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:
    disruptor和ArrayBlockingQueue和LinkedBlockingQueue队列性能对比
    守护线程的作用和前台线程的区别
    tomcat导入idea作为maven项目
    百度网盘不限速
    netty ChannelOption参数 backlog 和 somaxconn同时设置才会生效
    dubbo的初探
    IDEA的常用快捷键
    Lucene简单了解和使用
    Hadoop的简单了解与安装
    Nginx的简单了解与使用
  • 原文地址:https://www.cnblogs.com/tonghao/p/4712207.html
Copyright © 2011-2022 走看看