hdu 1358 Period 最小循环节

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

分析：已知字符串，求其由最小循环节构成的前缀字符串。

/*Period

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 3507    Accepted Submission(s): 1766

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 AK , 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 <cstdio>

#include <cstring>

const int maxn =  + ;

char a[maxn];

int next[maxn], n;

void getNext()

{

    int k = -, j = ;

    next[] = -;

    while(j < n){//遍历字符串

        if(k == - || a[j] == a[k]) next[++j] = ++k;

        else k = next[k];

    }

}

int main()

{

    int C = ;

    while(~scanf("%d", &n) && n){

        scanf("%s", a);

        getNext();

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

        for(int i = ; i <= n; i++){//遍历所有循环节长度

            int len = i;

            int cnt = len-next[len];

            if(cnt == ){//循环节长度为1，直接输出

                printf("%d %d\n", len, len);

            }

            else if(cnt != len && (len % cnt) == ){

                printf("%d %d\n", len, len/cnt);

            }

        }

        printf("\n");

    }

    return ;

}