POJ 2406

传送门 ： Power String

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.

 从小到大依次暴力枚举 子串，然后 KMP，毫无疑问，超时啊啊啊！！！当时还挺傻的，一直以为这样做没问题。。。

代码如下：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include<algorithm>
using namespace std;
int ne[10000000];
char s[10000000];
void GetNext(char *p,int len)
{
    int i=0,j = -1;
    ne[0]=-1;
    while(i < len){
        if(j==-1||p[i]==p[j]) ne[++i]=++j;
        else j=ne[j];
    }
}
int main()
{
    while(cin>>s)
    {
        if(s[0]=='.') break;
        int len = strlen(s);
        GetNext(s,len);
        if(len%(len-ne[len])==0){
            printf("%d\n",len/(len-ne[len]));
        }
        else
            printf("1\n");
    }
    return 0;
}

 这应该就是求循环节的方法吧 -_-!