN	!One compare.          N = VALUE(I);VALUE(I)=VALUE(I+H);VALUE(I+H)=N	!One swap.          CURSE = .TRUE.			!One curse.        END IF				!One test.      END DO			!One loop.      IF (CURSE .OR. H.GT.1) GO TO 310	!Work remains?Chase after some results.      H = 0		!Hunt the first omitted positive number.      VE = 0		!No equal values have been seen.      ME = 0		!So, their maximum run length is short.      PV = VALUE(1)	!Grab the first value,      DO I = 2,LOOP	!And scan the successors.        V = VALUE(I)		!The value of the moment.        IF (V.LE.0) CYCLE	!Only positive numbers are of interest.        IF (V.GT.PV + 1) THEN	!Is there a gap?          IF (H.LE.0) H = PV + 1	!Recall the first such.        END IF			!Perhaps a list of the first dew?        IF (V.EQ.PV) THEN	!Is it the same as the one before?          VE = VE + 1			!Yes. Count up the length of the run.          IF (VE.GT.ME) THEN		!Is this a longer run?            ME = VE				!Yes. Remember its length.            VV = V 				!And its value.          END IF			!So much for runs of equal values.         ELSE			!But if it is not the same,          VE = 0			!A fresh count awaits.        END IF			!So much for comparing one value to its predecessor.        PV = V			!Be ready for the next time around.      END DO		!On to the next. Cast forth the results.      IF (ME.GT.1) WRITE (MSG,320) VV,ME + 1	!Counting started with the second occurrence.  320 FORMAT (I0," has the maximum number of attainments:",I0)      IF (H.GT.0) WRITE (MSG,321) H		!Surely there will be one.  321 FORMAT ("The lowest positive sum that can't be expressed is ",I0)      WRITE (MSG,322) VALUE(LOOP - 9:LOOP)	!Surely LOOP > 9.  322 FORMAT ("The ten highest sums: ",10(I0:","))      END	!That was fun!`

Results:

```To find expressions of 9 digits in order, interspersed with + or -, adding to 100

1: -1+2-3+4+5+6+78+9
2: 12-3-4+5-6+7+89
3: 123-4-5-6-7+8-9
4: 123-45-67+89
5: 123+4-5+67-89
6: 123+45-67+8-9
7: 12+3-4+5+67+8+9
8: 12+3+4+5-6-7+89
9: 1+23-4+56+7+8+9
10: 1+23-4+5+6+78-9
11: 1+2+3-4+5+6+78+9
12: 1+2+34-5+67-8+9

13122 evaluations, 12 hit the target.
9 has the maximum number of attainments:46
The lowest positive sum that can't be expressed is 211
The ten highest sums: 3456786,3456788,3456790,3456792,3456801,12345669,12345687,23456788,23456790,123456789
```

## F#

` (*Generate the data setNigel Galloway February 22nd., 2017*)type N = {n:string; g:int}let N = seq {  let rec fn n i g e l = seq {    match i with    |9 -> yield {n=l + "-9"; g=g+e-9}          yield {n=l + "+9"; g=g+e+9}          yield {n=l +  "9"; g=g+e*10+9*n}    |_ -> yield! fn -1 (i+1) (g+e) -i (l + string -i)          yield! fn  1 (i+1) (g+e)  i (l + "+" + string i)          yield! fn  n (i+1) g (e*10+i*n) (l + string i)  }  yield! fn  1 2 0  1  "1"  yield! fn -1 2 0 -1 "-1"} `
Output:
` N |> Seq.filter(fun n->n.g=100) |> Seq.iter(fun n->printfn "%s" n.n) `
```1+2+3-4+5+6+78+9
1+2+34-5+67-8+9
1+23-4+5+6+78-9
1+23-4+56+7+8+9
12-3-4+5-6+7+89
12+3-4+5+67+8+9
12+3+4+5-6-7+89
123-4-5-6-7+8-9
123-45-67+89
123+4-5+67-89
123+45-67+8-9
-1+2-3+4+5+6+78+9
```
` let n,g = N |> Seq.filter(fun n->n.g>=0) |> Seq.countBy(fun n->n.g) |> Seq.maxBy(snd)printfn "%d has %d solutions" n g `
```9 has 46 solutions
```
` match N |> Seq.filter(fun n->n.g>=0) |> Seq.distinctBy(fun n->n.g) |> Seq.sortBy(fun n->n.g) |> Seq.pairwise |> Seq.tryFind(fun n->(snd n).g-(fst n).g > 1) with  |Some(n) -> printfn "least non-value is %d" ((fst n).g+1)  |None    -> printfn "No non-values found" `
```least non-value is 211
```
` N |> Seq.filter(fun n->n.g>=0) |> Seq.distinctBy(fun n->n.g) |> Seq.sortBy(fun n->(-n.g)) |> Seq.take 10 |> Seq.iter(fun n->printfn "%d" n.g ) `
```123456789
23456790
23456788
12345687
12345669
3456801
3456792
3456790
3456788
3456786
```

## Go

Translation of: C
`package main import (	"fmt"	"sort") const pow3_8 = 3 * 3 * 3 * 3 * 3 * 3 * 3 * 3 // 3^8const pow3_9 = 3 * pow3_8                    // 3^9const maxExprs = 2 * pow3_8                  // not 3^9 since first op can't be Join type op uint8 const (	Add  op = iota // insert a "+"	Sub            //     or a "-"	Join           //     or just join together) // code is an encoding of op, the nine "operations"// we do on each each digit. The op for 1 is in// the highest bits, the op for 9 in the lowest.type code uint16 // evaluate 123456789 with + - or "" prepended to each as indicated by `c`.func (c code) evaluate() (sum int) {	num, pow := 0, 1	for k := 9; k >= 1; k-- {		num += pow * k		switch op(c % 3) {		case Add:			sum += num			num, pow = 0, 1		case Sub:			sum -= num			num, pow = 0, 1		case Join:			pow *= 10		}		c /= 3	}	return sum} func (c code) String() string {	buf := make([]byte, 0, 18)	a, b := code(pow3_9), code(pow3_8)	for k := 1; k <= 9; k++ {		switch op((c % a) / b) {		case Add:			if k > 1 {				buf = append(buf, '+')			}		case Sub:			buf = append(buf, '-')		}		buf = append(buf, '0'+byte(k))		a, b = b, b/3	}	return string(buf)} type sumCode struct {	sum  int	code code}type sumCodes []sumCode type sumCount struct {	sum   int	count int}type sumCounts []sumCount // For sorting (could also use sort.Slice with just Less).func (p sumCodes) Len() int            { return len(p) }func (p sumCodes) Swap(i, j int)       { p[i], p[j] = p[j], p[i] }func (p sumCodes) Less(i, j int) bool  { return p[i].sum < p[j].sum }func (p sumCounts) Len() int           { return len(p) }func (p sumCounts) Swap(i, j int)      { p[i], p[j] = p[j], p[i] }func (p sumCounts) Less(i, j int) bool { return p[i].count > p[j].count } // For printing.func (sc sumCode) String() string {	return fmt.Sprintf("% 10d = %v", sc.sum, sc.code)}func (sc sumCount) String() string {	return fmt.Sprintf("% 10d has %d solutions", sc.sum, sc.count)} func main() {	// Evaluate all expressions.	expressions := make(sumCodes, 0, maxExprs/2)	counts := make(sumCounts, 0, 1715)	for c := code(0); c < maxExprs; c++ {		// All negative sums are exactly like their positive		// counterpart with all +/- switched, we don't need to		// keep track of them.		sum := c.evaluate()		if sum >= 0 {			expressions = append(expressions, sumCode{sum, c})		}	}	sort.Sort(expressions) 	// Count all unique sums	sc := sumCount{expressions.sum, 1}	for _, e := range expressions[1:] {		if e.sum == sc.sum {			sc.count++		} else {			counts = append(counts, sc)			sc = sumCount{e.sum, 1}		}	}	counts = append(counts, sc)	sort.Sort(counts) 	// Extract required results 	fmt.Println("All solutions that sum to 100:")	i := sort.Search(len(expressions), func(i int) bool {		return expressions[i].sum >= 100	})	for _, e := range expressions[i:] {		if e.sum != 100 {			break		}		fmt.Println(e)	} 	fmt.Println("\nThe positive sum with maximum number of solutions:")	fmt.Println(counts) 	fmt.Println("\nThe lowest positive number that can't be expressed:")	s := 1	for _, e := range expressions {		if e.sum == s {			s++		} else if e.sum > s {			fmt.Printf("% 10d\n", s)			break		}	} 	fmt.Println("\nThe ten highest numbers that can be expressed:")	for _, e := range expressions[len(expressions)-10:] {		fmt.Println(e)	}}`
Output:
```All solutions that sum to 100:
100 = -1+2-3+4+5+6+78+9
100 = 1+23-4+5+6+78-9
100 = 123+45-67+8-9
100 = 123+4-5+67-89
100 = 1+2+3-4+5+6+78+9
100 = 1+2+34-5+67-8+9
100 = 12+3-4+5+67+8+9
100 = 1+23-4+56+7+8+9
100 = 123-45-67+89
100 = 12-3-4+5-6+7+89
100 = 12+3+4+5-6-7+89
100 = 123-4-5-6-7+8-9

The positive sum with maximum number of solutions:
9 has 46 solutions

The lowest positive number that can't be expressed:
211

The ten highest numbers that can be expressed:
3456786 = -1-2+3456789
3456788 = 1-2+3456789
3456790 = -1+2+3456789
3456792 = 1+2+3456789
3456801 = 12+3456789
12345669 = 12345678-9
12345687 = 12345678+9
23456788 = -1+23456789
23456790 = 1+23456789
123456789 = 123456789
```

## Haskell

`import Data.Monoid ((<>))import Data.Ord (comparing)import Control.Arrow ((&&&))import Data.Char (intToDigit)import Control.Monad (replicateM)import Data.List (nub, group, sort, sortBy, find, intercalate) data Sign  = Unsigned  | Plus  | Minus  deriving (Eq, Show) universe :: [[(Int, Sign)]]universe =  zip [1 .. 9] <\$>  filter ((/= Plus) . head) (replicateM 9 [Unsigned, Plus, Minus]) allNonNegativeSums :: [Int]allNonNegativeSums = sort \$ filter (>= 0) (asSum <\$> universe) uniqueNonNegativeSums :: [Int]uniqueNonNegativeSums = nub allNonNegativeSums asSum :: [(Int, Sign)] -> IntasSum xs =  n +  (case s of     [] -> 0     _ -> read s :: Int)  where    (n, s) = foldr readSign (0, []) xs    readSign :: (Int, Sign) -> (Int, String) -> (Int, String)    readSign (i, x) (n, s)      | x == Unsigned = (n, intToDigit i : s)      | otherwise =        ( (case x of             Plus -> (+)             _ -> (-))            n            (read (show i <> s) :: Int)        , []) asString :: [(Int, Sign)] -> StringasString = foldr signedDigit []  where    signedDigit (i, x) s      | x == Unsigned = intToDigit i : s      | otherwise =        (case x of           Plus -> " +"           _ -> " -") <>        [intToDigit i] <>        s main :: IO ()main =  putStrLn \$  unlines    [ "Sums to 100:"    , unlines (asString <\$> filter ((100 ==) . asSum) universe)    , "\n10 commonest sums (sum, number of routes to it):"    , show        ((head &&& length) <\$>         take 10 (sortBy (flip (comparing length)) (group allNonNegativeSums)))    , "\nFirst positive integer not expressible as a sum of this kind:"    , maybeReport (find (uncurry (/=)) (zip [0 ..] uniqueNonNegativeSums))    , "\n10 largest sums:"    , show (take 10 (sortBy (flip compare) uniqueNonNegativeSums))    ]  where    maybeReport      :: Show a      => Maybe (a, b) -> String    maybeReport (Just (x, _)) = show x    maybeReport _ = "No gaps found"`
Output:

(Run in Atom editor, through Script package)

```Sums to 100:
123 +45 -67 +8 -9
123 +4 -5 +67 -89
123 -45 -67 +89
123 -4 -5 -6 -7 +8 -9
12 +3 +4 +5 -6 -7 +89
12 +3 -4 +5 +67 +8 +9
12 -3 -4 +5 -6 +7 +89
1 +23 -4 +56 +7 +8 +9
1 +23 -4 +5 +6 +78 -9
1 +2 +34 -5 +67 -8 +9
1 +2 +3 -4 +5 +6 +78 +9
-1 +2 -3 +4 +5 +6 +78 +9

10 commonest sums [sum, number of routes to it]:
[(9,46),(27,44),(1,43),(15,43),(21,43),(45,42),(3,41),(5,40),(7,39),(17,39)]

First positive integer not expressible as a sum of this kind:
211

10 largest sums:
[123456789,23456790,23456788,12345687,12345669,3456801,3456792,3456790,3456788,3456786]

[Finished in 1.204s]```

## J

Since J has no verb precedence, -1-2 would evaluate to 1 and not to -3. That's why I decided to multiply each of the partitions of '123456789' (like '123','45', '6', '78', '9') with each possible +1/-1 vectors of length 9 (like 1 1 -1 1 -1 -1 1 1 -1) and to add up the results. This leads to 512*256 results, that of course include a lot of duplicates. To use directly ~. (nub) on the 512x256x9 vector is very slow and that's why I computed a sort of a hash to use it to get only the unique expressions. The rest is trivial - I check which expressions add up to 100; sort the sum vector and find the longest sequence ot repeating sums; get the 10 largest sums and finnaly check which sum differs with more then 1 from the previous one.

` p =: ,"2".>(#: (+ i.)2^8) <;.1 '123456789'm =. (9\$_1x)^"1#:i.2^9s =. 131072 9 \$ ,m *"1/ ps2 =: (~: (10x^i._9)#.s)#sss =: +/"1 s2'100=';<'bp<+>' 8!:2 (I.100=ss){s2pos =: (0<ss)#ss =: /:~ss({.;'times';{:)>{.\:~(#,{.) each </.~ ss'Ten largest:';,.(->:i.10){ss 'First not expressible:';>:pos{~ 1 i.~ 1<|2-/\pos `
Output:
```┌───┬────────────────────────┐
│100│+12 +3 +4 +5 -6 -7 +89  │
│   │+1  +2 +3 -4 +5 +6 +78+9│
│   │+1  +2 +34-5 +67-8 +9   │
│   │+12 +3 -4 +5 +67+8 +9   │
│   │+1  +23-4 +56+7 +8 +9   │
│   │+1  +23-4 +5 +6 +78-9   │
│   │+123+45-67+8 -9         │
│   │+123+4 -5 +67-89        │
│   │+123-45-67+89           │
│   │+12 -3 -4 +5 -6 +7 +89  │
│   │+123-4 -5 -6 -7 +8 -9   │
│   │-1  +2 -3 +4 +5 +6 +78+9│
└───┴────────────────────────┘
┌──┬─────┬─┐
│46│times│9│
└──┴─────┴─┘
┌────────────┬─────────┐
│Ten largest:│123456789│
│            │ 23456790│
│            │ 23456788│
│            │ 12345687│
│            │ 12345669│
│            │  3456801│
│            │  3456792│
│            │  3456790│
│            │  3456788│
│            │  3456786│
└────────────┴─────────┘
┌───────────────────────┬───┐
│First not expressible :│211│
└───────────────────────┴───┘

```

## Java

Works with: Java version 8
Translation of: C++

For each expression of sum s, there is at least one expression whose sum is -s. If the sum s can be represented by n expressions, the sum -s can also be represented by n expressions. The change of all signs in an expression change the sign of the sum of this expression. For example, -1+23-456+789 has the opposite sign than +1-23+456-789. Therefore only the positive sum with the maximum number of solutions is shown. The program does not check uniqueness of this sum. We can easily check (modifying the program) that: sum 9 has 46 solutions; sum -9 has 46 solutions; any other sum has less than 46 solutions.

`/*  * RossetaCode: Sum to 100, Java 8.  * * Find solutions to the "sum to one hundred" puzzle. */package rosettacode; import java.io.PrintStream;import java.util.Arrays;import java.util.Collections;import java.util.HashMap;import java.util.HashSet;import java.util.Iterator;import java.util.Map;import java.util.Set; public class SumTo100 implements Runnable {     public static void main(String[] args) {        new SumTo100().run();    }     void print(int givenSum) {        Expression expression = new Expression();        for (int i = 0; i < Expression.NUMBER_OF_EXPRESSIONS; i++, expression.next()) {            if (expression.toInt() == givenSum) {                expression.print();            }        }    }     void comment(String commentString) {        System.out.println();        System.out.println(commentString);        System.out.println();    }     @Override    public void run() {        final Stat stat = new Stat();         comment("Show all solutions that sum to 100");        final int givenSum = 100;        print(givenSum);         comment("Show the sum that has the maximum number of solutions");        final int maxCount = Collections.max(stat.sumCount.keySet());        int maxSum;        Iterator<Integer> it = stat.sumCount.get(maxCount).iterator();        do {            maxSum = it.next();        } while (maxSum < 0);        System.out.println(maxSum + " has " + maxCount + " solutions");         comment("Show the lowest positive number that can't be expressed");        int value = 0;        while (stat.countSum.containsKey(value)) {            value++;        }        System.out.println(value);         comment("Show the ten highest numbers that can be expressed");        final int n = stat.countSum.keySet().size();        final Integer[] sums = stat.countSum.keySet().toArray(new Integer[n]);        Arrays.sort(sums);        for (int i = n - 1; i >= n - 10; i--) {            print(sums[i]);        }    }     private static class Expression {         private final static int NUMBER_OF_DIGITS = 9;        private final static byte ADD = 0;        private final static byte SUB = 1;        private final static byte JOIN = 2;         final byte[] code = new byte[NUMBER_OF_DIGITS];        final static int NUMBER_OF_EXPRESSIONS = 2 * 3 * 3 * 3 * 3 * 3 * 3 * 3 * 3;         Expression next() {            for (int i = 0; i < NUMBER_OF_DIGITS; i++) {                if (++code[i] > JOIN) {                    code[i] = ADD;                } else {                    break;                }            }            return this;        }         int toInt() {            int value = 0;            int number = 0;            int sign = (+1);            for (int digit = 1; digit <= 9; digit++) {                switch (code[NUMBER_OF_DIGITS - digit]) {                    case ADD:                        value += sign * number;                        number = digit;                        sign = (+1);                        break;                    case SUB:                        value += sign * number;                        number = digit;                        sign = (-1);                        break;                    case JOIN:                        number = 10 * number + digit;                        break;                }            }            return value + sign * number;        }         @Override        public String toString() {            StringBuilder s = new StringBuilder(2 * NUMBER_OF_DIGITS + 1);            for (int digit = 1; digit <= NUMBER_OF_DIGITS; digit++) {                switch (code[NUMBER_OF_DIGITS - digit]) {                    case ADD:                        if (digit > 1) {                            s.append('+');                        }                        break;                    case SUB:                        s.append('-');                        break;                }                s.append(digit);            }            return s.toString();        }         void print() {            print(System.out);        }         void print(PrintStream printStream) {            printStream.format("%9d", this.toInt());            printStream.println(" = " + this);        }    }     private static class Stat {         final Map<Integer, Integer> countSum = new HashMap<>();        final Map<Integer, Set<Integer>> sumCount = new HashMap<>();         Stat() {            Expression expression = new Expression();            for (int i = 0; i < Expression.NUMBER_OF_EXPRESSIONS; i++, expression.next()) {                int sum = expression.toInt();                countSum.put(sum, countSum.getOrDefault(sum, 0) + 1);            }            for (Map.Entry<Integer, Integer> entry : countSum.entrySet()) {                Set<Integer> set;                if (sumCount.containsKey(entry.getValue())) {                    set = sumCount.get(entry.getValue());                } else {                    set = new HashSet<>();                }                set.add(entry.getKey());                sumCount.put(entry.getValue(), set);            }        }    }}`
Output:
```Show all solutions that sum to 100

100 = 1+2+3-4+5+6+78+9
100 = 1+2+34-5+67-8+9
100 = 1+23-4+5+6+78-9
100 = 1+23-4+56+7+8+9
100 = 12+3+4+5-6-7+89
100 = 12+3-4+5+67+8+9
100 = 12-3-4+5-6+7+89
100 = 123+4-5+67-89
100 = 123+45-67+8-9
100 = 123-4-5-6-7+8-9
100 = 123-45-67+89
100 = -1+2-3+4+5+6+78+9

Show the sum that has the maximum number of solutions

9 has 46 solutions

Show the lowest positive number that can't be expressed

211

Show the ten highest numbers that can be expressed

123456789 = 123456789
23456790 = 1+23456789
23456788 = -1+23456789
12345687 = 12345678+9
12345669 = 12345678-9
3456801 = 12+3456789
3456792 = 1+2+3456789
3456790 = -1+2+3456789
3456788 = 1-2+3456789
3456786 = -1-2+3456789```

## JavaScript

### ES5

Translation of: Haskell
`(function () {    'use strict';     // GENERIC FUNCTIONS ----------------------------------------------------     // permutationsWithRepetition :: Int -> [a] -> [[a]]    var permutationsWithRepetition = function (n, as) {        return as.length > 0 ?            foldl1(curry(cartesianProduct)(as), replicate(n, as)) : [];    };     // cartesianProduct :: [a] -> [b] -> [[a, b]]    var cartesianProduct = function (xs, ys) {        return [].concat.apply([], xs.map(function (x) {            return [].concat.apply([], ys.map(function (y) {                return [                    [x].concat(y)                ];            }));        }));    };     // curry :: ((a, b) -> c) -> a -> b -> c    var curry = function (f) {        return function (a) {            return function (b) {                return f(a, b);            };        };    };     // flip :: (a -> b -> c) -> b -> a -> c    var flip = function (f) {        return function (a, b) {            return f.apply(null, [b, a]);        };    };     // foldl1 :: (a -> a -> a) -> [a] -> a    var foldl1 = function (f, xs) {        return xs.length > 0 ? xs.slice(1)            .reduce(f, xs) : [];    };     // replicate :: Int -> a -> [a]    var replicate = function (n, a) {        var v = [a],            o = [];        if (n < 1) return o;        while (n > 1) {            if (n & 1) o = o.concat(v);            n >>= 1;            v = v.concat(v);        }        return o.concat(v);    };     // group :: Eq a => [a] -> [[a]]    var group = function (xs) {        return groupBy(function (a, b) {            return a === b;        }, xs);    };     // groupBy :: (a -> a -> Bool) -> [a] -> [[a]]    var groupBy = function (f, xs) {        var dct = xs.slice(1)            .reduce(function (a, x) {                var h = a.active.length > 0 ? a.active : undefined,                    blnGroup = h !== undefined && f(h, x);                 return {                    active: blnGroup ? a.active.concat(x) : [x],                    sofar: blnGroup ? a.sofar : a.sofar.concat([a.active])                };            }, {                active: xs.length > 0 ? [xs] : [],                sofar: []            });        return dct.sofar.concat(dct.active.length > 0 ? [dct.active] : []);    };     // compare :: a -> a -> Ordering    var compare = function (a, b) {        return a < b ? -1 : a > b ? 1 : 0;    };     // on :: (b -> b -> c) -> (a -> b) -> a -> a -> c    var on = function (f, g) {        return function (a, b) {            return f(g(a), g(b));        };    };     // nub :: [a] -> [a]    var nub = function (xs) {        return nubBy(function (a, b) {            return a === b;        }, xs);    };     // nubBy :: (a -> a -> Bool) -> [a] -> [a]    var nubBy = function (p, xs) {        var x = xs.length ? xs : undefined;         return x !== undefined ? [x].concat(nubBy(p, xs.slice(1)            .filter(function (y) {                return !p(x, y);            }))) : [];    };     // find :: (a -> Bool) -> [a] -> Maybe a    var find = function (f, xs) {        for (var i = 0, lng = xs.length; i < lng; i++) {            if (f(xs[i], i)) return xs[i];        }        return undefined;    };     // Int -> [a] -> [a]    var take = function (n, xs) {        return xs.slice(0, n);    };     // unlines :: [String] -> String    var unlines = function (xs) {        return xs.join('\n');    };     // show :: a -> String    var show = function (x) {        return JSON.stringify(x);    }; //, null, 2);     // head :: [a] -> a    var head = function (xs) {        return xs.length ? xs : undefined;    };     // tail :: [a] -> [a]    var tail = function (xs) {        return xs.length ? xs.slice(1) : undefined;    };     // length :: [a] -> Int    var length = function (xs) {        return xs.length;    };     // SIGNED DIGIT SEQUENCES  (mapped to sums and to strings)     // data Sign :: [ 0 | 1 | -1 ] = ( Unsigned | Plus | Minus )    // asSum :: [Sign] -> Int    var asSum = function (xs) {        var dct = xs.reduceRight(function (a, sign, i) {            var d = i + 1; //  zero-based index to [1-9] positions            if (sign !== 0) {                // Sum increased, digits cleared                return {                    digits: [],                    n: a.n + sign * parseInt([d].concat(a.digits)                        .join(''), 10)                };            } else return { // Digits extended, sum unchanged                digits: [d].concat(a.digits),                n: a.n            };        }, {            digits: [],            n: 0        });        return dct.n + (            dct.digits.length > 0 ? parseInt(dct.digits.join(''), 10) : 0        );    };     // data Sign :: [ 0 | 1 | -1 ] = ( Unsigned | Plus | Minus )    // asString :: [Sign] -> String    var asString = function (xs) {        var ns = xs.reduce(function (a, sign, i) {            var d = (i + 1)                .toString();            return sign === 0 ? a + d : a + (sign > 0 ? ' +' : ' -') + d;        }, '');         return ns === '+' ? tail(ns) : ns;    };     // SUM T0 100 ------------------------------------------------------------     // universe :: [[Sign]]    var universe = permutationsWithRepetition(9, [0, 1, -1])        .filter(function (x) {            return x !== 1;        });     // allNonNegativeSums :: [Int]    var allNonNegativeSums = universe.map(asSum)        .filter(function (x) {            return x >= 0;        })        .sort();     // uniqueNonNegativeSums :: [Int]    var uniqueNonNegativeSums = nub(allNonNegativeSums);     return ["Sums to 100:\n", unlines(universe.filter(function (x) {                return asSum(x) === 100;            })            .map(asString)),         "\n\n10 commonest sums (sum, followed by number of routes to it):\n",        show(take(10, group(allNonNegativeSums)            .sort(on(flip(compare), length))            .map(function (xs) {                return [xs, xs.length];            }))),         "\n\nFirst positive integer not expressible as a sum of this kind:\n",        show(find(function (x, i) {            return x !== i;        }, uniqueNonNegativeSums.sort(compare)) - 1), // zero-based index         "\n10 largest sums:\n",        show(take(10, uniqueNonNegativeSums.sort(flip(compare))))    ].join('\n') + '\n';})();`
Output:

(Run in Atom editor, through Script package)

```Sums to 100:

123 +45 -67 +8 -9
123 +4 -5 +67 -89
123 -45 -67 +89
123 -4 -5 -6 -7 +8 -9
12 +3 +4 +5 -6 -7 +89
12 +3 -4 +5 +67 +8 +9
12 -3 -4 +5 -6 +7 +89
1 +23 -4 +56 +7 +8 +9
1 +23 -4 +5 +6 +78 -9
1 +2 +34 -5 +67 -8 +9
1 +2 +3 -4 +5 +6 +78 +9
-1 +2 -3 +4 +5 +6 +78 +9

10 commonest sums (sum, followed by number of routes to it):

[[9,46],[27,44],[1,43],[15,43],[21,43],[45,42],[3,41],[5,40],[17,39],[7,39]]

First positive integer not expressible as a sum of this kind:

211

10 largest sums:

[123456789,23456790,23456788,12345687,12345669,3456801,3456792,3456790,3456788,3456786]

[Finished in 0.381s]```

### ES6

Translation of: Haskell
`(() => {    'use strict';     // GENERIC FUNCTIONS ----------------------------------------------------     // permutationsWithRepetition :: Int -> [a] -> [[a]]    const permutationsWithRepetition = (n, as) =>        as.length > 0 ? (            foldl1(curry(cartesianProduct)(as), replicate(n, as))        ) : [];     // cartesianProduct :: [a] -> [b] -> [[a, b]]    const cartesianProduct = (xs, ys) =>        [].concat.apply([], xs.map(x =>        [].concat.apply([], ys.map(y => [[x].concat(y)]))));     // curry :: ((a, b) -> c) -> a -> b -> c    const curry = f => a => b => f(a, b);     // flip :: (a -> b -> c) -> b -> a -> c    const flip = f => (a, b) => f.apply(null, [b, a]);     // foldl1 :: (a -> a -> a) -> [a] -> a    const foldl1 = (f, xs) =>        xs.length > 0 ? xs.slice(1)        .reduce(f, xs) : [];     // replicate :: Int -> a -> [a]    const replicate = (n, a) => {        let v = [a],            o = [];        if (n < 1) return o;        while (n > 1) {            if (n & 1) o = o.concat(v);            n >>= 1;            v = v.concat(v);        }        return o.concat(v);    };     // group :: Eq a => [a] -> [[a]]    const group = xs => groupBy((a, b) => a === b, xs);     // groupBy :: (a -> a -> Bool) -> [a] -> [[a]]    const groupBy = (f, xs) => {        const dct = xs.slice(1)            .reduce((a, x) => {                const                    h = a.active.length > 0 ? a.active : undefined,                    blnGroup = h !== undefined && f(h, x);                 return {                    active: blnGroup ? a.active.concat(x) : [x],                    sofar: blnGroup ? a.sofar : a.sofar.concat([a.active])                };            }, {                active: xs.length > 0 ? [xs] : [],                sofar: []            });        return dct.sofar.concat(dct.active.length > 0 ? [dct.active] : []);    };     // compare :: a -> a -> Ordering    const compare = (a, b) => a < b ? -1 : (a > b ? 1 : 0);     // on :: (b -> b -> c) -> (a -> b) -> a -> a -> c    const on = (f, g) => (a, b) => f(g(a), g(b));     // nub :: [a] -> [a]    const nub = xs => nubBy((a, b) => a === b, xs);     // nubBy :: (a -> a -> Bool) -> [a] -> [a]    const nubBy = (p, xs) => {        const x = xs.length ? xs : undefined;         return x !== undefined ? [x].concat(            nubBy(p, xs.slice(1)                .filter(y => !p(x, y)))        ) : [];    };     // find :: (a -> Bool) -> [a] -> Maybe a    const find = (f, xs) => {        for (var i = 0, lng = xs.length; i < lng; i++) {            if (f(xs[i], i)) return xs[i];        }        return undefined;    }     // Int -> [a] -> [a]    const take = (n, xs) => xs.slice(0, n);     // unlines :: [String] -> String    const unlines = xs => xs.join('\n');     // show :: a -> String    const show = x => JSON.stringify(x); //, null, 2);     // head :: [a] -> a    const head = xs => xs.length ? xs : undefined;     // tail :: [a] -> [a]    const tail = xs => xs.length ? xs.slice(1) : undefined;     // length :: [a] -> Int    const length = xs => xs.length;      // SIGNED DIGIT SEQUENCES  (mapped to sums and to strings)     // data Sign :: [ 0 | 1 | -1 ] = ( Unsigned | Plus | Minus )    // asSum :: [Sign] -> Int    const asSum = xs => {        const dct = xs.reduceRight((a, sign, i) => {            const d = i + 1; //  zero-based index to [1-9] positions            if (sign !== 0) { // Sum increased, digits cleared                return {                    digits: [],                    n: a.n + (sign * parseInt([d].concat(a.digits)                        .join(''), 10))                };            } else return { // Digits extended, sum unchanged                digits: [d].concat(a.digits),                n: a.n            };        }, {            digits: [],            n: 0        });        return dct.n + (dct.digits.length > 0 ? (            parseInt(dct.digits.join(''), 10)        ) : 0);    };     // data Sign :: [ 0 | 1 | -1 ] = ( Unsigned | Plus | Minus )    // asString :: [Sign] -> String    const asString = xs => {        const ns = xs.reduce((a, sign, i) => {            const d = (i + 1)                .toString();            return (sign === 0 ? (                a + d            ) : (a + (sign > 0 ? ' +' : ' -') + d));        }, '');         return ns === '+' ? tail(ns) : ns;    };      // SUM T0 100 ------------------------------------------------------------     // universe :: [[Sign]]    const universe = permutationsWithRepetition(9, [0, 1, -1])        .filter(x => x !== 1);     // allNonNegativeSums :: [Int]    const allNonNegativeSums = universe.map(asSum)        .filter(x => x >= 0)        .sort();     // uniqueNonNegativeSums :: [Int]    const uniqueNonNegativeSums = nub(allNonNegativeSums);      return [        "Sums to 100:\n",        unlines(universe.filter(x => asSum(x) === 100)            .map(asString)),         "\n\n10 commonest sums (sum, followed by number of routes to it):\n",        show(take(10, group(allNonNegativeSums)            .sort(on(flip(compare), length))            .map(xs => [xs, xs.length]))),         "\n\nFirst positive integer not expressible as a sum of this kind:\n",        show(find(            (x, i) => x !== i,            uniqueNonNegativeSums.sort(compare)        ) - 1), // i is the the zero-based Array index.         "\n10 largest sums:\n",        show(take(10, uniqueNonNegativeSums.sort(flip(compare))))    ].join('\n') + '\n';})();`
Output:

(Run in Atom editor, through Script package)

```Sums to 100:

123 +45 -67 +8 -9
123 +4 -5 +67 -89
123 -45 -67 +89
123 -4 -5 -6 -7 +8 -9
12 +3 +4 +5 -6 -7 +89
12 +3 -4 +5 +67 +8 +9
12 -3 -4 +5 -6 +7 +89
1 +23 -4 +56 +7 +8 +9
1 +23 -4 +5 +6 +78 -9
1 +2 +34 -5 +67 -8 +9
1 +2 +3 -4 +5 +6 +78 +9
-1 +2 -3 +4 +5 +6 +78 +9

10 commonest sums (sum, followed by number of routes to it):

[[9,46],[27,44],[1,43],[15,43],[21,43],[45,42],[3,41],[5,40],[17,39],[7,39]]

First positive integer not expressible as a sum of this kind:

211

10 largest sums:

[123456789,23456790,23456788,12345687,12345669,3456801,3456792,3456790,3456788,3456786]

[Finished in 0.382s]```

### ES3 (JScript)

Translation of: AWK
`SumTo100(); function SumTo100(){               var         ADD  = 0,         SUB  = 1,         JOIN = 2;     var         nexpr = 13122;      function out(something)      {         WScript.Echo(something);     }     function evaluate(code)    {        var             value  = 0,             number = 0,             power  = 1;         for ( var k = 9; k >= 1; k-- )        {            number = power*k + number;            switch( code % 3 )            {                case ADD:  value = value + number; number = 0; power = 1; break;                case SUB:  value = value - number; number = 0; power = 1; break;                case JOIN: power = power * 10                           ; break;            }            code = Math.floor(code/3);        }        return value;        }     function print(code)    {        var             s = "";        var             a = 19683,            b = 6561;                 for ( var k = 1; k <= 9; k++ )        {            switch( Math.floor(  (code % a) / b  ) ){                case ADD: if ( k > 1 ) s = s + '+'; break;                case SUB:              s = s + '-'; break;            }            a = b;            b = Math.floor(b/3);            s = s + String.fromCharCode(0x30+k);        }          out(evaluate(code) + " = " + s);    }     function comment(commentString)    {        out("");        out(commentString);        out("");            }     comment("Show all solutions that sum to 100");    for ( var i = 0; i < nexpr; i++)             if ( evaluate(i) == 100 )             print(i);             comment("Show the sum that has the maximum number of solutions");     var stat = {};    for ( var i = 0; i < nexpr; i++ )    {        var sum = evaluate(i);        if (stat[sum])            stat[sum]++;        else            stat[sum] = 1;    }     var best = 0;    var nbest = -1;    for ( var i = 0; i < nexpr; i++ )    {        var sum = evaluate(i);        if ( sum > 0 )            if ( stat[sum] > nbest )            {                best = i;                            nbest = stat[sum];            }    }    out("" + evaluate(best) + " has " + nbest + " solutions");     comment("Show the lowest positive number that can't be expressed");    for ( var i = 0; i <= 123456789; i++ )    {        for ( var j = 0; j < nexpr; j++)             if ( i == evaluate(j) ) break;         if ( i != evaluate(j) ) break;    }    out(i);     comment("Show the ten highest numbers that can be expressed");    var limit = 123456789 + 1;    for ( i = 1; i <= 10; i++ )     {        var best = 0;        for ( var j = 0; j < nexpr; j++)        {            var test = evaluate(j);            if ( test < limit && test > best )                 best = test;        }        for ( var j = 0; j < nexpr; j++)            if ( evaluate(j) == best ) print(j);        limit = best;    } } `
Output:
```Show all solutions that sum to 100

100 = 1+2+3-4+5+6+78+9
100 = 1+2+34-5+67-8+9
100 = 1+23-4+5+6+78-9
100 = 1+23-4+56+7+8+9
100 = 12+3+4+5-6-7+89
100 = 12+3-4+5+67+8+9
100 = 12-3-4+5-6+7+89
100 = 123+4-5+67-89
100 = 123+45-67+8-9
100 = 123-4-5-6-7+8-9
100 = 123-45-67+89
100 = -1+2-3+4+5+6+78+9

Show the sum that has the maximum number of solutions

9 has 46 solutions

Show the lowest positive number that can't be expressed

211

Show the ten highest numbers that can be expressed

123456789 = 123456789
23456790 = 1+23456789
23456788 = -1+23456789
12345687 = 12345678+9
12345669 = 12345678-9
3456801 = 12+3456789
3456792 = 1+2+3456789
3456790 = -1+2+3456789
3456788 = 1-2+3456789
3456786 = -1-2+3456789```

## jq

For ease of understanding, the problems will be solved separately, using the machinery defined in the following section.

All possible sums

`# Generate a "sum" in the form:  [I, 1, X, 2, X, 3, ..., X, n] where I is "-" or "", and X is "+", "-", or ""def generate(n):  def pm: ["+"], ["-"], [""];   if n == 1 then (["-"], [""]) +   else generate(n-1) + pm +  [n]  end; # The numerical value of a "sum"def addup:  reduce .[] as \$x ({sum:0, previous: "0"};     if   \$x == "+" then .sum += (.previous|tonumber) | .previous = ""     elif \$x == "-" then .sum += (.previous|tonumber) | .previous = "-"     elif \$x == "" then .     else .previous += (\$x|tostring)     end)     | .sum + (.previous | tonumber) ; # Pretty-print a "sum", e.g. ["",1,"+", 2] => 1 + 2def pp: map(if . == "+" or . == "-" then " " + . else tostring end) | join(""); `

Solutions to "Sum to 100" problem

`generate(9) | select(addup == 100) | pp`
Output:
```1 +23 -4 +56 +7 +8 +9
12 +3 -4 +5 +67 +8 +9
1 +2 +34 -5 +67 -8 +9
-1 +2 -3 +4 +5 +6 +78 +9
1 +2 +3 -4 +5 +6 +78 +9
123 -4 -5 -6 -7 +8 -9
123 +45 -67 +8 -9
1 +23 -4 +5 +6 +78 -9
12 -3 -4 +5 -6 +7 +89
12 +3 +4 +5 -6 -7 +89
123 -45 -67 +89
123 +4 -5 +67 -89```

Helper Functions

For brevity, we define an efficient function for computing a histogram in the form of a JSON object, and a helper function for identifying the values with the n highest frequencies.

`def histogram(s): reduce s as \$x ({}; (\$x|tostring) as \$k | .[\$k] += 1); # Emit an array of [ value, frequency ] pairsdef greatest(n):  to_entries  | map( [.key, .value] )  | sort_by(.)  | .[(length-n):]  | reverse ; `

Maximum number of solutions

`histogram(generate(9) | addup | select(.>0)) | greatest(1)`
Output:
`[["9",46]]`

Ten most frequent sums

`histogram(generate(9) | addup | select(.>0)) | greatest(1)`
Output:
`[["9",46],["27",44],["1",43],["21",43],["15",43],["45",42],["3",41],["5",40],["7",39],["17",39]]`

First unsolvable

`def first_missing(s):    first( foreach s as \$i (null;           if . == null or \$i == . or \$i == .+1 then \$i else [.+1] end;           select(type == "array") | .)); first_missing( [generate(9) | addup | select(.>0) ] | unique[])`
Output:
```   211
```

Ten largest sums

`[generate(9) | addup | select(.>0)] | unique | .[(length-10):]`
Output:
```   [3456786,3456788,3456790,3456792,3456801,12345669,12345687,23456788,23456790,123456789]
```

## Julia

`# v0.6 using IterTools expr(p::String...)::String = @sprintf("%s1%s2%s3%s4%s5%s6%s7%s8%s9", p...)function genexpr()::Vector{String}    op = ["+", "-", ""]    return collect(expr(p...) for (p) in product(op, op, op, op, op, op, op, op, op) if p != "+")end using DataStructures function allexpr()::Dict{Int,Int}    rst = DefaultDict{Int,Int}(0)    for e in genexpr()        val = eval(parse(e))        rst[val] += 1    end    return rstend sumto(val::Int)::Vector{String} = filter(e -> eval(parse(e)) == val, genexpr())function maxsolve()::Dict{Int,Int}    ae = allexpr()    vmax = maximum(values(ae))    smax = filter(ae) do v, f        f == vmax    end    return smaxendfunction minsolve()::Int    ae = keys(allexpr())    for i in 1:typemax(Int)        if i ∉ ae            return i        end    endendfunction highestsums(n::Int)::Vector{Int}    sums = collect(keys(allexpr()))    return sort!(sums; rev=true)[1:n]end solutions = sumto(100)max   = maxsolve()min   = minsolve()hsums = highestsums(10) println("100 =")foreach(println, solutions) println("\nMax number of solutions:")for (v, f) in max    @printf("%3i -> %2i\n", v, f)end println("\nMin number with no solutions: \$min") println("\nHighest sums representable:")foreach(println, hsums)`
Output:
```100 =
1+23-4+56+7+8+9
12+3-4+5+67+8+9
1+2+34-5+67-8+9
-1+2-3+4+5+6+78+9
1+2+3-4+5+6+78+9
123-4-5-6-7+8-9
123+45-67+8-9
1+23-4+5+6+78-9
12-3-4+5-6+7+89
12+3+4+5-6-7+89
123-45-67+89
123+4-5+67-89

Max number of solutions:
9 -> 46
-9 -> 46

Min number with no solutions: 211

Highest sums representable:
123456789
23456790
23456788
12345687
12345669
3456801
3456792
3456790
3456788
3456786```

## Kotlin

Translation of: C++
`// version 1.1.51 class Expression {     private enum class Op { ADD, SUB, JOIN }    private val code = Array<Op>(NUMBER_OF_DIGITS) { Op.ADD }     companion object {        private const val NUMBER_OF_DIGITS = 9        private const val THREE_POW_4 = 3 * 3 * 3 * 3        private const val FMT = "%9d"        const val NUMBER_OF_EXPRESSIONS = 2 * THREE_POW_4 * THREE_POW_4         fun print(givenSum: Int) {            var expression = Expression()            repeat(Expression.NUMBER_OF_EXPRESSIONS) {                if (expression.toInt() == givenSum) println("\${FMT.format(givenSum)} = \$expression")                expression++            }        }    }     operator fun inc(): Expression {        for (i in 0 until code.size) {            code[i] = when (code[i]) {                Op.ADD  -> Op.SUB                Op.SUB  -> Op.JOIN                Op.JOIN -> Op.ADD            }            if (code[i] != Op.ADD) break        }        return this    }     fun toInt(): Int {        var value = 0        var number = 0        var sign = +1        for (digit in 1..9) {            when (code[NUMBER_OF_DIGITS - digit]) {                Op.ADD  -> { value += sign * number; number = digit; sign = +1 }                Op.SUB  -> { value += sign * number; number = digit; sign = -1 }                Op.JOIN -> { number = 10 * number + digit }            }        }        return value + sign * number    }     override fun toString(): String {        val sb = StringBuilder()        for (digit in 1..NUMBER_OF_DIGITS) {            when (code[NUMBER_OF_DIGITS - digit]) {                Op.ADD  -> if (digit > 1) sb.append(" + ")                Op.SUB  -> sb.append(" - ")                Op.JOIN -> {}            }            sb.append(digit)        }        return sb.toString().trimStart()    }} class Stat {     val countSum = mutableMapOf<Int, Int>()    val sumCount = mutableMapOf<Int, MutableSet<Int>>()     init {        var expression = Expression()        repeat (Expression.NUMBER_OF_EXPRESSIONS) {            val sum = expression.toInt()            countSum.put(sum, 1 + (countSum[sum] ?: 0))            expression++        }        for ((k, v) in countSum) {            val set = if (sumCount.containsKey(v))                sumCount[v]!!            else                mutableSetOf<Int>()            set.add(k)            sumCount.put(v, set)        }    }} fun main(args: Array<String>) {    println("100 has the following solutions:\n")    Expression.print(100)     val stat = Stat()    val maxCount = stat.sumCount.keys.max()    val maxSum = stat.sumCount[maxCount]!!.max()    println("\n\$maxSum has the maximum number of solutions, namely \$maxCount")     var value = 0    while (stat.countSum.containsKey(value)) value++    println("\n\$value is the lowest positive number with no solutions")     println("\nThe ten highest numbers that do have solutions are:\n")    stat.countSum.keys.toIntArray().sorted().reversed().take(10).forEach { Expression.print(it) }}`
Output:
```100 has the following solutions:

100 = 1 + 2 + 3 - 4 + 5 + 6 + 78 + 9
100 = 1 + 2 + 34 - 5 + 67 - 8 + 9
100 = 1 + 23 - 4 + 5 + 6 + 78 - 9
100 = 1 + 23 - 4 + 56 + 7 + 8 + 9
100 = 12 + 3 + 4 + 5 - 6 - 7 + 89
100 = 12 + 3 - 4 + 5 + 67 + 8 + 9
100 = 12 - 3 - 4 + 5 - 6 + 7 + 89
100 = 123 + 4 - 5 + 67 - 89
100 = 123 + 45 - 67 + 8 - 9
100 = 123 - 4 - 5 - 6 - 7 + 8 - 9
100 = 123 - 45 - 67 + 89
100 = - 1 + 2 - 3 + 4 + 5 + 6 + 78 + 9

9 has the maximum number of solutions, namely 46

211 is the lowest positive number with no solutions

The ten highest numbers that do have solutions are:

123456789 = 123456789
23456790 = 1 + 23456789
23456788 = - 1 + 23456789
12345687 = 12345678 + 9
12345669 = 12345678 - 9
3456801 = 12 + 3456789
3456792 = 1 + 2 + 3456789
3456790 = - 1 + 2 + 3456789
3456788 = 1 - 2 + 3456789
3456786 = - 1 - 2 + 3456789
```

## Lua

Translation of: C
`local expressionsLength = 0function compareExpressionBySum(a, b)    return a.sum - b.sumend local countSumsLength = 0function compareCountSumsByCount(a, b)    return a.counts - b.countsend function evaluate(code)    local value = 0    local number = 0    local power = 1    for k=9,1,-1 do        number = power*k + number        local mod = code % 3        if mod == 0 then            -- ADD            value = value + number            number = 0            power = 1        elseif mod == 1 then            -- SUB            value = value - number            number = 0            power = 1        elseif mod == 2 then            -- JOIN            power = 10 * power        else            print("This should not happen.")        end        code = math.floor(code / 3)    end    return valueend function printCode(code)    local a = 19683    local b = 6561    local s = ""    for k=1,9 do        local temp = math.floor((code % a) / b)        if temp == 0 then            -- ADD            if k>1 then                s = s .. '+'            end        elseif temp == 1 then            -- SUB            s = s .. '-'        end        a = b        b = math.floor(b/3)        s = s .. tostring(k)    end    print("\t"..evaluate(code).." = "..s)end -- Mainlocal nexpr = 13122 print("Show all solutions that sum to 100")for i=0,nexpr-1 do    if evaluate(i) == 100 then        printCode(i)    endendprint() print("Show the sum that has the maximum number of solutions")local nbest = -1for i=0,nexpr-1 do    local test = evaluate(i)    if test>0 then        local ntest = 0        for j=0,nexpr-1 do            if evaluate(j) == test then                ntest = ntest + 1            end            if ntest > nbest then                best = test                nbest = ntest            end        end    endendprint(best.." has "..nbest.." solutions\n") print("Show the lowest positive number that can't be expressed")local code = -1for i=0,123456789 do    for j=0,nexpr-1 do        if evaluate(j) == i then            code = j            break        end    end    if evaluate(code) ~= i then        code = i        break    endendprint(code.."\n") print("Show the ten highest numbers that can be expressed")local limit = 123456789 + 1for i=1,10 do    local best=0    for j=0,nexpr-1 do        local test = evaluate(j)        if (test<limit) and (test>best) then            best = test        end    end    for j=0,nexpr-1 do        if evaluate(j) == best then            printCode(j)        end    end    limit = bestend`
Output:
```Show all solutions that sum to 100
100 = 1+2+3-4+5+6+78+9
100 = 1+2+34-5+67-8+9
100 = 1+23-4+5+6+78-9
100 = 1+23-4+56+7+8+9
100 = 12+3+4+5-6-7+89
100 = 12+3-4+5+67+8+9
100 = 12-3-4+5-6+7+89
100 = 123+4-5+67-89
100 = 123+45-67+8-9
100 = 123-4-5-6-7+8-9
100 = 123-45-67+89
100 = -1+2-3+4+5+6+78+9

Show the sum that has the maximum number of solutions
9 has 46 solutions

Show the lowest positive number that can't be expressed
211

Show the ten highest numbers that can be expressed
123456789 = 123456789
23456790 = 1+23456789
23456788 = -1+23456789
12345687 = 12345678+9
12345669 = 12345678-9
3456801 = 12+3456789
3456792 = 1+2+3456789
3456790 = -1+2+3456789
3456788 = 1-2+3456789
3456786 = -1-2+3456789```

## Mathematica

Defining all possible sums:

`operations =   DeleteCases[Tuples[{"+", "-", ""}, 9], {x_, y__} /; x == "+"]; sums =   Map[StringJoin[Riffle[#, CharacterRange["1", "9"]]] &, operations];`

Sums to 100:

` [email protected][sums, [email protected]# == 100 &] `
Output:
```-1+2-3+4+5+6+78+9
1+2+3-4+5+6+78+9
1+2+34-5+67-8+9
1+23-4+5+6+78-9
1+23-4+56+7+8+9
12+3+4+5-6-7+89
12+3-4+5+67+8+9
12-3-4+5-6+7+89
123+4-5+67-89
123+45-67+8-9
123-4-5-6-7+8-9
123-45-67+89```

Maximum number of solutions:

` MaximalBy[[email protected]@sums, Identity] `
Output:
` <|9 -> 46, -9 -> 46|> `

First unsolvable:

` pos = Cases[[email protected], _?Positive];n = 1; While[MemberQ[pos, n], ++n]; `
Output:
`211`

Ten largest sums:

` {#, [email protected]#}&/@TakeLargestBy[sums, ToExpression, 10]//TableForm `
Output:
``` 123456789	123456789
1+23456789	23456790
-1+23456789	23456788
12345678+9	12345687
12345678-9	12345669
12+3456789	3456801
1+2+3456789	3456792
-1+2+3456789	3456790
1-2+3456789	3456788
-1-2+3456789	3456786 ```

## Modula-2

Translation of: C
`MODULE SumTo100;FROM FormatString IMPORT FormatString;FROM Terminal IMPORT WriteString,WriteLn,ReadChar; PROCEDURE Evaluate(code : INTEGER) : INTEGER;VAR    value,number,power,k : INTEGER;BEGIN    value := 0;    number := 0;    power := 1;     FOR k:=9 TO 1 BY -1 DO        number := power * k + number;        IF code MOD 3 = 0 THEN            (* ADD *)            value := value + number;            number := 0;            power := 1        ELSIF code MOD 3 = 1 THEN            (* SUB *)            value := value - number;            number := 0;            power := 1        ELSE            (* CAT *)            power := power * 10        END;        code := code / 3    END;     RETURN valueEND Evaluate; PROCEDURE Print(code : INTEGER);VAR    expr,buf : ARRAY[0..63] OF CHAR;    a,b,k,p : INTEGER;BEGIN    a := 19683;    b := 6561;    p := 0;     FOR k:=1 TO 9 DO        IF (code MOD a) / b = 0 THEN            IF k > 1 THEN                expr[p] := '+';                INC(p)            END        ELSIF (code MOD a) / b = 1 THEN            expr[p] := '-';            INC(p)        END;         a := b;        b := b / 3;        expr[p] := CHR(k + 30H);        INC(p)    END;    expr[p] := 0C;     FormatString("%9i = %s\n", buf, Evaluate(code), expr);    WriteString(buf)END Print; (* Main *)CONST nexpr = 13122;VAR    i,j : INTEGER;    best,nbest,test,ntest,limit : INTEGER;    buf : ARRAY[0..63] OF CHAR;BEGIN    WriteString("Show all solution that sum to 100");    WriteLn;    FOR i:=0 TO nexpr-1 DO        IF Evaluate(i) = 100 THEN            Print(i)        END    END;    WriteLn;     WriteString("Show the sum that has the maximum number of solutions");    WriteLn;    nbest := -1;    FOR i:=0 TO nexpr-1 DO        test := Evaluate(i);        IF test > 0 THEN            ntest := 0;            FOR j:=0 TO nexpr-1 DO                IF Evaluate(j) = test THEN                    INC(ntest)                END;                IF ntest > nbest THEN                    best := test;                    nbest := ntest                END            END        END    END;    FormatString("%i has %i solutions\n\n", buf, best, nbest);    WriteString(buf);     WriteString("Show the lowest positive number that can't be expressed");    WriteLn;    FOR i:=0 TO 123456789 DO        FOR j:=0 TO nexpr-1 DO            IF i = Evaluate(j) THEN                BREAK            END        END;        IF i # Evaluate(j) THEN            BREAK        END    END;    FormatString("%i\n\n", buf, i);    WriteString(buf);     WriteString("Show the ten highest numbers that can be expressed");    WriteLn;    limit := 123456789 + 1;    FOR i:=1 TO 10 DO        best := 0;        FOR j:=0 TO nexpr-1 DO            test := Evaluate(j);            IF (test < limit) AND (test > best) THEN                best := test            END        END;        FOR j:=0 TO nexpr-1 DO            IF Evaluate(j) = best THEN                Print(j)            END        END;        limit := best    END;     ReadCharEND SumTo100.`
Output:
```Show all solutions that sum to 100
100 = 1+2+3-4+5+6+78+9
100 = 1+2+34-5+67-8+9
100 = 1+23-4+5+6+78-9
100 = 1+23-4+56+7+8+9
100 = 12+3+4+5-6-7+89
100 = 12+3-4+5+67+8+9
100 = 12-3-4+5-6+7+89
100 = 123+4-5+67-89
100 = 123+45-67+8-9
100 = 123-4-5-6-7+8-9
100 = 123-45-67+89
100 = -1+2-3+4+5+6+78+9

Show the sum that has the maximum number of solutions
9 has 46 solutions

Show the lowest positive number that can't be expressed
211

Show the ten highest numbers that can be expressed
123456789 = 123456789
23456790 = 1+23456789
23456788 = -1+23456789
12345687 = 12345678+9
12345669 = 12345678-9
3456801 = 12+3456789
3456792 = 1+2+3456789
3456790 = -1+2+3456789
3456788 = 1-2+3456789
3456786 = -1-2+3456789```

## Nim

` import strutils var  ligne: string = ""  sum: int  opera: array[0..9, int] = [0,0,1,1,1,1,1,1,1,1]  curseur: int = 9  boucle: bool  tot: array[1..123456789, int]  pG: int  plusGrandes: array[1..10, string] let  ope: array[0..3, string] = ["-",""," +"," -"]  aAtteindre = 100 proc calcul(li: string): int =  var liS: seq[string]  liS = split(li," ")  for i in liS:    result += parseInt(i) echo "Valeur à atteindre : ",aAtteindre while opera<2:  ligne.add(ope[opera])  ligne.add("1")  for i in 2..9:    ligne.add(ope[opera[i]])    ligne.add(\$i)  sum = calcul(ligne)  if sum == aAtteindre:    stdout.write(ligne)    echo " = ",sum  if sum>0:    tot[sum] += 1    pG = 1    while pG<10:      if sum>calcul(plusGrandes[pG]):        for k in countdown(10,pG+1):          plusGrandes[k]=plusGrandes[k-1]        plusGrandes[pG]=ligne        pG = 11      pG += 1  ligne = ""  boucle = true  while boucle:    opera[curseur] += 1    if opera[curseur] == 4:      opera[curseur]=1      curseur -= 1    else:      curseur = 9      boucle = false echo "Valeur atteinte ",tot[aAtteindre]," fois."echo "" var  min0: int = 0  max: int = 0  valmax: int = 0 for i in 1..123456789:  if tot[i]==0 and min0 == 0:    min0 = i  if tot[i]>max:    max = tot[i]    valmax = i echo "Plus petite valeur ne pouvant pas être atteinte : ",min0echo "Valeur atteinte le plus souvent : ",valmax,", atteinte ",max," fois."echo ""echo "Plus grandes valeurs pouvant être atteintes :"for i in 1..10:  echo calcul(plusGrandes[i])," = ",plusGrandes[i]`
Output:
```Valeur à atteindre : 100
-1 +2 -3 +4 +5 +6 +78 +9 = 100
123 +45 -67 +8 -9 = 100
123 +4 -5 +67 -89 = 100
123 -45 -67 +89 = 100
123 -4 -5 -6 -7 +8 -9 = 100
12 +3 +4 +5 -6 -7 +89 = 100
12 +3 -4 +5 +67 +8 +9 = 100
12 -3 -4 +5 -6 +7 +89 = 100
1 +23 -4 +56 +7 +8 +9 = 100
1 +23 -4 +5 +6 +78 -9 = 100
1 +2 +34 -5 +67 -8 +9 = 100
1 +2 +3 -4 +5 +6 +78 +9 = 100
Valeur atteinte 12 fois.

Plus petite valeur ne pouvant pas être atteinte : 211
Valeur atteinte le plus souvent : 9, atteinte 46 fois.

Plus grandes valeurs pouvant être atteintes :
123456789 = 123456789
23456790 = 1 +23456789
23456788 = -1 +23456789
12345687 = 12345678 +9
12345669 = 12345678 -9
3456801 = 12 +3456789
3456792 = 1 +2 +3456789
3456790 = -1 +2 +3456789
3456788 = 1 -2 +3456789
3456786 = -1 -2 +3456789
```

## Pascal

Works with: Lazarus
Translation of: C
`{ RossetaCode: Sum to 100, Pascal.   Find solutions to the "sum to one hundred" puzzle.   We don't use arrays, but recompute all values again and again.   It is a little surprise that the time efficiency is quite acceptable. } program sumto100; const  ADD = 0; SUB = 1; JOIN = 2; { opcodes inserted between digits }  NEXPR = 13122;              { the total number of expressions }var  i, j: integer;  loop: boolean;  test, ntest, best, nbest, limit: integer;   function evaluate(code: integer): integer;  var    k: integer;    value, number, power: integer;  begin    value  := 0;    number := 0;    power  := 1;    for  k := 9 downto 1 do    begin      number := power * k + number;      case code mod 3 of        ADD: begin value := value + number; number := 0; power := 1; end;        SUB: begin value := value - number; number := 0; power := 1; end;        JOIN:                                            power := power * 10      end;      code := code div 3    end;    evaluate := value  end;   procedure print(code: integer);  var    k: integer;    a, b: integer;  begin    a := 19683;    b := 6561;    write( evaluate(code):9 );    write(' = ');    for  k := 1 to 9 do    begin      case ((code mod a) div b) of        ADD: if k > 1 then write('+');        SUB: { always }    write('-');      end;      a := b;      b := b div 3;      write( k:1 )    end;    writeln  end; begin  writeln;  writeln('Show all solutions that sum to 100');  writeln;  for i := 0 to NEXPR - 1 do    if evaluate(i) = 100 then      print(i);   writeln;  writeln('Show the sum that has the maximum number of solutions');  writeln;  nbest := (-1);  for i := 0 to NEXPR - 1 do  begin    test := evaluate(i);    if test > 0 then    begin      ntest := 0;      for j := 0 to NEXPR - 1 do        if evaluate(j) = test then          ntest := ntest + 1;      if ntest > nbest then      begin        best := test;        nbest := ntest;      end    end  end;  writeln(best, ' has ', nbest, ' solutions');   writeln;  writeln('Show the lowest positive number that can''t be expressed');  writeln;  i := 0;  loop := TRUE;  while (i <= 123456789) and loop do  begin    j := 0;    while (j < NEXPR - 1) and (i <> evaluate(j)) do      j := j + 1;    if i <> evaluate(j) then      loop := FALSE    else      i := i + 1;  end;  writeln(i);   writeln;  writeln('Show the ten highest numbers that can be expressed');  writeln;  limit := 123456789 + 1;  for i := 1 to 10 do  begin    best := 0;    for j := 0 to NEXPR - 1 do    begin      test := evaluate(j);      if (test < limit) and (test > best) then        best := test;    end;    for j := 0 to NEXPR - 1 do      if evaluate(j) = best then        print(j);    limit := best;  endend.`
Output:
```Show all solutions that sum to 100

100 = 1+2+3-4+5+6+78+9
100 = 1+2+34-5+67-8+9
100 = 1+23-4+5+6+78-9
100 = 1+23-4+56+7+8+9
100 = 12+3+4+5-6-7+89
100 = 12+3-4+5+67+8+9
100 = 12-3-4+5-6+7+89
100 = 123+4-5+67-89
100 = 123+45-67+8-9
100 = 123-4-5-6-7+8-9
100 = 123-45-67+89
100 = -1+2-3+4+5+6+78+9

Show the sum that has the maximum number of solutions

9 has 46 solutions

Show the lowest positive number that can't be expressed

211

Show the ten highest numbers that can be expressed

123456789 = 123456789
23456790 = 1+23456789
23456788 = -1+23456789
12345687 = 12345678+9
12345669 = 12345678-9
3456801 = 12+3456789
3456792 = 1+2+3456789
3456790 = -1+2+3456789
3456788 = 1-2+3456789
3456786 = -1-2+3456789
```

## Perl

Works with: Perl version 5.10
`#!/usr/bin/perluse warnings;use strict;use feature qw{ say }; my \$string = '123456789';my \$length = length \$string;my @possible_ops = ("" , '+', '-'); {    my @ops;    sub Next {        return @ops = (0) x (\$length) unless @ops;         my \$i = 0;        while (\$i < \$length) {            if (\$ops[\$i]++ > \$#possible_ops - 1) {                \$ops[\$i++] = 0;                next            }            # + before the first number            next if 0 == \$i && '+' eq \$possible_ops[ \$ops ];             return @ops        }        return    }} sub evaluate {    my (\$expression) = @_;    my \$sum;    \$sum += \$_ for \$expression =~ /([-+]?[0-9]+)/g;    return \$sum} my %count = ( my \$max_count = 0 => 0 ); say 'Show all solutions that sum to 100'; while (my @ops = Next()) {    my \$expression = "";    for my \$i (0 .. \$length - 1) {        \$expression .= \$possible_ops[ \$ops[\$i] ];        \$expression .= substr \$string, \$i, 1;    }    my \$sum = evaluate(\$expression);    ++\$count{\$sum};    \$max_count = \$sum if \$count{\$sum} > \$count{\$max_count};    say \$expression if 100 == \$sum;} say 'Show the sum that has the maximum number of solutions';say "sum: \$max_count; solutions: \$count{\$max_count}"; my \$n = 1;++\$n until ! exists \$count{\$n};say "Show the lowest positive sum that can't be expressed";say \$n; say 'Show the ten highest numbers that can be expressed';say for (sort { \$b <=> \$a } keys %count)[0 .. 9];`
Output:
```Show all solutions that sum to 100
123-45-67+89
12-3-4+5-6+7+89
12+3+4+5-6-7+89
123+4-5+67-89
-1+2-3+4+5+6+78+9
1+2+3-4+5+6+78+9
12+3-4+5+67+8+9
1+23-4+56+7+8+9
1+2+34-5+67-8+9
1+23-4+5+6+78-9
123+45-67+8-9
123-4-5-6-7+8-9
Show the sum that has the maximum number of solutions
sum: 9; solutions: 46
Show the lowest positive sum that can't be expressed
211
Show the ten highest numbers that can be expressed
123456789
23456790
23456788
12345687
12345669
3456801
3456792
3456790
3456788
3456786```

### oneliner version

The first task posed can be solved simply with (pay attention to doublequotes around the program: adjust for you OS):

` perl -E "say for grep{eval \$_ == 100} glob '{-,}'.join '{+,-,}',1..9" `

While the whole task can be solved by:

` perl -MList::Util="first" -E "@c[0..10**6]=(0..10**6);say for grep{\$e=eval;\$c[\$e]=undef if \$e>=0;\$h{\$e}++;eval \$_==100}glob'{-,}'.join'{+,-,}',1..9;END{say for(sort{\$h{\$b}<=>\$h{\$a}}grep{\$_>=0}keys %h),first{defined \$_}@c;say for(sort{\$b<=>\$a}grep{\$_>0}keys %h)[0..9]}" `

which outputs

```-1+2-3+4+5+6+78+9
1+2+3-4+5+6+78+9
1+2+34-5+67-8+9
1+23-4+5+6+78-9
1+23-4+56+7+8+9
12+3+4+5-6-7+89
12+3-4+5+67+8+9
12-3-4+5-6+7+89
123+4-5+67-89
123+45-67+8-9
123-4-5-6-7+8-9
123-45-67+89
9
211
123456789
23456790
23456788
12345687
12345669
3456801
3456792
3456790
3456788
3456786
```

## Perl 6

Works with: Rakudo version 2017.03
`my \$sum = 100;my \$N   = 10;my @ops = ['-', ''], |( [' + ', ' - ', ''] xx 8 );my @str = [X~] map { .Slip }, ( @ops Z 1..9 );my %sol = @str.classify: *.subst( ' - ', ' -', :g )\                          .subst( ' + ',  ' ', :g ).words.sum; my %count.push: %sol.map({ .value.elems => .key }); my \$max-solutions    = %count.max( + *.key );my \$first-unsolvable = first { %sol{\$_} :!exists }, 1..*;sub n-largest-sums (Int \$n) { %sol.sort(-*.key)[^\$n].fmt: "%8s => %s\n" } given %sol{\$sum}:p {    say "{.value.elems} solutions for sum {.key}:";    say "    \$_" for .value.list;} .say for :\$max-solutions, :\$first-unsolvable, "\$N largest sums:", n-largest-sums(\$N);`
Output:
```12 solutions for sum 100:
-1 + 2 - 3 + 4 + 5 + 6 + 78 + 9
1 + 2 + 3 - 4 + 5 + 6 + 78 + 9
1 + 2 + 34 - 5 + 67 - 8 + 9
1 + 23 - 4 + 5 + 6 + 78 - 9
1 + 23 - 4 + 56 + 7 + 8 + 9
12 + 3 + 4 + 5 - 6 - 7 + 89
12 + 3 - 4 + 5 + 67 + 8 + 9
12 - 3 - 4 + 5 - 6 + 7 + 89
123 + 4 - 5 + 67 - 89
123 + 45 - 67 + 8 - 9
123 - 4 - 5 - 6 - 7 + 8 - 9
123 - 45 - 67 + 89
max-solutions => 46 => [-9 9]
first-unsolvable => 211
10 largest sums:
123456789 => 123456789
23456790 => 1 + 23456789
23456788 => -1 + 23456789
12345687 => 12345678 + 9
12345669 => 12345678 - 9
3456801 => 12 + 3456789
3456792 => 1 + 2 + 3456789
3456790 => -1 + 2 + 3456789
3456788 => 1 - 2 + 3456789
3456786 => -1 - 2 + 3456789```

## Phix

This is just a trivial count in base 3, with a leading '+' being irrelevant, so from 0(3)000_000_000 to 0(3)122_222_222 which is only (in decimal) 13,122 ...
Admittedly, categorising them into 3429 bins is slightly more effort, but otherwise I am somewhat bemused by all the applescript/javascript/Haskell shenanegins.

`enum SUB=-1, NOP=0, ADD=1 function eval(sequence s)integer res = 0, this = 0, op = ADD    for i=1 to length(s) do        if s[i]=NOP then            this = this*10+i        else            res += op*this            this = i            op = s[i]        end if    end for    return res + op*thisend function procedure show(sequence s)string res = ""    for i=1 to length(s) do        if s[i]!=NOP then            res &= ','-s[i]        end if                  res &= '0'+i    end for    puts(1,res&" = ")end procedure -- Logically this intersperses -/nop/+ between each digit, but you do not actually need the digit.sequence s = repeat(SUB,9)  -- (==> ..nop+add*8) bool done = falseinteger maxl = 0, maxrinteger count = 0while not done do    count += 1    integer r = eval(s), k = getd_index(r)    sequence solns = iff(k=0?{s}:append(getd_by_index(k),s))    setd(r,solns)    if r>0 and maxl<length(solns) then        maxl = length(solns)        maxr = r    end if    for i=length(s) to 1 by -1 do        if i=1 and s[i]=NOP then            done = true            exit        elsif s[i]!=ADD then            s[i] += 1            exit        end if        s[i] = SUB    end forend while printf(1,"%d solutions considered (dictionary size: %d)\n",{count,dict_size()}) sequence s100 = getd(100)printf(1,"There are %d sums to 100:\n",{length(s100)})for i=1 to length(s100) do    show(s100[i])    ?100end for printf(1,"The positive sum of %d has the maximum number of solutions: %d\n",{maxr,maxl}) integer prev = 0function missing(integer key, sequence /*data*/, integer /*pkey*/, object /*user_data=-2*/)    if key!=prev+1 then        return 0    end if    prev = key    return 1end functiontraverse_dict_partial_key(routine_id("missing"),1)printf(1,"The lowest positive sum that cannot be expressed: %d\n",{prev+1}) sequence highest = {}function top10(integer key, sequence /*data*/, object /*user_data*/)    highest &= key    return length(highest)<10end functiontraverse_dict(routine_id("top10"),rev:=1)printf(1,"The 10 highest sums: ") ?highest`
Output:
```13122 solutions considered (dictionary size: 3429)
There are 12 sums to 100:
-1+2-3+4+5+6+78+9 = 100
12-3-4+5-6+7+89 = 100
123-4-5-6-7+8-9 = 100
123-45-67+89 = 100
123+4-5+67-89 = 100
123+45-67+8-9 = 100
12+3-4+5+67+8+9 = 100
12+3+4+5-6-7+89 = 100
1+23-4+56+7+8+9 = 100
1+23-4+5+6+78-9 = 100
1+2+3-4+5+6+78+9 = 100
1+2+34-5+67-8+9 = 100
The positive sum of 9 has the maximum number of solutions: 46
The lowest positive sum that cannot be expressed: 211
The 10 highest sums: {123456789,23456790,23456788,12345687,12345669,3456801,3456792,3456790,3456788,3456786}
```

## Python

`from itertools import product, islice  def expr(p):    return "{}1{}2{}3{}4{}5{}6{}7{}8{}9".format(*p)  def gen_expr():    op = ['+', '-', '']    return [expr(p) for p in product(op, repeat=9) if p != '+']  def all_exprs():    values = {}    for expr in gen_expr():        val = eval(expr)        if val not in values:            values[val] = 1        else:            values[val] += 1    return values  def sum_to(val):    for s in filter(lambda x: x == val, map(lambda x: (eval(x), x), gen_expr())):        print(s)  def max_solve():    print("Sum {} has the maximum number of solutions: {}".          format(*max(all_exprs().items(), key=lambda x: x)))  def min_solve():    values = all_exprs()    for i in range(123456789):        if i not in values:            print("Lowest positive sum that can't be expressed: {}".format(i))            return  def highest_sums(n=10):    sums = map(lambda x: x,               islice(sorted(all_exprs().items(), key=lambda x: x, reverse=True), n))    print("Highest Sums: {}".format(list(sums)))  sum_to(100)max_solve()min_solve()highest_sums()`
Output:
```(100, '-1+2-3+4+5+6+78+9')
(100, '1+2+3-4+5+6+78+9')
(100, '1+2+34-5+67-8+9')
(100, '1+23-4+5+6+78-9')
(100, '1+23-4+56+7+8+9')
(100, '12+3+4+5-6-7+89')
(100, '12+3-4+5+67+8+9')
(100, '12-3-4+5-6+7+89')
(100, '123+4-5+67-89')
(100, '123+45-67+8-9')
(100, '123-4-5-6-7+8-9')
(100, '123-45-67+89')
Sum 9 has the maximum number of solutions: 46
Lowest positive sum that can't be expressed: 211
Highest Sums: [123456789, 23456790, 23456788, 12345687, 12345669, 3456801, 3456792, 3456790, 3456788, 3456786]```

### Alternate solution

Mostly the same algorithm, but both shorter and faster.

`import itertoolsfrom collections import defaultdict, Counter s = "123456789"h = defaultdict(list)for v in itertools.product(["+", "-", ""], repeat=9):    if v != "+":        e = "".join("".join(u) for u in zip(v, s))        h[eval(e)].append(e) print("Solutions for 100")for e in h:    print(e) c = Counter({k: len(v) for k, v in h.items() if k >= 0}) k, m = c.most_common(1)print("Maximum number of solutions for %d (%d solutions)" % (k, m)) v = sorted(c.keys()) for i in range(v[-1]):    if i not in c:        print("Lowest impossible sum: %d" % i)        break print("Ten highest sums")for k in reversed(v[-10:]):    print(k)`
Output:
```Solutions for 100
-1+2-3+4+5+6+78+9
1+2+3-4+5+6+78+9
1+2+34-5+67-8+9
1+23-4+5+6+78-9
1+23-4+56+7+8+9
12+3+4+5-6-7+89
12+3-4+5+67+8+9
12-3-4+5-6+7+89
123+4-5+67-89
123+45-67+8-9
123-4-5-6-7+8-9
123-45-67+89
Maximum number of solutions for 9 (46 solutions)
Lowest impossible sum: 211
Ten highest sums
123456789
23456790
23456788
12345687
12345669
3456801
3456792
3456790
3456788
3456786```

## Racket

`#lang racket (define list-partitions  (match-lambda    [(list) (list null)]    [(and L (list _)) (list (list L))]    [(list L ...)     (for*/list          ((i (in-range 1 (add1 (length L))))           (r (in-list (list-partitions (drop L i)))))        (cons (take L i) r))])) (define digits->number (curry foldl (λ (dgt acc) (+ (* 10 acc) dgt)) 0)) (define partition-digits-to-numbers  (let ((memo (make-hash)))    (λ (dgts)      (hash-ref! memo dgts                 (λ ()                   (map (λ (p) (map digits->number p))                        (list-partitions dgts))))))) (define (fold-sum-to-ns digits kons k0)  (define (get-solutions nmbrs acc chain k)    (match nmbrs      [(list)       (kons (cons acc (let ((niahc (reverse chain)))                                      (if (eq? '+ (car niahc)) (cdr niahc) niahc)))             k)]      [(cons a d)       (get-solutions d (- acc a) (list* a '- chain)                      (get-solutions d (+ acc a) (list* a '+ chain) k))]))  (foldl (λ (nmbrs k) (get-solutions nmbrs 0 null k)) k0 (partition-digits-to-numbers digits))) (define sum-to-ns/hash-promise  (delay (fold-sum-to-ns          '(1 2 3 4 5 6 7 8 9)          (λ (a.s d) (hash-update d (car a.s) (λ (x) (cons (cdr a.s) x)) list))          (hash)))) (module+ main  (define S (force sum-to-ns/hash-promise))  (displayln "Show all solutions that sum to 100")  (pretty-print (hash-ref S 100))   (displayln "Show the sum that has the maximum number of solutions (from zero to infinity*)")  (let-values (([k-max v-max]                (for/fold ((k-max #f) (v-max 0))                          (([k v] (in-hash S)) #:when (> (length v) v-max))                  (values k (length v)))))    (printf "~a has ~a solutions~%" k-max v-max))   (displayln "Show the lowest positive sum that can't be expressed (has no solutions), using the rules for this task")  (for/first ((n (in-range 1 (add1 123456789))) #:unless (hash-has-key? S n)) n)   (displayln "Show the ten highest numbers that can be expressed using the rules for this task")  (take (sort (hash-keys S) >) 10)) (module+ test  (require rackunit)  (check-equal? (list-partitions null) '(()))  (check-equal? (list-partitions '(1)) '(((1))))  (check-equal? (list-partitions '(1 2)) '(((1) (2)) ((1 2))))  (check-equal? (partition-digits-to-numbers '()) '(()))  (check-equal? (partition-digits-to-numbers '(1)) '((1)))  (check-equal? (partition-digits-to-numbers '(1 2)) '((1 2) (12))))`
Output:
```Show all solutions that sum to 100
'((123 - 45 - 67 + 89)
(123 + 45 - 67 + 8 - 9)
(123 + 4 - 5 + 67 - 89)
(123 - 4 - 5 - 6 - 7 + 8 - 9)
(12 + 3 - 4 + 5 + 67 + 8 + 9)
(12 - 3 - 4 + 5 - 6 + 7 + 89)
(12 + 3 + 4 + 5 - 6 - 7 + 89)
(1 + 23 - 4 + 56 + 7 + 8 + 9)
(1 + 23 - 4 + 5 + 6 + 78 - 9)
(1 + 2 + 34 - 5 + 67 - 8 + 9)
(- 1 + 2 - 3 + 4 + 5 + 6 + 78 + 9)
(1 + 2 + 3 - 4 + 5 + 6 + 78 + 9))
Show the sum that has the maximum number of solutions (from zero to infinity*)
9 has 46 solutions
Show the lowest positive sum that can't be expressed (has no solutions),
using the rules for this task
211
Show the ten highest numbers that can be expressed using the rules for this task
'(123456789 23456790 23456788 12345687 12345669 3456801 3456792 3456790 3456788 3456786)```

## REXX

`/*REXX pgm solves a puzzle:  using the string 123456789, insert  -  or  +  to sum to 100*/parse arg LO HI .                                /*obtain optional arguments from the CL*/if LO=='' | LO==","  then LO=100                 /*Not specified?  Then use the default.*/if HI=='' | HI==","  then HI=LO                  /* "      "         "   "   "     "    */if LO==00            then HI=123456789           /*LOW specified as zero with leading 0.*/ops= '+-';             L=length(ops) + 1         /*define operators (and their length). */@.=;        do i=1  to L-1;  @.i=substr(ops,i,1) /*   "   some handy-dandy REXX literals*/            end   /*i*/                          /*   "   individual operators for speed*/mx=0;  mn=999999                                 /*initialize the minimums and maximums.*/mxL=;  mnL=;       do j=LO  to HI  until LO==00  &  mn==0   /*solve with a range of sums*/                   z=solve(j)                               /*find # of solutions for J.*/                   if z> mx  then mxL=                      /*see if this is a new max. */                   if z>=mx  then do; mxL=mxL j; mx=z; end  /*remember this new maximum.*/                   if z< mn  then mnL=                      /*see if this is a new min. */                   if z<=mn  then do; mnL=mnL j; mn=z; end  /*remember this new minimum.*/                   end   /*j*/if LO==HI then exit                                         /*don't display max & min ? */@@= 'number of solutions: ';   say_=words(mxL);  say 'sum's(_)   "of"   mxL  ' 's(_,"have",'has')   'the maximum'    @@   mx_=words(mnL);  say 'sum's(_)   "of"   mnL  ' 's(_,"have",'has')   'the minimum'    @@   mnexit                                             /*stick a fork in it,  we're all done. *//*──────────────────────────────────────────────────────────────────────────────────────*/s:     if arg(1)==1  then return arg(3);  return word(arg(2) "s",1)  /*simple pluralizer*//*──────────────────────────────────────────────────────────────────────────────────────*/solve: parse arg answer;         # =0            /*obtain the answer (sum) to the puzzle*/          do a=L-1  to L;        aa=      @.a'1' /*choose one  of  ─       or  nothing. */           do b=1  for L;        bb=aa || @.b'2' /*   "    "    "  ─   +,  or  abutment.*/            do c=1  for L;       cc=bb || @.c'3' /*   "    "    "  "   "    "      "    */             do d=1  for L;      dd=cc || @.d'4' /*   "    "    "  "   "    "      "    */              do e=1  for L;     ee=dd || @.e'5' /*   "    "    "  "   "    "      "    */               do f=1  for L;    ff=ee || @.f'6' /*   "    "    "  "   "    "      "    */                do g=1  for L;   gg=ff || @.g'7' /*   "    "    "  "   "    "      "    */                 do h=1  for L;  hh=gg || @.h'8' /*   "    "    "  "   "    "      "    */                  do i=1  for L; ii=hh || @.i'9' /*   "    "    "  "   "    "      "    */                  interpret '\$=' ii              /*calculate the sum of modified string.*/                  if \$\==answer  then iterate    /*Is sum not equal to answer? Then skip*/                  #=#+1;         if LO==HI  then say 'solution: '    \$    " ◄───► "     ii                  end   /*i*/                 end    /*h*/                end     /*g*/               end      /*f*/              end       /*e*/             end        /*d*/            end         /*c*/           end          /*b*/          end           /*a*/       y=#                                       /* [↓]  adjust the number of solutions?*/       if y==0  then y='no'                      /* [↓]  left justify plural of solution*/       if LO\==00  then say right(y, 9)           'solution's(#, , " ")   'found for'  ,                            right(j, length(HI) )                         left('', #, "─")       return #                                  /*return the number of solutions found.*/`
output   when using the default input:
```solution:  100  ◄───►  -1+2-3+4+5+6+78+9
solution:  100  ◄───►  1+2+3-4+5+6+78+9
solution:  100  ◄───►  1+2+34-5+67-8+9
solution:  100  ◄───►  1+23-4+5+6+78-9
solution:  100  ◄───►  1+23-4+56+7+8+9
solution:  100  ◄───►  12+3+4+5-6-7+89
solution:  100  ◄───►  12+3-4+5+67+8+9
solution:  100  ◄───►  12-3-4+5-6+7+89
solution:  100  ◄───►  123+4-5+67-89
solution:  100  ◄───►  123+45-67+8-9
solution:  100  ◄───►  123-4-5-6-7+8-9
solution:  100  ◄───►  123-45-67+89
12 solutions found for 100
```
output   when the following input is used:   00
```sum of  9  has the maximum number of solutions:  46
sum of  211  has the minimum number of solutions:  0
```

## Ruby

Translation of: Elixir
`def gen_expr  x = ['-', '']  y = ['+', '-', '']  x.product(y,y,y,y,y,y,y,y)   .map do |a,b,c,d,e,f,g,h,i|      "#{a}1#{b}2#{c}3#{d}4#{e}5#{f}6#{g}7#{h}8#{i}9"    endend def sum_to(val)  gen_expr.map{|expr| [eval(expr), expr]}.select{|v,expr| v==val}.each{|x| p x}end def max_solve  n,size = gen_expr.group_by{|expr| eval(expr)}                   .select{|val,_| val>=0}                   .map{|val,exprs| [val, exprs.size]}                   .max_by{|_,size| size}  puts "sum of #{n} has the maximum number of solutions : #{size}"end def min_solve  solves = gen_expr.group_by{|expr| eval(expr)}  n = 0.step{|i| break i unless solves[i]}  puts "lowest positive sum that can't be expressed : #{n}"end def highest_sums(n=10)  n = gen_expr.map{|expr| eval(expr)}.uniq.sort.reverse.take(n)  puts "highest sums : #{n}"end sum_to(100)max_solvemin_solvehighest_sums`
Output:
```[100, "-1+2-3+4+5+6+78+9"]
[100, "1+2+3-4+5+6+78+9"]
[100, "1+2+34-5+67-8+9"]
[100, "1+23-4+5+6+78-9"]
[100, "1+23-4+56+7+8+9"]
[100, "12+3+4+5-6-7+89"]
[100, "12+3-4+5+67+8+9"]
[100, "12-3-4+5-6+7+89"]
[100, "123+4-5+67-89"]
[100, "123+45-67+8-9"]
[100, "123-4-5-6-7+8-9"]
[100, "123-45-67+89"]
sum of 9 has the maximum number of solutions : 46
lowest positive sum that can't be expressed : 211
highest sums : [123456789, 23456790, 23456788, 12345687, 12345669, 3456801, 3456792, 3456790, 3456788, 3456786]
```

## Sidef

Translation of: Ruby
`func gen_expr() is cached {    var x = ['-', '']    var y = ['+', '-', '']     gather {        cartesian([x,y,y,y,y,y,y,y,y], {|a,b,c,d,e,f,g,h,i|            take("#{a}1#{b}2#{c}3#{d}4#{e}5#{f}6#{g}7#{h}8#{i}9")        })    }} func eval_expr(expr) is cached {    expr.scan(/([-+]?\d+)/).sum_by { Num(_) }} func sum_to(val) {    gen_expr().grep { eval_expr(_) == val }} func max_solve() {    gen_expr().grep     { eval_expr(_) >= 0 } \              .group_by { eval_expr(_)      } \              .max_by   {|_,v| v.len        }} func min_solve() {    var h = gen_expr().group_by { eval_expr(_) }    for i in (0..Inf) { h.exists(i) || return i }} func highest_sums(n=10) {    gen_expr().map { eval_expr(_) }.uniq.sort.reverse.first(n)} sum_to(100).each { say "100 = #{_}" } var (n, solutions) = max_solve()...say "Sum of #{n} has the maximum number of solutions: #{solutions.len}"say "Lowest positive sum that can't be expressed : #{min_solve()}"say "Highest sums: #{highest_sums()}"`
Output:
```100 = -1+2-3+4+5+6+78+9
100 = 1+2+3-4+5+6+78+9
100 = 1+2+34-5+67-8+9
100 = 1+23-4+5+6+78-9
100 = 1+23-4+56+7+8+9
100 = 12+3+4+5-6-7+89
100 = 12+3-4+5+67+8+9
100 = 12-3-4+5-6+7+89
100 = 123+4-5+67-89
100 = 123+45-67+8-9
100 = 123-4-5-6-7+8-9
100 = 123-45-67+89
Sum of 9 has the maximum number of solutions: 46
Lowest positive sum that can't be expressed : 211
Highest sums: [123456789, 23456790, 23456788, 12345687, 12345669, 3456801, 3456792, 3456790, 3456788, 3456786]
```

## Tcl

`proc sum_to_100 {} {    for {set i 0} {\$i <= 13121} {incr i} {	set i3 [format %09d [dec2base 3 \$i]]	set form ""	set subs {"" - +}	foreach a [split \$i3 ""] b [split 123456789 ""] {	    append form [lindex \$subs \$a] \$b	}	lappend R([expr \$form]) \$form    }    puts "solutions for sum=100:\n[join [lsort \$R(100)] \n]"    set max -1    foreach key [array names R] {	if {[llength \$R(\$key)] > \$max} {	    set max [llength \$R(\$key)]	    set maxkey \$key	}    }    puts "max solutions: \$max for \$maxkey"    for {set i 0} {\$i <= 123456789} {incr i} {	if ![info exists R(\$i)] {	    puts "first unsolvable: \$i"	    break	}    }    puts "highest 10:\n[lrange [lsort -integer -decr [array names R]] 0 9]"}proc dec2base {base dec} {    set res ""    while {\$dec > 0} {	set res [expr \$dec%\$base]\$res	set dec [expr \$dec/\$base]    }    if {\$res eq ""} {set res 0}    return \$res}sum_to_100`
```~ \$ ./sum_to_100.tcl
solutions for sum=100:
-1+2-3+4+5+6+78+9
1+2+3-4+5+6+78+9
1+2+34-5+67-8+9
1+23-4+5+6+78-9
1+23-4+56+7+8+9
12+3+4+5-6-7+89
12+3-4+5+67+8+9
12-3-4+5-6+7+89
123+4-5+67-89
123+45-67+8-9
123-4-5-6-7+8-9
123-45-67+89
max solutions: 46 for 9
first unsolvable: 211
highest 10:
123456789 23456790 23456788 12345687 12345669 3456801 3456792 3456790 3456788 3456786
```

## Visual Basic .NET

Of course, one could just code-convert the existing C# example, but I thought this could be written with some simpler constructs. The point of doing this is to make the code more compatible with other BASIC languages. Not every language has something similar to the Enumerable Range construct. I also found the Dictionary construct could be implemented with something more primitive.

Another interesting thing this program can do is solve for other sets of numbers easily, as neither the number of digits, nor the digit sequence itself, is hard-coded. You could solve for the digits 1 through 8, for example, or the digits starting at 9 and going down to 1. One can even override the target sum (of 100) parameter, if you happen to be interested in another number.

`' Recursively iterates (increments) iteration array, returns -1 when out of "digits".Function plusOne(iAry() As Integer, spot As Integer) As Integer    Dim spotLim As Integer = If(spot = 0, 1, 2) ' The first "digit" has a lower limit.    If iAry(spot) = spotLim Then ' Check if spot has reached limit        If spot = 0 Then Return -1 ' No previous spot to increment, so indicate completed.        iAry(spot) = 0 ' Reset current spot, and        Return plusOne(iAry, spot - 1) ' Increment previous spot.    Else        iAry(spot) += 1 ' Increment current spot.    End If    Return spotEnd Function ' Returns string sequence of operations from iAry and terms stringFunction generate(iAry() As Integer, terms As String) As String    Dim operations As String() = {"", "-", "+"} ' Possible operations.    generate = ""    For i As Integer = 0 To iAry.Count - 1        generate &= operations(iAry(i)) & Mid(terms, i + 1, 1).ToString()    NextEnd Function ' Returns evaluation of string sequenceFunction eval(sequence As String) As Integer    eval = 0    Dim term As Integer = 0, operation As Integer = 1    For Each ch As Char In sequence        Select Case ch            Case "-", "+" ' New operation detected, apply previous operation to term,                eval += If(operation = 0, -term, term) : term = 0 ' and reset term.                operation = If(ch = "-", 0, 1) ' Note next operation.            Case Else ' Digit detected, increase term.                term = term * 10 + Val(ch)        End Select    Next    eval += If(operation = 0, -term, term) ' Apply final term.End Function ' Sorts a pair of List(Of Integer) by the firstSub reSort(ByRef first As List(Of Integer), ByRef second As List(Of Integer))    Dim lou As New List(Of ULong) ' Temporary list of ULong for sorting.    For i As Integer = 0 To first.Count - 1        lou.Add((CULng(first(i)) << 32) + second(i)) ' "Pack" list items.    Next    lou.Sort()    For k As Integer = 0 To first.Count - 1        first(k) = lou(k) >> 32 ' "Unpack" first list item.        second(k) = lou(k) And &H7FFFFFFF ' "Unpack" second list item.    NextEnd Sub ' Returns first result not in sequence, assumes passed list is sorted before call,'  uses binary search algo.Function firstMiss(loi As List(Of Integer))    Dim low As Integer = 0, high As Integer = loi.Count - 1, middle = (low + high) \ 2    Do        If loi(middle) = middle Then low = middle + 1 Else high = middle - 1        middle = (low + high) \ 2    Loop Until high <= low    Return middle + If(loi(middle) = middle, 1, 0)End Function ' Iterates through all possible operations, '  uses a pair of List (of Integer) to tabulate solutions.Sub Solve100(Optional terms As String = "123456789",             Optional targSum As Integer = 100,             Optional highNums As Integer = 10)    Dim lastDig As Integer = Len(terms) - 1 ' The final "digit".    Dim iAry() As Integer = New Integer(lastDig) {} ' Iterations array.    Dim seq As String ' Sequence of numbers and operations.    Dim sVal As Integer ' Sequence value.    Dim sCnt As Integer = 1 ' Solution count (targSum).    Dim res As New List(Of Integer) ' List of results.    Dim tally As New List(Of Integer) ' Tally of results.    Console.WriteLine("List of solutions that evaluate to 100:")    Do ' Tabulate results until digits are exhausted.        seq = generate(iAry, terms) ' Obtain next expression.        sVal = eval(seq) ' Obtain next evaluation.        If sVal >= 0 Then ' Don't bother saving the negative results.            If res.Contains(sVal) Then tally(res.IndexOf(sVal)) += 1 _                                  Else res.Add(sVal) : tally.Add(1)            If sVal = targSum Then _                Console.WriteLine(" {0,2} {1}", sCnt, seq) : sCnt += 1        End If    Loop Until plusOne(iAry, lastDig) < 0    reSort(tally, res) ' Sort by tally to find result with the most solutions.    Console.WriteLine("The sum that has the the most solutions is {0}, (at {1}).",                      res.Last, tally.Last)    reSort(res, tally) ' Sort by result to find first missing result and top results.    Console.WriteLine("The lowest positive sum that can't be expressed is {0}.",                      firstMiss(res))    Console.WriteLine("The ten highest numbers that can be expressed are:")    res.Reverse() ' To let us take the last items for output.    sCnt = 0 ' Keep track of items displayed (for formatting).    For Each item As Integer In res.Take(highNums)        Console.Write("{0, -11}", item)        sCnt = (sCnt + 1) Mod 5 : If sCnt = 0 Then Console.WriteLine()    NextEnd Sub Sub Main()    Solve100() ' if interested, try this: Solve100("987654321")End Sub`
Output:
```List of solutions that evaluate to 100:
1 123-45-67+89
2 123-4-5-6-7+8-9
3 123+45-67+8-9
4 123+4-5+67-89
5 12-3-4+5-6+7+89
6 12+3-4+5+67+8+9
7 12+3+4+5-6-7+89
8 1+23-4+56+7+8+9
9 1+23-4+5+6+78-9
10 1+2+34-5+67-8+9
11 1+2+3-4+5+6+78+9
12 -1+2-3+4+5+6+78+9
The sum that has the the most solutions is 9, (at 46).
The lowest positive sum that can't be expressed is 211.
The ten highest numbers that can be expressed are:
123456789  23456790   23456788   12345687   12345669
3456801    3456792    3456790    3456788    3456786```

## zkl

Taking a big clue from Haskell and just calculate the world.

`var all =  // ( (1,12,123...-1,-12,...), (2,23,...) ...)   (9).pump(List,fcn(n){ split("123456789"[n,*]) })       // 45   .apply(fcn(ns){ ns.extend(ns.copy().apply('*(-1))) }); // 90fcn calcAllSums{  // calculate all 6572 sums (1715 unique)   fcn(n,sum,soFar,r){      if(n==9) return();      foreach b in (all[n]){	 if(sum+b>=0 and b.abs()%10==9) r.appendV(sum+b,"%s%+d".fmt(soFar,b));	 self.fcn(b.abs()%10,sum + b,"%s%+d".fmt(soFar,b),r);      }   }(0,0,"",r:=Dictionary());   r}    // "123" --> (1,12,123)fcn split(nstr){ (1).pump(nstr.len(),List,nstr.get.fp(0),"toInt") }`
`fcn showSums(allSums,N=100,printSolutions=2){   slns:=allSums.find(N,T);   if(printSolutions)    println("%d solutions for N=%d".fmt(slns.len(),N));   if(printSolutions==2) println(slns.concat("\n"));   println();} allSums:=calcAllSums();showSums(allSums);showSums(allSums,0,1); println("Smallest postive integer with no solution: ",   [1..].filter1('wrap(n){ Void==allSums.find(n) })); println("5 commonest sums (sum, number of ways to calculate to it):");ms:=allSums.values.apply("len").sort()[-5,*];	        // 5 mostest sumsallSums.pump(List,					// get those pairs   'wrap([(k,v)]){ v=v.len(); ms.holds(v) and T(k.toInt(),v) or Void.Skip }).sort(fcn(kv1,kv2){ kv1>kv2 })			// and sort.println();`
Output:
```12 solutions for N=100
+1+2+3-4+5+6+78+9
+1+2+34-5+67-8+9
+1+23-4+5+6+78-9
+1+23-4+56+7+8+9
+12+3+4+5-6-7+89
+12+3-4+5+67+8+9
+12-3-4+5-6+7+89
+123+4-5+67-89
+123+45-67+8-9
+123-4-5-6-7+8-9
+123-45-67+89
-1+2-3+4+5+6+78+9

22 solutions for N=0

Smallest postive integer with no solution: 211

5 commonest sums (sum, number of ways to calculate to it):
L(L(9,46),L(27,44),L(15,43),L(1,43),L(21,43))
```