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最新推荐文章于 2020-10-07 16:37:01 发布

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#integer #colors #each #input #ini

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文章探讨了使用博弈论原理解决M&M糖果分配游戏的问题，通过输入不同颜色糖果的数量，预测游戏的最终胜者是John还是他的弟弟。

John

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 945 Accepted Submission(s): 495

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

Sample Output

   
    John
Brother

Source

Southeastern Europe 2007

#include<iostream>
using namespace std;
int main()
{
int T;
cin>>T;
while(T>0)
{
T--;
int n;
cin>>n;
int u,k=0,t;
cin>>t;
if(t>1)k++;
for(int i=1;i<n;i++)
{
cin>>u;
t=t^u;
if(u>1)k++;
}
if(k==0)
{
if(n%2==0)cout<<"John"<<endl;
else cout<<"Brother"<<endl;
}
else
{
if(t!=0)cout<<"John"<<endl;
else cout<<"Brother"<<endl;
}
}
return 0;
}