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Poj1961--Period(Kmp, Next数组求循环节长度 && 出现次数)

时间:2015-08-08 17:58:18      阅读:275      评论:0      收藏:0      [点我收藏+]

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Period
Time Limit: 3000MS   Memory Limit: 30000K
Total Submissions: 14657   Accepted: 6967

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 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

Source

RE: 求长度为(0 ~ i)的循环节出现的次数; (注意循环节和(前、后缀)的区别);
 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 using namespace std;
 5 char a[1000100]; int lena, p[1000100];
 6 void Getp()
 7 {
 8     int i = 0, j = -1;
 9     p[i] = j;
10     while(i < lena)
11     {
12         if(j == -1 || a[i] == a[j])
13         {
14             i++; j++;
15             p[i] = j;
16         }
17         else
18             j = p[j];
19     }
20 }
21 int main()
22 {
23     int t, csse = 1;; 
24     while(~scanf("%d", &t), t)
25     {
26         scanf("%s", a);
27         lena = strlen(a); 
28         Getp();
29         printf("Test case #%d\n", csse++);    
30         for(int i = 1; i <= lena; i++)
31         {
32             //int length = i - p[i];              //循环节长度;
33             if(i % (i - p[i]) == 0 && p[i] != 0)  //判断是否为循环节;
34                 printf("%d %d\n", i, i/(i-p[i]));
35         }
36         printf("\n");
37     }
38     return 0;
39 }

 

 

Poj1961--Period(Kmp, Next数组求循环节长度 && 出现次数)

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原文地址:http://www.cnblogs.com/fengshun/p/4713364.html

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