poj--1961--KMP

Period

Time Limit: 3000MS Memory Limit: 30000KTotal Submissions: 14046 Accepted: 6647

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

3aaa
12
aabaabaabaab0

Sample Output

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

/*一道比较水的字符处理的题目，题意大概是给你一个长度为N的字符串，全是由小写的英文字母组成，需要你找出该字符串的所有前缀中所有的满足

形势的前缀，输出A, K.. */

/*因为正在复习KMP,所以找了这个题目。刚开始看的时候觉的和KMP没有什么关系。。。。。。 但是好好琢磨之后才发现这是用了KMP中next数组的性质*/

<span style="font-size:14px;">#include <stdio.h>#include <iostream>#include <algorithm>#include <string.h>using namespace std;int Next[1000010];int m;char x[1000010];void kmp_pre(){int i,j;j=Next[0]=-1;i=0;while(i<m){while (-1!=j && x[i]!=x[j])j=Next[j];Next[++i]=++j;}}int main(){//freopen("test.in","r",stdin);int j=1;while(scanf("%d",&m)){if(m==0)break;scanf("%s",x);kmp_pre();printf("Test case #%d\n",j++ );for(int i=1;i<=m;i++){int k=i-Next[i];if(i/k>1&&i%k==0)printf("%d %d\n",i,i/k);}printf("\n");}return 0;}</span>