hdu 1907 尼姆博弈

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

Southeastern Europe 2007

 

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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<stdio.h>
#include<string.h>
#include<math.h>
int a[];

int main(){
   int t;
   scanf("%d",&t);
   int n;
   while(t--){
        scanf("%d",&n);
        bool flag=true;
        for(int i=;i<=n;i++){
            scanf("%d",&a[i]);
            if(a[i]!=)
                flag=false;
        }
        int ans=a[];
        for(int i=;i<=n;i++)
            ans=ans^a[i];

            if(flag){
               if(n%)
                     printf("Brother\n");
               else
                 printf("John\n");

            }
            else{
        if(ans)
            printf("John\n");
        else
            printf("Brother\n");
            }

   }
   return ;
}