poj1961
思路：

满足 $i % (i-prefix[i]) == 0$ 且i != i-prefix[i] ,说明当前子串S[1~i]的长度是（最长公共前后缀）的前缀长度的倍数，那么 i/(i-prefix[i])就是循环节的个数

#include <iostream>
#include <string.h>
using namespace std;
const int maxn=1e6+10;
char pattern[maxn];
int prefix[maxn];

void prefix_table(char pattern[],int prefix[],int n) {
	prefix[0] = 0;//字符串的第一个公共前后缀长度为0
	int i = 1, len = 0;
	while ( i < n ) {
		if ( pattern[i] == pattern[len]) {
			len++;
			prefix[i] = len;
			i++;
		}
		else {
			if ( len > 0 ) {
				len = prefix[ len - 1 ];
			}
			else {
				prefix[i] = len;
				i++;
			}
		}
	}
}
void move_prefix(int prefix[],int n) {
	for (int i = n; i>0; i--) {
		prefix[i] = prefix[i-1];
	}
	prefix[0] = -1;
}

int main() {
//	freopen("a.txt","r",stdin);
	int k;
	int cnt = 0;
	while ( scanf("%d",&k) && k ) {
		scanf("%s",pattern);

		int n = strlen(pattern);

		prefix_table(pattern,prefix,n);

        move_prefix(prefix,n);

		printf("Test case #%d\n",++cnt);

		for (int i=1; i<=n;i++) {
            int dif = i - prefix[i];
			if ( i!=dif && (i%dif == 0) ) { // i!=dif 避免了 prefix[i]==0 没有公共前后缀的情况
				printf("%d %d\n", i, i/dif );
			}
		}
		printf("\n");
	}
	return 0;
}