John

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 4704    Accepted Submission(s): 2720

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
 
 
 #include <bits/stdc++.h>
using namespace std; int main()
{
int t;
int n;
int a;
int i;
int sum, ones; scanf("%d", &t); while (t--) {
scanf("%d", &n);
sum = , ones = ;
for (i = ; i < n; ++i) {
scanf("%d", &a);
sum ^= a;
if (a == ) {
++ones;
}
} //win : t0, s1, s2
//lose : s0, t2
if (sum == ) {//t
if (n - ones == ) {//t0
printf("John\n");
} else {//t2
printf("Brother\n");
}
} else {//s
if (n - ones == ) {//s0
printf("Brother\n");
} else {//s1, s2
printf("John\n");
}
}
} return ;
}

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