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;
}