POJ 3480 & HDU 1907 John（尼姆博弈变形）

题目链接：

PKU:http://poj.org/problem?id=3480

HDU:http://acm.hdu.edu.cn/showproblem.php?pid=1907

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 A_i, separated by spaces – amount of M&Ms of i-th color.

Constraints:

1 <= A_i <= 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

PS:

尼姆博弈变形！

代码如下：

#include <cstdio>
#define maxn 5017
int main()
{
    int t;
    int a[maxn];
    int n, s, k;
    scanf("%d",&t);
    while(t--)
    {
        s = 0, k = 0;
        scanf("%d",&n);
        for(int i = 0; i < n; i++)
        {
            scanf("%d",&a[i]);
            if(a[i] > 1)
                k++;
            s = s^a[i];
        }
        if(k == 0)//全为1
        {
            if(n%2 == 0)//偶数堆
                printf("John\n");
            else
                printf("Brother\n");
        }
        else
        {
            if(s)
                printf("John\n");
            else
                printf("Brother\n");
        }
    }
    return 0;
}