POJ2961_kmp

时间：2017-03-12 11:39:59 阅读：179 评论：0 收藏：0 [点我收藏+]

标签：src exp art splay 原理 sample widget lap ima

Time Limit: 3000MS

Memory Limit: 30000KB

64bit IO Format: %I64d & %I64u

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

__________________________________________________________________________________________

题目大意：求字符串的前缀中循环节的出现次数（大于1的）。

原理同上题一样，只是要注意next[]==-1时的判断。

__________________________________________________________________________________________

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 
 5 using namespace std;
 6 const int maxl=1e6+10;
 7 int t,l,cas=0;
 8 char s[maxl];
 9 int next[maxl];
10 void getnext()
11 {
12     next[0]=-1;
13     for(int j,i=1;i<l;i++)
14     {
15         j=next[i-1];
16         while(s[i]!=s[j+1] && j>=0)j=next[j];
17         next[i]=s[i]==s[j+1]?j+1:-1;
18     }
19 }
20 int main()
21 {
22     while(scanf("%d",&l)==1 && l!=0)
23     {
24         printf("Test case #%d\n",++cas);
25         scanf("%s",s);
26         getnext();
27         for(int i=1;i<l;i++)
28         {
29             if(next[i]!=-1 && (i+1)%(i-next[i])==0)printf("%d %d\n",i+1,(i+1)/(i-next[i]));
30         }
31         printf("\n");
32     }
33     return 0;
34 }

View Code

POJ2961_kmp

标签：src exp art splay 原理 sample widget lap ima

原文地址：http://www.cnblogs.com/gryzy/p/6537101.html

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