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;
}