Joseph - POJ 1012 打表

最新推荐文章于 2018-12-14 11:16:17 发布

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本文探讨了一个基于经典约瑟夫问题的变形版本，目标是在特定条件下确保所有“坏人”被先行淘汰，而“好人”得以幸存。通过分析与编程实现，找到了使前k个“好人”存活的最小数值m。

Joseph

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 47228		Accepted: 17791

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：顺时针数，让第几个人出去，最后要留下前k个。

思路：打表。

AC代码如下：

#include<cstdio>
#include<cstring>
using namespace std;
int main()
{ int num[]={0,2,7,5,30,169,441,1872,7632,1740,93313,459901,1358657,2504881};
  int i,j,k;
  while(~scanf("%d",&k) && k>0)
   printf("%d\n",num[k]);
}

打表代码如下：

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
int r[14];
bool solve(int k,int i)
{ int n=k*2,x=0;
  while(n>k)
  { x=(x+i-1)%n;
	if(x<k) return false;
    n--;
  }
  return true;
}
int main(){
	int i,j,k;
	for(k=1;k<14;k++)
	 for(i=k+1;;i+=k+1)
     { if(solve(k,i))
       { r[k]=i;
         break;
       }
       else if(solve(k,i+1))
       { r[k]=i+1;
		 break;
       }
      }
	for(i=1;i<=13;i++)
	 printf("%d %d\n",i,r[i]);
}