hust 1010 最小循环节

There is a string A. The length of A is less than 1,000,000. I rewrite it again and again. Then I got a new string: AAAAAA…… Now I cut it from two different position and get a new string B. Then, give you the string B, can you tell me the length of the shortest possible string A. For example, A=”abcdefg”. I got abcd efgabcdefgabcdefgabcdefg…. Then I cut the red part: efgabcdefgabcde as string B. From B, you should find out the shortest A.
Input
Multiply Test Cases. For each line there is a string B which contains only lowercase and uppercase charactors. The length of B is no more than 1,000,000.
Output
For each line, output an integer, as described above.
Sample Input
bcabcab
efgabcdefgabcde
Sample Output
3
7

有一个字符串A，一次次的重写A，会得到一个新的字符串AAAAAAAA…..,现在将这个字符串从中切去一部分得到一个字符串B，例如有一个字符串A=”abcdefg”.，复制几次之后得到abcdefgabcdefgabcdefgabcdefg….，现在切去中间红色的部分，得到字符串B，现在只给出字符串B，求出字符串A的长度
就是求最小循环节

#include <bits/stdc++.h>using namespace std;char s[1000100];int f[1000100];void getfill(char *s,int len){    memset(f,0,sizeof(f));    f[0]=-1;    for(int i=1;i<=len;i++)    {        int j=f[i];        while(j&&s[i]!=s[j])            j=f[j];        f[i+1]=(s[i]==s[j])?j+1:0;    }}int main(){    while(scanf("%s",s+1)!=EOF)    {        int len=strlen(s+1);        getfill(s+1,len);        int le=len-f[len];        printf("%d\n",le );    }}