HDU 1907 John

最新推荐文章于 2018-04-23 15:37:21 发布

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本文介绍了一个基于Nim博弈的游戏变种，其中玩家需从包含多种颜色M&M豆的盒子中取豆，最后取者为败。文章提供了解决此游戏策略问题的代码实现，并通过样例输入输出展示如何确定胜者。

题目很简单，比起普通的Nim博弈只是多了一种情况。因为普通的Nim问题是最后一个取的人赢，而这道题是最后一个取的人输，所以结论要反过来，但是还有一种情况必须考虑，就是如果每一堆的数目都是1的话，那么要考虑一下堆数。

John

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 3545 Accepted Submission(s): 2009

Problem Description

Little John is playing very funny game with his younger brother. There is one big box filled with M&Ms of different colors. At first John has to eat several M&Ms of the same color. Then his opponent has to make a turn. And so on. Please note that each player has to eat at least one M&M during his turn. If John (or his brother) will eat the last M&M from the box he will be considered as a looser and he will have to buy a new candy box.

Both of players are using optimal game strategy. John starts first always. You will be given information about M&Ms and your task is to determine a winner of such a beautiful game.

Input

The first line of input will contain a single integer T – the number of test cases. Next T pairs of lines will describe tests in a following format. The first line of each test will contain an integer N – the amount of different M&M colors in a box. Next line will contain N integers Ai, separated by spaces – amount of M&Ms of i-th color.

Constraints:
1 <= T <= 474,
1 <= N <= 47,
1 <= Ai <= 4747

Output

Output T lines each of them containing information about game winner. Print “John” if John will win the game or “Brother” in other case.

Sample Input

Sample Output

  
   John
Brother

题目链接： HDU 1907 John

AC代码：

#include <cstdio>
#include <iostream>
using namespace std;
int main()
{
    int T,m,i,x,sum;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&m);
        sum=0;
        int k=0;
        for(i=0;i<m;i++)
        {
            scanf("%d",&x);
            sum=sum^x;
            if(x>1)
                k=1;
        }
        if(k==0)
        {
            if(m%2!=0)
                printf("Brother\n");
            else
                printf("John\n");
        }
        else
        {
            if(!sum)
                printf("Brother\n");
            else
                printf("John\n");
        }
    }
    return 0;
}