poj 2406 最小循环节kmp

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

求周期最多的循环字符串，求出次数
总长度/最小循环节
特判没有循环节的情况

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;char s[1000100];int f[1000100];void getfill(char *s,int len){    memset(f,0,sizeof(f));    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)    {        if(s[1]=='.') break;        int len=strlen(s+1);        getfill(s+1,len);        int le=len-f[len];        if(le&&len%le==0&&f[len])        printf("%d\n",len/le );        else printf("1\n");    }}