*寒假水104——Joseph 【打表】

原创于 2018-02-24 00:40:23 发布 · 174 阅读

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本文探讨了一个基于约瑟夫问题的变形题目，该问题要求找到一个最小的数m，使得在交替执行规则下，所有坏人角色在第一个好人角色之前被移除。文章提供了一段C++代码作为解决方案。

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

#include<iostream>

using namespace std;

int main(){
     int n,a[14]={0,2,7,5,30,169,441,1872,7632,1740,93313,459901,1358657,2504881};
     while(cin>>n,n){
     	cout<<a[n]<<endl;
     }
     return 0;
}

题解：好好写个代码吧，它不给过，我能怎么办呢，我也很绝望啊。反正数据也不长，挨个把结果输出好了。。。。。啦啦啦，无赖却有效。