HDU 1358 Period(KMP+求某个前缀含几个循环)

来源:互联网 发布:济南js复合保温模板 编辑:程序博客网 时间:2024/07/24 00:56

Period

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7903    Accepted Submission(s): 3794


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
3aaa12aabaabaabaab0
 

Sample Output
Test case #12 23 3Test case #22 26 29 312 4
 

Recommend

JGShining   |   We have carefully selected several similar problems for you:  1686 3068 2203 2087 1277 


#include<stdio.h>#include<cstring>char b[1000001];int p[1000001];int m;void get(){p[1] = 0;int i,j=0;for(i=2;i<=m;i++){while(j > 0 && b[j+1] != b[i]){j = p[j];}if(b[j+1] == b[i]){j += 1; }p[i] = j;} }   int main() { int cases = 1; while(scanf("%d",&m),m) { scanf("%s",b+1); printf("Test case #%d\n",cases++); get(); for(int i = 2; i <= m; i++) { if(i%(i-p[i])==0&&p[i]!=0)    //i个前缀含几个循环 { printf("%d %d\n",i,i/(i-p[i])); }    }    puts(""); } }  


0 0
原创粉丝点击