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.
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 1Sample 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; } }