POJ 2406 Power Strings （kmp数组）

题目链接：http://poj.org/problem?id=2406点击打开链接

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 53824		Accepted: 22425

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

求循环节长度 判断是否可整除即求得个数

#include <stdio.h>
#include <string.h>
using namespace std;
char t[1111111];
int n,m;
int nnext[1111111];
void getnext()
{
	int i=0,j=-1;
	nnext[0]=-1;
	while(i<n)
	{
		if(j==-1||t[i]==t[j])
		{
			i++;
			j++;
			nnext[i]=j;
		}
		else
		{
			j=nnext[j];
		}
	}
}
int  main()
{
	while(scanf("%s",&t)&&strcmp(t,"."))
	{
		int ans=1;
		n=strlen(t);
		getnext();
		int mid=n-nnext[n];
		//cout << mid  << n <<endl;
		if((n%mid)==0)
		{
			ans=(n/mid);
		}

		printf("%d\n",ans);
	}
}