Period


Time Limit: 3000MS

 

Memory Limit: 30000K

Total Submissions: 17207

 

Accepted: 8293


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 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：给定一个n长度的字符串，询问在从第2个开始到第i个有多少循环节。

题解：KMP的应用，对next数组的理解


#include <stdio.h>
#include <string.h>
#include <iostream>

using namespace std;

char str[1000010];
int next[1000010];
int n;

void get_next()
{
    int i = 0;
    int j;
    next[i] = -1;
    j = -1;
    while(i < n)
    {
        if(str[i] == str[j] || j == -1)
        {
            i++;
            j++;
            next[i] = j;
        }
        else
        {
            j = next[j];
        }
    }
}
int main()
{
    int i;
    int h = 0;
    while(~scanf("%d", &n) && n != 0)
    {
        getchar();
        scanf("%s", str);
        get_next();
        printf("Test case #%d\n", ++h);
        for(i = 1; i <= n; i++)
        {
            int len = i - next[i]; 
            if(i != len && i % len == 0) 
                printf("%d %d\n", i, i / len); 
        }
        printf("\n");
    }
    return 0;
}