John
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 6527 Accepted Submission(s): 3759
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
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#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int t;
int main()
{
scanf("%d",&t);
while(t--)
{
int n;
scanf("%d",&n);
int x;
int ans=0,sum=0;
for(int i=1;i<=n;i++)
{scanf("%d",&x);
if(x>1)
sum++;
ans^=x;
}
if((ans==0&&sum==0)||(ans!=0&&sum>0))
printf("John\n");
else
printf("Brother\n");
}
return 0;
}