poj 2406 Power Strings

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

Hint

This problem has huge input, use scanf instead of cin to avoid time limit exceed.

找出最小循环节且总串是由多个最小循环节拼接而成，输出这个数字，如果不存在那么答案就是1。

代码：

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

char str[1000001];
int Next[1000001],len,k;
void findNext() {
    int j = -1,i = 0;
    Next[0] = -1;
    while(i < len) {
        if(j == -1 || str[i] == str[j]) {
            Next[++ i] = ++ j;
        }
        else j = Next[j];
    }
}
int main() {
    while(gets(str) && strcmp(str,".")) {
        len = strlen(str);
        findNext();///先确立Next数组
        int d = len - Next[len];
        int ans;
        if(len % d) ans = 1;
        else ans = len / d;
        printf("%d
",ans);
    }
}