R

 

R - Joseph

 

题目如下：

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

大意是写一个程序，在好人一个都不死的情况下，循环杀死后面的坏人。

首先如果每次都循环求解，会TL，所以需要先写一个打表程序，输入数据<=14，所以可以先写程序判定每个n的答案，然后再建表写真正的答案程序。

代码如下：

#include<iostream>

usingnamespace std;

intmain()

{

        intnum[14] = { 0,2,7,5,30,169,441,1872,7632,1740,93313,459901,1358657,2504881 };

        intk;

        while(cin>>k&& k) {

               cout<< num[k]<< endl;

        }

        return0;

}

/*int main()//打表算法

{

        int a[30],b[30],n,m;

        while (cin >> n&&n)

        {

               memset(a, 0, sizeof(a));

               memset(b, 0, sizeof(b));

               for (int i = 1; i <= 2 * n; i++)

                       a[i]= i;

               m = n+1;

               for (m;; m++)

               {

                       int t = 2*n,count=0,temp=0,flag=0,goodguy=0;

                       for (int i = 1; i <= 2 * n; i++)

                              b[i] = a[i];

                       while(!goodguy)

                       {

                              int sum0 = 0;

                              while (sum0 != m)

                              {

                                      temp++;

                                      if (temp > t)temp %= t;

                                      if (b[temp])sum0++;

                              }

                              if (temp <= n)

                              {

                                      goodguy++; break;

                              }

                              if (temp > n)

                              {

                                      count++; b[temp] = 0;

                              }

                              if (count == n) {

                                      flag = 1; break;

                              }

                       }

                       if (flag) {

                              cout << m; break;

                       }

               }

        }

        return 0;

}*/