Largest five adjacent number
- Task
Generate random 1000-digit number.
Find the five adjacent digits in the 1000-digit number that form the largest 5-digit number.
ALGOL 68
Adding the minimum number for good measure... <lang algol68>BEGIN # generate 1000 random digits and find the largest/smallest numbers formed from 5 consecutive digits #
[ 1 : 1000 ]CHAR digits; FOR i TO UPB digits DO digits[ i ] := REPR ( ENTIER ( next random * 10 ) + ABS "0" ) OD; STRING max number := digits[ 1 : 5 ]; STRING min number := digits[ 1 : 5 ]; FOR i FROM 2 TO UPB digits - 4 DO STRING next number = digits[ i : i + 4 ]; IF next number > max number THEN # found a new higher number # max number := next number FI; IF next number < min number THEN # found a new lower number # min number := next number FI OD; print( ( "Largest 5 consecutive digits from 1000 random digits: ", max number, newline ) ); print( ( "Smallest 5 consecutive digits from 1000 random digits: ", min number, newline ) )
END</lang>
- Output:
Largest 5 consecutive digits from 1000 random digits: 99987 Smallest 5 consecutive digits from 1000 random digits: 00119
APL
<lang apl>⌈/((⊣+10×⊢)/(⌽↓))⌺5⊢(-⎕IO)+?1000/10</lang>
- Output:
(example)
99994
Factor
<lang factor>USING: grouping io math.functions prettyprint random sequences ;
1000 10^ random unparse 5 <clumps> supremum print</lang>
- Output:
99987
Julia
<lang julia>dig = rand(0:9, 1000) @show maximum(evalpoly(10, dig[i:i+4]) for i in 1:length(dig)-4)
</lang>
- Output:
maximum((evalpoly(10, dig[i:i + 4]) for i = 1:length(dig) - 4)) = 99993
Or, using strings, and see Nigel's comment in the discussion:
julia> setprecision(3324) 3324 julia> s = string(BigFloat(pi))[3:end] "141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117067982148086513282306647093844609550582231725359408128481117450284102701938521105559644622948954930381964428810975665933446128475648233786783165271201909145648566923460348610454326648213393607260249141273724587006606315588174881520920962829254091715364367892590360011330530548820466521384146951941511609433057270365" ⋯ 180 bytes ⋯ "66940513200056812714526356082778577134275778960917363717872146844090122495343014654958537105079227968925892354201995611212902196086403441815981362977477130996051870721134999999837297804995105973173281609631859502445945534690830264252230825334468503526193118817101000313783875288658753320838142061717766914730359825349042875546873115956286388235378759375195778185778053217122680661300192787661119590921642019898" julia> m, pos = maximum((s[i:i+4], i) for i in 1:length(s)-4) ("99999", 763) julia> println("Maximum is $m at position $pos.") Maximum is 99999 at position 763.
Nim
<lang Nim>import random, strutils
randomize()
const N = 1000 type Digit = 0..9
- Build a 1000-digit number.
var number: array[1..N, Digit] number[1] = rand(1..9) # Make sure the first digit is not 0. for i in 1..N: number[i] = rand(9)
func `>`(s1, s2: seq[Digit]): bool =
## Compare two digit sequences. ## Defining `<` rather than `>` would work too. assert s1.len == s2.len for i in 0..s1.high: let comp = cmp(s1[i], s2[i]) if comp != 0: return comp == 1 result = false
var max = @[Digit 0, 0, 0, 0, 0] for i in 5..N:
let n = number[i-4..i] if n > max: max = n
echo "Largest 5-digit number extracted from random 1000-digit number: ", max.join()</lang>
- Output:
Largest 5-digit number extracted from random 1000-digit number: 99855
Pascal
inspired by Wren
Assumes that there at least is a "1" 4 digits before end of all digits.Else I have to include sysutils and s := Format('%.5d',[i]); for leading zeros.
<lang pascal> var
digits, s : AnsiString; i : LongInt;
begin
randomize; setlength(digits,1000); for i := 1 to 1000 do digits[i] := chr(random(10)+ord('0')); for i := 99999 downto 0 do begin str(i:5,s); if Pos(s,digits) > 0 then break; end; writeln(s, ' found as largest 5 digit number ')
end.</lang>
- Output:
99889 found as largest 5 digit number
Perl
<lang perl>#!/usr/bin/perl
use strict; # https://rosettacode.org/wiki/Largest_five_adjacent_number use warnings;
$_ = join , map int rand 10, 1 .. 1e3; my @n; @n[ /(?=(\d{5}))/g ] = (); print "$#n\n";</lang>
- Output:
99899
Raku
Show minimum too because... why not?
Use some Tamil Unicode numbers for brevity, and for amusement purposes.
௰ - Tamil number ten ௲ - Tamil number one thousand
Do it 5 times for variety, it's random after all.
<lang perl6>(^௰).roll(௲).rotor(5 => -4)».join.minmax.bounds.put xx 5</lang>
- Sample output:
00371 99975 00012 99982 00008 99995 00012 99945 00127 99972
Ring
<lang ring> digit = "" max = 0 maxOld = 0 limit = 1000
for n = 1 to limit
rand = random(9) randStr = string(rand) digit += randStr
next
for n = 1 to len(digit)-5
res = substr(digit,n,5) resNum = number(res) if resNum > maxold max = resNum maxOld = max ok
next
see max + nl </lang>
- Output:
The largest number is: 99638
Wren
Very simple approach as there's little need for speed here. <lang ecmascript>import "random" for Random import "/fmt" for Fmt
var rand = Random.new() var digits = List.filled(1000, 0) for (i in 0...999) digits[i] = rand.int(10) var number = digits.join() for (i in 99999...0) {
var quintet = Fmt.swrite("$05d", i) if (number.contains(quintet)) { Fmt.print("The largest number formed from 5 adjacent digits is: $,d", i) return }
}</lang>
- Output:
Sample output:
The largest number formed from 5 adjacent digits is: 99,850
XPL0
<lang XPL0>char Number(1000); int Num, Max, I, J; [for I:= 0 to 1000-1 do \generate 1000-digit number
Number(I):= Ran(10);
Max:= 0; \find its largest 5-digit number for I:= 0 to 1000-5 do
[Num:= 0; for J:= 0 to 5-1 do Num:= Num*10 + Number(I+J); if Num > Max then Max:= Num; ];
IntOut(0, Max); ]</lang>
- Output:
99930