Power Strings||POJ2406

link:http://poj.org/problem?id=2406
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.

题解：令n为字符串长度，如果n可以被n-next[n]整除，那么这个商就是结果，反之输出1
AC代码：

#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
char s[1000010];
int next[1000010];
int kmp()
{
    int n=strlen(s);
    int i,j;
    i=0,j=-1;
    next[0]=-1;
    while(s[i])
    {
        if(j==-1||s[i]==s[j])
        {
            i++,j++;
            next[i]=j;
        }
        else
        {
            j=next[j];
        }
    }
    int s;
    s=n-j;//此时的j就和next[n]相等 
    if(n%s==0)
        return n/s;
    else
        return 1;
}
int main()
{
    while(scanf("%s",s),s[0]!='.')
    {
        printf("%d\n",kmp());
    }
return 0;
}