John

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 4440    Accepted Submission(s):
2541

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

2

3

3 5 1

1

1

哎呀卧槽。。。

之前看nim游戏的论文，性子太急，看了一半（拿最后一个胜的情况），然后以为自己弄懂了，然后遇到这个题就有点懵逼，再把论文看完，发现之前自己理解得并不是很透彻。

无论拿最后一个胜还是拿最后一个败，都是在争夺S1状态（只有一个充裕堆）。仔细一想，确实是这样，当到达S1状态，玩家可以决定下一个状态，即把下一个状态变为T0或S0

也就是说谁争夺到S1，谁就必胜。

在拿最后一个败的情况中，当所有堆为孤单堆时，需要特殊判断一下，原因容易想到。

博弈这方面，还需多多练习，多多揣摩！！！

#include<iostream>
#include<cstdio>
#include<cstring>
#include<stdlib.h>
#include<algorithm>
#include<cmath>
using namespace std;

int main()
{
    int t,n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        int res=0,num;
        int flag=0;
        for(int i=0; i<n; i++)
        {
            scanf("%d",&num);
            if(num>1)
                flag=1;
            res^=num;
        }
        if(flag)
        {
            if(res>0)
                printf("John\n");
            else
                printf("Brother\n");
        }
        else
        {
            if(n%2==0)
                printf("John\n");
            else
                printf("Brother\n");
        }
    }
    return 0;
}