poj 2406

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 64350		Accepted: 26531

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

本题利用next数组的性质，如果n%(n-f[n]) ==0 ，那么循环节长度就是n-f[n],个数就是n/(n-f[n])，否则循环节就是1个。 

#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<algorithm>
using namespace std;  
int const N=1000000+3;  
char s[N]; 
int n,f[N]; 
int getnext(){
    int j=0; 
    for(int i=2;i<=n;i++){
        while (j && s[i-1]!=s[j]) j=f[j]; 
        f[i]=s[i-1]==s[j] ? ++j:0; 
    }
}
int main(){
    while (gets(s) && s[0]!='.'){
        n=strlen(s); 
        getnext(); 
        if(n % (n-f[n]) ==0) printf("%d\n",n/(n-f[n])); 
        else printf("1\n"); 
    }
    return 0; 
}