POJ 1961 Period KMP

Period

Time Limit: 3000MS		Memory Limit: 30000K
Total Submissions: 9758		Accepted: 4461

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

 1 /* 功能Function Description:     poj-1961
 2    开发环境Environment:          DEV C++ 4.9.9.1
 3    技术特点Technique:
 4    版本Version:
 5    作者Author:                   可笑痴狂
 6    日期Date:                      20120814
 7    备注Notes:
 8     题意：给定一个字符串，问这个字符串的所有前缀中，有哪些前缀可以由某个串重复k次组成，
 9     并且这个k最大是多少。这个k需要大于1。
10     --------KMP
11 */
12 #include<cstdio>
13 #include<string>
14 #include<cstdlib>
15 
16 void get_next(int *next,char *s,int n)
17 {
18     int i,j;
19     i=0;
20     j=-1;
21     next[0]=-1;
22     while(s[i])
23     {
24         if(j==-1||s[i]==s[j])
25         {
26             ++i;
27             ++j;
28             next[i]=j;
29         }
30         else
31             j=next[j];
32     }
33 }
34 
35 int main()
36 {
37     int n,i,T=1;
38     int *next;
39     char *s;
40     while(scanf("%d",&n),n)
41     {
42         next=new int[n+1];
43         s=new char[n+2];
44         scanf("%s",s);
45         get_next(next,s,n);
46         printf("Test case #%d\n",T++);
47         for(i=1;i<=n;++i)
48         {
49             if(i%(i-next[i])==0&&i!=i-next[i])         //i-next[i]为循环节长度
50                 printf("%d %d\n",i,i/(i-next[i]));
51         }
52         printf("\n");
53     }
54     return 0;
55 }

功不成，身已退