poj 2406 Power Strings，poj2406

poj 2406 Power Strings，poj2406
Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 33636		Accepted: 13973

Description

Given two 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 is defined in the normal way: a^0 = "" (the empty string) and a^(n+1) = a*(a^n).

Input

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

Output

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

Sample Input

abcdaaaaababab.

Sample Output

14
3
題目大意：給出一個字串，求這個串是有幾個子串構成的思路：用KMP演算法中的next數組，這個字串的next數組裡邊，字串的總長度減去最後一個字元所對應的next值就是子串的長度具體為什麼，應該是next數組的值是根據這個字串本身，用一定規則得出的，這個真的好巧妙的說，kmp演算法，好好看看。2014,12,5
#include<stdio.h>#include<string.h>char x[1100000];int next[1100000];void getnext(char s[]){int len1,i,j;len1=strlen(s);i=0;j=-1;next[i]=j;while(i<len1){if(j==-1||s[i]==s[j]){++i;++j;next[i]=j;}else j=next[j];}}int main(){int t,len;while(scanf("%s",x)){getnext(x);if(x[0]=='.') break;else {len=strlen(x);t=len-next[len];if(len%t==0)printf("%d\n",len/t);else printf("1\n");}}return 0;}