题目描述
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.
输入
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.
输出
Output T lines each of them containing information about game winner. Print “John” if John will win the game or “Brother” in other case. Constrains: 1 <= T <= 474, 1 <= N <= 47, 1 <= Ai <= 4747
样例输入
2 3 3 5 1 1 1
样例输出
John Brother
尼姆博弈，很好玩的博弈问题，我就会这种和单个的博弈问题
#include<iostream> #include<cstdio> using namespace std; int main() { int sum=0,n,t,x; bool k; scanf("%d",&t); while(t--) { scanf("%d",&n); k=false; sum=0; for (int i=0;i<n;i++) { scanf("%d",&x); sum^=x; if (x>1) k=true; } if (!k) { if (n%2==0) printf("John "); else printf("Brother "); } else { if (sum==0) printf("Brother "); else printf("John "); } } return 0; }