hdu1907（Nim博弈）

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

John

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)

Total Submission(s): 2061    Accepted Submission(s): 1115


Problem 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

 

做这一题时想多了，只需要注意一个地方，就是当所有堆数都为一时，本该john胜得情况变为Brother胜利，反之，本该Brother胜利的变为John胜利。

#include<cstdio>
#include<cmath>
#include<iostream>
using namespace std;
int main()
{
int t,i,m,n,ans,a[50];
scanf("%d",&t);
while(t--)
{
ans=n=0;
scanf("%d",&m);
for(i=1;i<=m;i++)
{
scanf("%d",&a[i]);
if(a[i]>1) n=1;
ans^=a[i];
}
if((ans&&n)||(!ans&&!n)) puts("John");
else puts("Brother");
}
}