John

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.

InputThe 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

OutputOutput 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

代码：

///1.如果所有堆都为1（即c为0） 且为偶数堆（即flag为0）后手拿最后一个  比如 1 1 1 1
///2.如果c不为零，存在d大于1 ，如果这样的d只有一个  （flag肯定不为0） 比如 1 1 4 先手拿3个造成对方必败 1 1 1 4 先手拿4个 对方必败
///3.如果这样的d不只一个 先手拿一次将flag变为0即可 且总还存在某堆石子数大于1（如果不存在就必败了）。转换成二进制来看就是拿走flag个 剩下的每个位的1都是偶数个 是必败态
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#define Max 1001
#define PI 3.1415927
using namespace std;

int main()
{
    int t,n,d;
    cin>>t;
    while(t --)
    {
        int flag = 0,c = 0;
        cin>>n;
        while(n --)
        {
            cin>>d;
            flag ^= d;
            if(d > 1)c ++;
        }
        if(!flag&&!c || flag&&c)cout<<"John"<<endl;
        else cout<<"Brother"<<endl;
    }
}

 

转载于:https://www.cnblogs.com/8023spz/p/7966762.html