POJ 题目1012Joseph（数学，约瑟夫环）

Joseph

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 48185		Accepted: 18184

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

Source

Central Europe 1995

a题目大意：NYOJ题目191

思路：

a(n) = [a(n-1)+m-1]mod(2k-n+1)

要求a(n) > k；n = 1,2,3,...,k

其中2k-n+1是第i-1次踢人后剩下的人数。

可以设计如下算法：


bool Joseph(int k, int m) // 这个算法确定对于给定的k,m是否满足上面的要求




...{


    int n;


    for(n=1;n<=k;n++)




    ...{


        a = (a+m-1)%(k2-n+1);


        if(a == 0) a = k2-n+1;


        if(a<=k && a>=1) return false;


    }


    return true;


}

然后，我们注意到，第一次踢的人是 m%2k，我们要求 m%2k > k，也就是 m = 2k*r+h，h>k，那么就可以设计如下的算法找出最小的m：
for(r=0;;r++)

{

    for(h=k+1;h<=2*k;h++)

    {

         m = 2*k*r+h;

         if(Joseph(k,m)) goto end; // 找到m跳出
     }

}

end: 

ac代码

#include<stdio.h>
#include<string.h>
int k,m;
int a[220];
int find(int k,int m)
{
int n,a=1;
for(n=1;n<=k;n++)
{
a=(a+m-1)%(2*k-n+1);
if(!a)
a=2*k-n+1;
if(a<=k&&a>=1)
return 0;
}
return 1;
}
int main()
{
int n;
while(scanf("%d",&k)!=EOF,k)
{
int r,h,w=0;
if(!a[k])
{
for(r=0;;r++)
{
for(h=k+1;h<=2*k;h++)
{
m=2*k*r+h;
if(find(k,m))
{
w=1;
break;
}
}
if(w==1)
break;
}
}
else
w=1;
if(w)
{
if(!a[k])
a[k]=m;
printf("%d\n",a[k]);
}
}
}