Nim游戏——Rabbit and Grass HDU - 1849

题解：简单的Nim游戏，每个棋子距离做短点的位置看做石子的数量，然后进行亦或就可以

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int main()
{
    int n,a,sum = 0;
    while(~scanf("%d",&n)&&n)
    {
        sum =0;
        for(int i = 0 ; i < n ; i++)
        {
            scanf("%d",&a);
            sum^=a;
        }
        if(sum == 0) cout<<"Grass Win!"<<endl;
        else cout<<"Rabbit Win!"<<endl;
    }
    return 0;
}