HDU 1907 John（尼姆博弈）

John

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 3223    Accepted Submission(s): 1811

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

Sample Output

John Brother

Source

Southeastern Europe 2007

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

int main()
{
    int n;
    int T;
    scanf("%d",&T);
    while (T--)
    {
        scanf("%d", &n);
        __int64 ans = 0;
        __int64 tmp;
        int cnt = 0;
        for (int i = 0; i < n; i++)
        {
            scanf ("%I64d", &tmp);
            if(tmp == 1)
            {
                cnt++;
            }
            ans = ans^tmp;
        }
        if(cnt == n)
        {
            if(n%2 == 1)
            {
                printf("Brother\n");
            }
            else
            {
                printf("John\n");
            }
        }
        else
        {
            if (ans == 0)
            {
                printf("Brother\n");
            }
            else
            {
                printf("John\n");
            }
        }

    }
}