（KMP 根据循环节来计算）Period -- hdu -- 1358

http://acm.hdu.edu.cn/showproblem.php?pid=1358

 

 

Period

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4849    Accepted Submission(s): 2353

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

 

#include<stdio.h>
#include<string.h>
#define N 1000007

char S[N];
int Next[N];

void FindNext(int Slen)
{
    int i=0, j=-1;
    Next[0] = -1;

    while(i<Slen)
    {
        if(j==-1 || S[i]==S[j])
            Next[++i] = ++j;
        else
            j = Next[j];
    }
}


int main()
{
    int n, iCase=1;

    while(scanf("%d", &n), n)
    {
        int i;

        scanf("%s", S);
        FindNext(n);

        printf("Test case #%d\n", iCase++);
        for(i=2; i<=n; i++)
        {
            if(i%(i-Next[i])==0 && Next[i])  ///这里是个坑，当Next[i]==0的时候虽然满足前面的条件，但是它的前缀和后缀的最大匹配度是0
                printf("%d %d\n", i, i/(i-Next[i]));
        }

        printf("\n");
    }
    return 0;
}