zoukankan      html  css  js  c++  java
  • poj 1961 Period【求前缀的长度,以及其中最小循环节的循环次数】

    Period
    Time Limit: 3000MS   Memory Limit: 30000K
    Total Submissions: 14653   Accepted: 6965

    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
    题意:求前缀的长度,以及其中最小循环节的循环次数 但是循环次数必须大于1
    #include<stdio.h>
    #include<string.h>
    #define MAX 1000100
    int next[MAX];
    char str[MAX];
    int a[MAX],b[MAX];
    int n,m,k;
    void getfail()
    {
    	int i,j;
    	next[0]=next[1]=0;
    	for(i=1;i<n;i++)
    	{
    		j=next[i];
    		while(j&&str[i]!=str[j])
    		    j=next[j];
    		next[i+1]=str[i]==str[j]?j+1:0;
    	}
    }
    void kmp()
    {
    	int i,j;
    	for(i=1;i<=n;i++)
    	{
    		k=i;
    		if(k==(k-next[k]))//k-next[k]最小循环节长度 
    		continue;
    		if(k%(k-next[k])==0)
    		printf("%d %d
    ",i,k/(k-next[k]));
    	}
    	printf("
    ");
    }
    int main()
    {
    	int i,t=1;
    	while(scanf("%d",&n),n)
    	{
    		getchar();
    		for(i=0;i<n;i++)
    		scanf("%c",&str[i]);
    		getfail();
    		printf("Test case #%d
    ",t++);
    		kmp();
    	}
    	return 0;
    }
    

      

  • 相关阅读:
    架设证书服务器 及 让IIS启用HTTPS服务
    实验二. 使用LoadRunner进行压力测试
    loadrunner简单使用——HTTP,WebService,Socket压力测试脚本编写
    康威定律,可以学习一下
    各国iPhone5系列最新裸机价格
    .yml是什么文件
    win32多线程程序设计笔记(第五章)
    kvm libvirt: hostdev passthrough support 解决加密狗冲突问题
    九度OnlineJudge之1023:EXCEL排序
    [置顶] Android开发之XML文件的解析
  • 原文地址:https://www.cnblogs.com/tonghao/p/4712207.html
Copyright © 2011-2022 走看看