Hdu 1443 Joseph

Joseph

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1512    Accepted Submission(s):
948

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

 

3

4

0

 

Sample Output

 

5

30

 

这是一道关于约瑟夫环的问题，题目的大意是总共有2k个伙计围成一圈，1~k是好人，k+1~2k是坏人，报数m，使得k+1~2k的坏人必须在好人之前出去，求出这个m。

 

很显然，这道题要模拟整个过程，因为题目要求k小于14，所以定义一个一位数组保存所求出的m值。

 

 #include <iostream>
 using namespace std;
 int Joseph(int k,int m)
 {
     int len = *k;
     int i;
     int n = ;        //n是要出去的位置，起始位置初始化为1
     for (i=;i<=k;i++)
     {
         n = (n+m-)%len;    //第i次出去的人
         if(n==)        //最后一个人出去
             n = len;
         if(n<=k)            //n小于或等于k时候，返回false
             return false;
         len--;        //出去一个人，则长度减1
     }
     return true;
 }

 int main()
 {
     int a[];
     int i,j;
     for (i=;i<;i++)
         for(j=i+;;j++)
             if (Joseph(i,j))
             {
                 a[i] = j;
                 break;
             }
     int n;
     while(cin>>n && n)
         cout<<a[n]<<endl;
     return ;
 }