poj 2406 Power Strings【字符串+最小循环节的个数】

                                                                                                  Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 38038		Accepted: 15740

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

abcd
aaaa
ababab
.

Sample Output

1
4
3

题目分析：给你一个字符串，求出这个字符串的最小循环节的个数，如果该串没有子循环节，就只有自己本身一个循环节，当然结果就是1啦！
像ababab：循环节就是ab，共有3个ab;  像abc：循环节就是abc，只有本身一个。

code：

#include <stdio.h>
#include <string.h>

char s[1000002];

int main()
{
    int len;
    while(scanf("%s", s)!=EOF)
    {
        if(s[0]=='.') break;
        len = strlen(s);
        for(int i=1; i<=len; i++)
            if(len%i==0)
            {
                int ok = 1;
                for(int j=i; j<len; j++){
                    if( s[j] != s[j%i] )
                    {
                        ok = 0;
                        break;
                    }
                }
                if(ok!=0){
                    printf("%d
", len/i);
                    break;
                }
            }
    }
    return 0;
}