HDU 1907 John

John

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

Total Submission(s): 3954    Accepted Submission(s): 2244


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

 

Source

Southeastern Europe 2007

 
题目大意：

john 和他的弟弟一起吃糖果，每次可以吃一堆中的至少一个，谁吃到最后一个谁输。输出谁必胜。

思路：

  尼姆博奕的应用，关于尼姆博弈的证明，大家可以去kuangbin 哪里看一下，给你们贴出传送门，他那里还给出了一些基本的博弈题目供大家练习。

根据常识，如果开局都是孤单堆的话，那么谁先手，谁输。
根据定理，首先设 奇异局势  为 p 非 奇异局势 为 n  奇异局势是输局
 那么，由定理的，p  可以由适当方式变为 n  ，  在 n 态下，任意做法都会使之变成 p 态。
那么，如果开局 是  p 态，先手输，否则先手胜。

感想：

这个题和南京理工那个校赛题目完全相同。啦啦啦~~很简单的对不对？？

AC代码：

#include <iostream>
using namespace std;

int main()
{
int T,N,i,a[50],sum,k;
cin>>T;
while (T--)
{
cin>>N;
k=0;//孤单堆的数量
sum=0;
for (i=0;i<N;i++)
{
cin>>a[i];
sum^=a[i];
if (a[i]>1)
{
k=1;//如果不全部是孤单堆
}
}
if (k==0)    //对于全是孤单堆的情况
{
if (N%2==0)
{
cout<<"John"<<endl;
}
else
{
cout<<"Brother"<<endl;
}
}
else
{
if (sum==0)   //对于奇异局势的情况下
{
cout<<"Brother"<<endl;
}
else          //对于非奇异局势的情况下
{
cout<<"John"<<endl;
}
}
}
return 0;
}