HDU1907：John(Nim) - c++编程基础

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

Sample Output

John
Brother  今天看了下博弈的文章，长篇大论啊。。。姑且分享下那篇文章吧。。。http://blog.csdn.net/acm_cxlove/article/details/7854530 这题就是最基础的nim博弈了只要按照那里总结出来的必胜态与必败态来处理即可 #include 
   
    
#include 
    
      #include 
     
       using namespace std; int main() { int t,n,i; int sum1,sum2,ans; int a[55]; scanf("%d",&t); while(t--) { scanf("%d",&n); sum1 = sum2 = ans = 0; for(i = 0;i
      
       =2) sum2++; else sum1++; } if((ans && sum2) || (!ans && !sum2)) printf("John\n"); if((ans && sum1%2 && !sum2) || (!ans && sum2>=2)) printf("Brother\n"); } return 0; }