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最新推荐文章于 2021-03-15 09:42:22 发布

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最新推荐文章于 2021-03-15 09:42:22 发布

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分类专栏：博弈 hdu

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本文介绍了一个关于M&M糖果的游戏，玩家需轮流吃掉同色糖果直至无法进行为止。通过分析糖果数量，采用Nim博弈策略，判断最终胜者。

John

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 2770 Accepted Submission(s): 1538

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

Source

Southeastern Europe 2007

Recommend

lcy

题目的规则改成了谁吃最后的糖果谁就输，所以分两种情况，第一种就是所有的colors都只有一种，这个时候就是奇数brother win，偶数jhon win;第二种情况就是nim了。

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

int a[48];

int main()
{
    int t;
    scanf("%d",&t);
    while(t--){
        int n;
        scanf("%d",&n);
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        int x=0,sum=0;
        for(int i=0;i<n;i++){
            x^=a[i];
            if(a[i]==1) sum++;
        }
        if(sum==n){
            if(n%2==0) puts("John");
            else puts("Brother");
        }
        else if(x==0) puts("Brother");
        else puts("John");
    }
    return 0;
}