John

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others) Total Submission(s): 6017    Accepted Submission(s): 3499

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

題意

兩個人取n堆石子，每個人至少去一個，最多把一堆石子取完，取到最後一個石子的人失敗

思路

先手勝的情況：

1. n堆石子全部都只有一個石子，且n堆石子的異或值為0
2. n堆石子不全是一個石子，且異或值不為0

證明：

1. 若所有堆石子數都為1且SG值為0，則共有偶數堆石子，故先手勝。
2. 只有一堆石子數大於1時，我們總可以對該堆石子操作，使操作後石子堆數為奇數且所有堆得石子數均為1
3. 有超過一堆石子數大於1時，先手將SG值變為0即可，且總還存在某堆石子數大於1

AC程式碼

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <math.h>
#include <limits.h>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <set>
#include <string>
#define ll long long
#define ull unsigned long long
#define ms(a) memset(a,0,sizeof(a))
#define pi acos(-1.0)
#define INF 0x7f7f7f7f
#define lson o<<1
#define rson o<<1|1
const double E=exp(1);
const int maxn=1e6+10;
const int mod=1e9+7;
using namespace std;
int main(int argc, char const *argv[])
{
	ios::sync_with_stdio(false);
	int t;
	int n;
	int x;
	cin>>t;
	while(t--)
	{
		cin>>n;
		int sum=0;
		int res=0;
		while(n--)
		{
			cin>>x;
			sum^=x;
			if(x>1)
				res++;
		}
		if(!res)
		{
			if(!sum)
				cout<<"John"<<endl;
			else
				cout<<"Brother"<<endl;
		}
		else
		{
			if(!sum)
				cout<<"Brother"<<endl;
			else
				cout<<"John"<<endl;

		}
	}
	return 0;
}