HDUOJ 1443 - Joseph

Problem 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

340

 

Sample Output

530

 

/*m的取值：我们考察一下只剩下k+1个人时候情况，即坏人还有一个未被处决，那么在这一轮中结束位置必定在最后一个坏人，那么开始位置在哪呢？这就需要找K+2个人的结束位置，然而K+2个人的结束位置必定是第K+2个人或者第K+1个人，这样就出现两种顺序情况：GGGG.....GGGXB 或  GGGG......GGGBX (X表示有K+2个人的那一轮退出的人）所以有K+1个人的那一轮的开始位置有两种可能即最后一个位置或K+1的那个位置，限定m有两种可能：GGGG......GGGBX 若K+2个人的结束位置在最后一个(第K+2个)，则m%(k+1)==0GGGG......GGGXB 若K+2个人的结束位置在倒数第二个(第K+1个)，则m%(k+1)==1*/#include <cstdio>int joseph[15];int confirm(int m, int k){    int pos = 0;    int len = k*2;    for (int i = 0; i < k; ++i)    {        pos = (pos+m-1) % (len-i);        if (pos < k)            return 0;    }    return 1;}int main(){    int m;    int k;    for (k = 1; k < 15; ++k)    {        m = k+1;        while (true)        {            if (confirm(m, k) == 1)            {                joseph[k] = m;                break;            }            if (confirm(m+1, k) == 1)            {                joseph[k] = m+1;                break;            }            m += k+1;        }    }    while (scanf("%d", &k) != EOF && k != 0)    {        printf("%d\n", joseph[k]);    }    return 0;}