hoj1017 Joseph's problem II

本文详细介绍了如何通过算法解决约瑟夫问题的一个变种,即在特定条件下确定杀死所有坏人前的最小m值。文章包含了解题思路、代码实现及验证过程,旨在为读者提供深入理解该问题及其解决方案的路径。

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Joseph's problem II

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  Source : ACM ICPC Central European Regional 1995
  Time limit : 2 sec   Memory limit : 32 M

Submitted : 1048, Accepted : 509

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
 
 
约瑟夫问题
一共有2*k个人,前k个人不能死,问在杀死最后一个后k个人的时候,m最小是多少
 
 
由于,k是小于14的,可以先打表。。
从m开始模拟,如果在区间前k内,则m++,维护当前的前k区间,注意x'=(x-m)%i,此时x'可能小于0,那就i 知道满足为止
 
 
code:
#include <iostream>
#include <queue>
#include <stdio.h>
#include <stack>
#include <algorithm>
#include <string.h>
#include <map>
#include <stdlib.h>
using namespace std;
bool solve(int k,int m)
{
    int start = 0,end = k - 1;
    bool flag = true;
    for(int i=2*k;i>k;i--)
    {
        int kill = (m-1)%i;
        if(kill>=start && kill<=end)
        {
            flag = false;
            break;
        }
        start = ((start - m)%i+i)%i;
        end = ((end - m)%i+i)%i;
    }
    return flag;
}
int kk[15];
int main()
{
	int k;
	for(int i=1;i<14;i++)
	{
		for(int m=i+1;;m++)
			if(solve(i,m))
		{
			kk[i]=m;
			break;
		}
	}
	while(cin>>k)
	{
		if(k==0) break;
		cout<<kk[k]<<endl;
	}
	return 0;
}
 
 
 
 
 
 
 
 
 
 
 
 
 
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