POJ 2406 (字符串的匹配) 最小循环节问题

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

#include 
#include 
#include 
using namespace std;
char array[1000005];
int Next[1000005];
int len;
void GetNext(){
	int i=0;
	int j=-1;
	Next[0]=-1;
	while(i-1&&array[i]!=array[j])
			j=Next[j];
		Next[++i]=++j;
	}
	return;
}
int main(){
	int result;
	while(scanf("%s",array)!=EOF){
		if(array[0]=='.')
			break;
		result=1;
		len=strlen(array);
		GetNext();
		if(len%(len-Next[len])==0)//情况三
			result=len/(len-Next[len]);
		printf("%d\n",result);
	}
	return 0;
}

也是三种情况：

1：abcde

2:abcdefrtgabc

3:abcabcabc