[HDOJ1358]Period

题目链接：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): 4711    Accepted Submission(s): 2285

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

 

 

求最短循环节，注意是每一个都要求。

 

#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <climits>
#include <cmath>

using namespace std;

const int maxn = 1000010;
int nb;
char b[maxn];
int pre[maxn];

void getpre(char *b, int *pre) {
    int j, k;
    pre[0] = -1;
    j = 0;
    k = -1;
    while(j < nb) {
        if(k == -1 || b[j] == b[k]) {
            j++;
            k++;
            pre[j] = k;
        }
        else {
            k = pre[k];
        }
    }
}

int main() {
    // freopen("in", "r", stdin);
    int kase = 1;
    while(~scanf("%d", &nb) && nb) {
        getchar();
        memset(b, 0, sizeof(b));
        gets(b);
        getpre(b, pre);
        printf("Test case #%d\n", kase++);
        for(int i = 2; i <= nb; i++) {
            if(pre[i] > 0 && i %(i - pre[i]) == 0) {
                printf("%d %d\n", i, i / (i - pre[i]));
            }
        }
        printf("\n");
    }
    return 0;
}