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 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
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;
}
POJ 3480 & HDU 1907 John（尼姆博弈变形）
原文地址：http://blog.csdn.net/u012860063/article/details/40477809
