John（HDU 1907） —— 尼姆博弈

John

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 2962    Accepted Submission(s): 1664

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
   
   
   
   

题意：有n种石块，每种石块有a[i]个。两个人，John先拿，每次可拿任意个石块，但必须是同一种，谁拿到最后的石块输。

 

思路：列出各种情况，1堆、2堆、3堆...一开始觉得与奇偶有关系，但后来发现不是，于是想到了异或关系（这些情况很杂，不是奇偶，特别的规律也很难搞，所以根据经验只能想到异或，其它的东西我也不知道），然后发现异或可以解决我自己举出的案例，故一试，成功。

#include<stdio.h>
int main()
{
    int T, n, i, a[50];

    scanf("%d", &T);
    while(T--)
    {
        scanf("%d", &n);
        int x = 0, num = 0, flag = 0;
        for(i=1; i<=n; i++)
        {
            scanf("%d", a+i);
            x ^= a[i];
            if(a[i] == 1) num++;
        }
        if(n & 1 && (x == 0 || num == n)) flag = 1;
        if(n % 2 == 0 && x == 0 && num != n) flag = 1;
        printf(flag ? "Brother\n" : "John\n");
    }
    return 0;
}

代码中把n%2==0换成n&1==0结果不一样，不知为何，都是判断奇偶的啊。。