HDU1907：John(Nim)

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

233 5 111

 

Sample Output

JohnBrother
 
 
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 <stdio.h>#include <string.h>#include <algorithm>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<n;i++)        {            scanf("%d",&a[i]);            ans^=a[i];            if(a[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;}