(Find the number of circulation sections) Power Strings--POJ--2406

Link:

http://poj.org/problem?id=2406

Power StringsTime limit:3000MS Memory Limit:65536KB 64bit IO Format:%i64d &%i64 U SubmitStatus

Description

Given Strings A and b we define a*b to be their concatenation. For example, if a = "abc" and B = "Def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative integer are defined in the normal way:a^0 = "" (The empty string) and a^ (n+1) = A * (a^n).

Input

Each test case was a line of input representing S, a string of printable characters. The length of S'll be is at least 1 and would not exceed 1 million characters. A line containing a period follows the last test case.

Output

For each s should print the largest n such, that s = a^n for some string a.

Sample Input

Abcdaaaaababab.

Sample Output

143

Hint

This problem have huge input, use scanf instead of CIN to avoid time limit exceed. Code:

#include <stdio.h>#include<string.h>#include<stdlib.h>#defineN 1000007CharS[n];intNext[n]; ///Next, the maximum similarity between prefix and suffix is saved.voidFindNext (intSlen,intNext[])///Next[i] represents the maximum match of the first I character{    intI=0, j=-1; next[0] = -1;  while(i<Slen) {        if(j==-1|| s[i]==R[j]) next[++i] = + +J; ElseJ=Next[j]; }}intMain () { while(SCANF ("%s", s), strcmp (S,"."))    {        intslen=strlen (S);        FindNext (Slen, Next); if(slen% (slen-Next[slen])) printf ("1\n"); Elseprintf ("%d\n", slen/(slen-Next[slen])); }    return 0;}

View Code

(Find the number of circulation sections) Power Strings--POJ--2406