# POJ 2406 Power Strings

## Description

Given two strings a and b we define ab to be their concatenation. For example, if a = “abc” and b = “def” then ab = “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


## 思路

KMP匹配算法之 next[] ;

## AC 代码

#include <stdio.h>
#include <string.h>
char s[1000005];
int num[1000005];
int main()
{
while(scanf("%s",s),s[0]!='.')
{
int j=-1;
num[0]=-1;
int n=strlen(s);
for(int i=0; i<n;)
{
if(j==-1||s[i]==s[j])num[++i]=++j;
else j=num[j];
}
if(n%(n-num[n])==0)
printf("%d\n",n/(n-num[n]));
else printf("1\n");
}
return 0;
}