당신의 임무는이 번호를 입력으로 취하는 것입니다 (다른 번호와도 작동해야하지만).

18349570345710975183495703457109751834957034571097518349570345710975183495703457109761834957034571097518349570345710975183495703457109751834957034571097518349570345710976183495703457109751834957034571097518349570345710975183495703457109751834957034571097618349570345710975183495703457109751834957034571097518349570345710975183495703457109761834957034571097518349570345710975183495703457109751834957034571097518349570345710976183495703457109751834957034571097518349570345710975183495703457109751834957034571097618349570345710975183495703457109751834957034571097518349570345710975183495703457109761834957034571097518349570345710975183495703457109751834957034571097518349570345710976183495703457109751834957034571097518349570345710975183495703457109751834957034571097618349570345710975183495703457109751834957034571097518349570345710975183495703457109761834957034571097518349570345710975183495703457109751834957034571097518349570345710976183495703457109751834957034571097518349570345710975183495703457109751834957034571097618349570345710975183495703457109751834957

이 경우 가장 작은 기간을 찾으십시오.

1834957034571097518349570345710975183495703457109751834957034571097518349570345710976

행운을 빌고 재미있게 보내!

설명 :

• 입력 번호에는 최소 1 개의 기간과 1 개의 부분 기간이 있습니다
• 기간은 항상 입력 번호의 시작 부분에서 시작합니다
• 주기는이 경우 반복되는 일련의 숫자를 의미합니다.

CJam, 20 16 바이트

Ll:Q{+_Q,*Q#!}=;

STDIN에서 읽습니다. 온라인으로 사용해보십시오.

위의 코드에는 O (n ² ) 메모리 가 필요 합니다. 여기서 n 은 입력 길이입니다. 그것은 것입니다 이 개 작업을 ¹⁶ 만큼 당신이 충분한 메모리를 가지고, 숫자.

이것은 5 바이트의 비용을 고정시킬 수 있습니다 :

Ll:Q{+_Q,1$,/)*Q#!}=;

예제 실행

$ cjam <(echo 'Ll:Q{+_Q,*Q#!}=;') <<< 18349570345710975183495703457109751834957034571097518349570345710975183495703457109761834957034571097518349570345710975183495703457109751834957034571097518349570345710976183495703457109751834957034571097518349570345710975183495703457109751834957034571097618349570345710975183495703457109751834957034571097518349570345710975183495703457109761834957034571097518349570345710975183495703457109751834957034571097518349570345710976183495703457109751834957034571097518349570345710975183495703457109751834957034571097618349570345710975183495703457109751834957034571097518349570345710975183495703457109761834957034571097518349570345710975183495703457109751834957034571097518349570345710976183495703457109751834957034571097518349570345710975183495703457109751834957034571097618349570345710975183495703457109751834957034571097518349570345710975183495703457109761834957034571097518349570345710975183495703457109751834957034571097518349570345710976183495703457109751834957034571097518349570345710975183495703457109751834957034571097618349570345710975183495703457109751834957; echo
1834957034571097518349570345710975183495703457109751834957034571097518349570345710976
$ cjam <(echo 'Ll:Q{+_Q,*Q#!}=;') <<< 12345123451; echo
12345
$ cjam <(echo 'Ll:Q{+_Q,*Q#!}=;') <<< 1234512345; echo
12345
$ cjam <(echo 'Ll:Q{+_Q,*Q#!}=;') <<< 123451; echo
12345

작동 원리

입력 Q의 경우 첫 번째 문자 len (Q) 시간을 반복하고 결과에서 Q의 인덱스가 0인지 확인하는 것이 좋습니다. 그렇지 않은 경우 첫 두 문자 len (Q) 시간 등을 반복하십시오 .

L                   " Push L := [].                                                       ";
 l:Q                " Read one line from STDIN and save the result in Q.                  ";
    {        }=     " Find the first element q ∊ Q that yields a truthy value:            ";
     +              "   Execute L += [q].                                                 ";
      _Q,*Q#        "   Push (L * len(Q)).index(Q).                                       ";
            !       "   Compute the logical NOT of the index.                             ";
               ;    " Discard the last q. This leaves L on the stack.                     ";

정규식 (.NET 풍미), 23 22 바이트

.+?(?=(.*$)(?<=^\1.*))

필요한 기간을 하위 문자열로 일치시킵니다.

여기에서 테스트하십시오.

어떻게 작동합니까?

# The regex will always find a match, so there's no need to anchor it to
# the beginning of the string - the match will start there anyway.
.+?        # Try matching periods from shortest to longest
(?=        # Lookahead to ensure that what we've matched is actually
           # a period. By using a lookahead, we ensure that this is
           # not part of the match.
  (.*$)    # Match and capture the remainder of the input in group 1.
  (?<=     # Use a lookahead to ensure that this remainder is the same
           # as the beginning of the input. .NET lookaheads are best
           # read from right to left (because that's how they are matched)
           # so you might want to read the next three lines from the 
           # bottom up.
    ^      # Make sure we can reach the beginning of the string.
    \1     # Match group 1.
    .*     # Skip some characters, because the capture won't cover the
           # entire string.
  )
)

파이썬 60

s 자릿수입니다

[s[:i]for i in range(len(s))if(s[:i]*len(s))[:len(s)]==s][0]

예 :

>>> s = '18349570345710975183495703457109751834957034571097518349570345710975183495703457109761834957034571097518349570345710975183495703457109751834957034571097518349570345710976183495703457109751834957034571097518349570345710975183495703457109751834957034571097618349570345710975183495703457109751834957034571097518349570345710975183495703457109761834957034571097518349570345710975183495703457109751834957034571097518349570345710976183495703457109751834957034571097518349570345710975183495703457109751834957034571097618349570345710975183495703457109751834957034571097518349570345710975183495703457109761834957034571097518349570345710975183495703457109751834957034571097518349570345710976183495703457109751834957034571097518349570345710975183495703457109751834957034571097618349570345710975183495703457109751834957034571097518349570345710975183495703457109761834957034571097518349570345710975183495703457109751834957034571097518349570345710976183495703457109751834957034571097518349570345710975183495703457109751834957034571097618349570345710975183495703457109751834957'
>>> [s[:i]for i in range(len(s))if(s[:i]*len(s))[:len(s)]==s][0]
'1834957034571097518349570345710975183495703457109751834957034571097518349570345710976'

Pyth , 14 자

hf}z*lzTm<zdUz

설명:

implicit:      z = input()
h              head(
 f                  filter(lambda T:
  }z                                z in
    *lz                                  len(z) * 
       T                                          T,
  m                        map(lambda d:
   <zd                                  z[:d],
   Uz                                   range(len(d)))))

기본적으로 입력의 모든 초기 시퀀스를 생성하고 각각 한 len(z)번씩 반복 z하며 입력이 결과 문자열 내에 있는지 여부를 확인 합니다.

이것은 올바른 답변은 아니지만 질문이 제기 된 후 12 문자 솔루션을 허용하는 기능이 최근 Pyth에 추가되었습니다.

<zf}z*lz<zT1

정수 기능에 필터를 사용합니다.

Japt , 8 바이트

å+ æ@¶îX

시도 해봐

Shaggy 덕분에 -2 바이트!

번역 된 JS 설명 :

// U is the input string representation of the number
U
 // cumulative reduce using the '+' operator
 // the result is an array of strings length 1, 2, ..., N
 // all substrings start with the first character from input
 .å("+")
 // find the first match
 .æ(function(X, Y, Z) {
  // repeat the substring until it is as long as the input
  // and compare it to the input
  return U === U.î(X)
 })

자바 8, 125 바이트

Java 이외의 문자열 이외의 1000 개 이상의 숫자를 나타내는 합리적인 방법이 없기 때문에 입력을 문자열로 사용합니다 (No BigInteger please).

s->{String o="";for(int i=0;java.util.Arrays.stream(s.split(o+=s.charAt(i++))).filter(b->!b.isEmpty()).count()>1;);return o;}

온라인으로 사용해보십시오!