For each prefix of a given string S with N characters (each character has an ASCII code between 97 and 126, inclusive), we want to know whether the prefix is a periodic string. That is, for each i (2 <= i <= N) we want to know the largest K > 1 (if there is one) such that the prefix of S with length i can be written as A^K , that is A concatenated K times, for some string A. Of course, we also want to know the period K.

The input file consists of several test cases. Each test case consists of two lines. The first one contains N (2 <= N <= 1 000 000) – the size of the string S. The second line contains the string S. The input file ends with a line, having the number zero on it.

For each test case, output “Test case #” and the consecutive test case number on a single line; then, for each prefix with length i that has a period K > 1, output the prefix size i and the period K separated by a single space; the prefix sizes must be in increasing order. Print a blank line after each test case.

3
aaa
12
aabaabaabaab
0

Test case #1
2 2
3 3

Test case #2
2 2
6 2
9 3
12 4
题意：求出以某一下标结尾的，循环次数大于1的前缀，输出结尾下标，循环次数。
这个题考察对kmp的next数组的理解。
/* ***********************************************
Author :xianxingwuguan
Created Time :2014-2-2 0:56:58
File Name :2.cpp
************************************************ */

#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
const int maxn=1000100;
const double pi =acos(-1.0);
const double eps =1e-8;
char str[maxn];
int next[maxn];
void get_next(char *str)
{
      int i,j=0,k=-1;
      next[0]=-1;
      int len=strlen(str);
      while(j<len){
	     if(k==-1||str[j]==str[k])
		    next[++j]=++k;
	     else 
		    k=next[k];
      }
     // for(i=0;i<=len;i++)cout<<next[i]<<" ";cout<<endl;
}
int main()
{
      //freopen("data.in","r",stdin);
      //freopen("data.out","w",stdout);
      int i,j,k,m,n,count=1;
      while(~scanf("%d",&n)&&n)
      {
	     scanf("%s",str);
	     get_next(str);
	      printf("Test case #%d\n",count++);  
	     for(i=1;i<=n;i++)
	     {
		    int ret=i-next[i];
		    if(i%ret==0&&i/ret>1)
		    {
			   printf("%d %d\n",i,i/ret);
		    }
	     }
	     puts("");
      }
      return 0;
}