zoukankan      html  css  js  c++  java
  • poj-1012-约瑟夫问题

    Description

    The Joseph's problem is notoriously known. For those who are not familiar with the original problem: from among n people, numbered 1, 2, . . ., n, standing in circle every mth is going to be executed and only the life of the last remaining person will be saved. Joseph was smart enough to choose the position of the last remaining person, thus saving his life to give us the message about the incident. For example when n = 6 and m = 5 then the people will be executed in the order 5, 4, 6, 2, 3 and 1 will be saved.

    Suppose that there are k good guys and k bad guys. In the circle the first k are good guys and the last k bad guys. You have to determine such minimal m that all the bad guys will be executed before the first good guy.

    Input

    The input file consists of separate lines containing k. The last line in the input file contains 0. You can suppose that 0 < k < 14.

    Output

    The output file will consist of separate lines containing m corresponding to k in the input file.

    Sample Input

    3
    4
    0
    

    Sample Output

    5
    30
    题目大意:
    有k个好人和k个坏人,从第一个人开始循环报数(即报到m后从1开始报),报到m的人被处死,要求全部的坏人都处死,好人留下,求出m。
    这个代码就是有点慢,所以我打了个表


    #include<iostream> using namespace std; int main() { int k; while(cin>>k) { if(!k) break; int m=1; int ans[30]={0}; for(int i=1;i<=k;i++) { ans[i]=(ans[i-1]+m-1)%(2*k-i+1); if(ans[i]<k) { i=0;m++; } } cout<<m<<endl; } return 0; } } sort(yes,yes+n,cmp); sort(no,no+n,cmp); cout<<yes[n-1]<<" "<<no[n-1]<<endl; return 0; }

      

    #include<stdio.h>
    int main()
    {
    	int k;
    	int Joseph[14]={0,2,7,5,30,169,441,1872,7632,1740,93313,459901,1358657,2504881};
    	while(scanf("%d",&k),k)
    		printf("%d
    ",Joseph[k]);
    	return 0;
    }
    

      

  • 相关阅读:
    摆动排序 II
    二进制中1的个数
    n的幂
    丢掉的数字
    phpcms直接取子栏目的内容、调用点击量的方法
    phpcms导航栏当前栏目选中方法
    PHP中的全局变量global和$GLOBALS的区别
    phpcms添加子栏目后的读取
    phpcms取内容发布管理中的来源
    cms替换主页
  • 原文地址:https://www.cnblogs.com/jin-nuo/p/5281139.html
Copyright © 2011-2022 走看看