Positive decimal integers with the digit 1 occurring exactly twice
Positive decimal integers with the digit 1 occurring exactly twice is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.
- Task
- Find numbers n in which number 1 occur twice, where n < 1,000
ALGOL 68
Generates the numbers. In order to print them in order, a table of double 1 numbers yes/no is generated. <lang algol68>BEGIN # find numbers where the digit 1 occurs twice, up to 999 #
[ 1 : 999 ]BOOL double 1; FOR i TO UPB double 1 DO double 1[ i ] := FALSE OD; # generte the numbers # FOR i FROM 0 TO 9 DO IF i /= 1 THEN double 1[ 110 + i ] := TRUE; double 1[ 101 + ( i * 10 ) ] := TRUE; double 1[ ( i * 100 ) + 11 ] := TRUE FI OD; # print the numbers in order # INT double 1 count := 0; FOR i TO UPB double 1 DO IF double 1[ i ] THEN print( ( " ", whole( i, -3 ) ) ); IF ( double 1 count +:= 1 ) MOD 10 = 0 THEN print( ( newline ) ) FI FI OD
END</lang>
- Output:
11 101 110 112 113 114 115 116 117 118 119 121 131 141 151 161 171 181 191 211 311 411 511 611 711 811 911
Phix
function two_ones(integer i) return length(find_all('1',sprint(i)))=2 end function sequence res = filter(tagset(999),two_ones) printf(1,"%d found:\n %s\n",{length(res),join_by(apply(res,sprint),1,9," ","\n ")})
- Output:
27 found: 11 101 110 112 113 114 115 116 117 118 119 121 131 141 151 161 171 181 191 211 311 411 511 611 711 811 911
Ring
<lang ring> load "stdlib.ring" see "working..." + nl see "Numbers n in which number 1 occur twice:" + nl
row = 0 sum = 0 limit = 1000
for n = 1 to limit
strn = string(n) ind = count(strn,"1") if ind = 2 see "" + n + " " row = row + 1 if row%5 = 0 see nl ok ok
next
see nl + "Found " + row + " numbers" + nl see "done..." + nl
func count(cstring,dstring)
sum = 0 while substr(cstring,dstring) > 0 sum = sum + 1 cstring = substr(cstring,substr(cstring,dstring)+len(string(sum))) end return sum
</lang>
- Output:
working... Numbers n in which number 1 occur twice: 11 101 110 112 113 114 115 116 117 118 119 121 131 141 151 161 171 181 191 211 311 411 511 611 711 811 911 Found 27 numbers done...
XPL0
<lang XPL0>func Ones(N); \Return count of 1's in N int N, Count; [Count:= 0; repeat N:= N/10;
if rem(0) = 1 then Count:= Count+1;
until N = 0; return Count; ];
int N, Count; [for N:= 1 to 1000-1 do
if Ones(N) = 2 then [IntOut(0, N); Count:= Count+1; if rem(Count/10) = 0 then CrLf(0) else ChOut(0, 9\tab\); ];
CrLf(0); IntOut(0, Count); Text(0, " such numbers found below 1000. "); ]</lang>
- Output:
11 101 110 112 113 114 115 116 117 118 119 121 131 141 151 161 171 181 191 211 311 411 511 611 711 811 911 27 such numbers found below 1000.