Problem Description

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. 

Input

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. 

Output

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. 

Sample Input

3
aaa
12
aabaabaabaab
0

Sample Output

Test case #1
2 2
3 3

Test case #2
2 2
6 2
9 3
12 4

题意：每组给出一个长度为 n 的字符串，要求输出这个字符串每个前缀的最短循环节，简单来说，对于这个字符串的第 i 个前缀 ，求一个最大的整数，使得这个前缀是由某个字符串重复 k 次得到的，输出所有存在这个 k 值的 i 与对应的 k

例如：aabaabaabaab

• 当 i=2 时，前缀为 aa，是由于 a 重复两次得到的，故有 k=2
• 当 i=6 时，前缀为 aabaab，是由于 aab 重复两次得到的，故有 k=2
• 当 i=9 时，前缀为 aabaabaab，是由于 a 重复三次得到的，故有 k=3
• 当 i=12 时，前缀为 aabaabaabaab ，是由于 a 重复四次得到的，故有 k=4

思路：

本质上还是求最小循环节长度，使用 KMP 的 next 数组即可解决，对于每组数据的字符串，枚举前缀 i ，当 i%(i-next[i])=0 时即存在循环节，其个数 k=i/(i-next[i]) 

Source Program

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<utility>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#define PI acos(-1.0)
#define E 1e-9
#define INF 0x3f3f3f3f
#define N 100501
#define LL long long
const int MOD=20091226;
const int dx[]= {-1,1,0,0};
const int dy[]= {0,0,-1,1};
using namespace std;

int Next[N];
char p[N];
int n;
void getNext(){

    Next[0]=-1;

    int len=strlen(p);
    int j=0;
    int k=-1;

    while(j<len){
        if(k==-1||p[j]==p[k]) {
            k++;
            j++;
            Next[j]=k;
        }else{
            k=Next[k];
        }
    }
}
int main(){

    int Case=1;
    while(scanf("%d",&n)!=EOF&&n){
        scanf("%s",p);
        getNext();

        printf("Test case #%d\n",Case++);
        for(int i=1;i<=n;i++){
            int len=i-Next[i];
            if((i%len==0) && Next[i]>0)
                printf("%d %d\n",i,i/len);
        }
        printf("\n");
    }
    return 0;
}