zoukankan      html  css  js  c++  java
  • Joseph

    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
    

    题意:约瑟夫环的变形。

    先说约翰夫环,即n个人围成一圈报数,报到数的人离开,问最后离开的那人的编号

    数学公式:

    s = 0;
    for(int i = 2; i <= n; i++){
         s = (s + m )%i;
    }
    printf("%d", s + 1);

    链表实现:

    输出的s就是最后的编号(根据n改变改变),即从零个数开始逆推上去原来的编号,不是推出的人,是编号的转变。

    本题是前n个好人后n个坏人,要算出的是坏人的编号,即推出的人的编号,所以是%len开始

    #include<cstdio>
    #include<cstring>
    using namespace std;
    bool check(int n,int m)
    {
        int len = 2*n;
        int s = 1;
        for(int i = 1; i <= n ; i++){
                 s = (s + m - 1)%len;
                 if(s == 0) s = len;
                 len--;
                 if(s <= n) return false;
                 if(s > len) s = 1;
        }
        return true;
    }
    int main(){
        int a[100];
        int n,m;
        for( n = 1; n < 14;n++){
                for( m = n+1;;m++){
                        if(check(n,m)){
                            a[n] = m;
                            break;
                        }
                }
        }
        while(~scanf("%d",&n)&&n){
                printf("%d
    ",a[n]);
        }
        return 0;
    }
    View Code
  • 相关阅读:
    php中防止SQL注入的方法
    谈谈asp,php,jsp的优缺点
    SSH原理与运用(一):远程登录
    优化MYSQL数据库的方法
    json_encode和json_decode区别
    静态方法与非静态方法的区别
    Java 异常的Exception e中的egetMessage()和toString()方法的区别
    $GLOBALS['HTTP_RAW_POST_DATA'] 和$_POST的区别
    HTML5开发,背后的事情你知道吗?
    使用C语言来实现模块化
  • 原文地址:https://www.cnblogs.com/zero-begin/p/4314171.html
Copyright © 2011-2022 走看看