John

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)

Total Submission(s): 3745    Accepted Submission(s): 2116

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

 

 

Recommend

lcy   |   We have carefully selected several similar problems for you:            

 

题意大概是有n种颜色的糖豆，john和他的哥哥轮流吃糖，每次可以选择其中的一种颜色中的若干个，最后谁最后吃完谁就输了

思路：

经典的尼姆博弈问题

对于尼姆博弈，类似于威佐夫博弈，奇异局势与非奇异局势碾转变换，先发者如果面对的奇异局势则输，反之则胜

对于如果判断面对的局势是否为奇异局势，有两种情况需要考虑

1  如果所有项都为1的话，只需要判断奇偶的数量就可以了

2  若不为1的情况，如果所有数的异或值为0，那么就是奇异局势，否则不是奇异局势

 

代码1a，

可做模板

重试信心-------》-----》

#include
      
       #include
       
        #include
        
         int a[100];int main(){   int t;   scanf("%d",&t);   int n;   while(t--){        scanf("%d",&n);        bool flag=true;        for(int i=1;i<=n;i++){            scanf("%d",&a[i]);            if(a[i]!=1)                flag=false;        }        int ans=a[1];        for(int i=2;i<=n;i++)            ans=ans^a[i];            if(flag){               if(n%2)                     printf("Brother\n");               else                 printf("John\n");            }            else{        if(ans)            printf("John\n");        else            printf("Brother\n");            }   }   return 0;}