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最新推荐文章于 2019-04-17 20:02:58 发布

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本文介绍了一个基于NIM博弈的游戏，玩家需轮流吃掉同色糖果直至无法进行为止。通过分析游戏规则及输入输出样例，文章提供了一段C++代码实现游戏胜负判断，并讨论了背后的博弈策略。

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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

【分析】

简单的NIM博弈，虽然简单，但是原理对于我这样蒟蒻来说还是不能理解，等什么时候真正的懂了在来更新.

【代码】

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <map>
#include <list>
#include <algorithm>
#include <iostream>

using namespace std;
int ans(void)
{
    int len;
    cin>>len;
    int a=0,b=1,c;
    for(int i=0;i<len;i++)
    {
        cin>>c;
        if(c!=1)
            b=0;
        a^=c;
    }
    if(b) cout << ((len&1)?"Brother":"John")<<endl;
    else cout<<(a?"John":"Brother")<<endl;
}
int main() {
//    freopen("in.txt","r",stdin);
    int T;
    cin>>T;
    while(T--)
    {
        ans();
    }
    return 0;
}