POJ2406Power Strings[KMP 失配函数]

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 45005		Accepted: 18792

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

Hint

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

Source

Waterloo local 2002.07.01

---

只需要求n的错位部分n-f[n]是不是循环节

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1e6+;
int n,f[N];
char p[N];
int main(){
    int cas=;
    while(true){
        scanf("%s",p);
        n=strlen(p);
        if(p[]=='.') break;
        f[]=f[]=;
        for(int i=;i<n;i++){
            int j=f[i];
            while(j&&p[j]!=p[i]) j=f[j];
            f[i+]=p[j]==p[i]?j+:;
        }
        if(n%(n-f[n])==) printf("%d\n",n/(n-f[n]));
        else printf("1\n");
    }
}