John and his brother are playing a game. There are several piles of stones. At first John has to pick up some stones from one pile. Then his brother has to make a turn. And so on. Everyone has to pick up at least one stone
during his turn. If John (or his brother) will pick up the last stone, he will be considered as a looser.
Both of players are using optimal game strategy. John starts first always.
The first line of input contains a single integer T (1<= T <= 400) indicating the number of test cases. Each test case starts with number N of piles (1<= N <= 47). Then the next line will contain N integers Si (1<= Si <= 4700), separated by spaces – amount of stones of i-th pile.
Output "Jhon" if Jhon win the game, output "Brother" in other case
2
3
3 5 1
1
1
John
Brother
#include<stdio.h>
int main()
{
int a[50];
int T,n,num1,num2,ans;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
ans=num1=num2=0;
for(int i=0;i<n;i++)
{
scanf("%d",a+i);
if(a[i]==1)
num1++;
else
num2++;
ans^=a[i];
}
if((ans&&num2!=0)||(!ans&&num2==0))
printf("John\n");
else
printf("Brother\n");
}
return 0;
}1408: The Game（博弈论）尼姆博弈,布布扣,bubuko.com
原文：http://blog.csdn.net/acvcla/article/details/22613359