HDU1907 Jhon

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. 

InputThe 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 

OutputOutput 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
题解:考虑多个1的情况
参考代码:
 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int T,n,ans,x,num; 
 4 
 5 int main()
 6 {
 7     scanf("%d",&T);
 8     while(T--)
 9     {
10         scanf("%d",&n);ans=num=0;
11         for(int i=1;i<=n;++i) {scanf("%d",&x);ans^=x;if(x>1) num++;}
12         if((!ans&&num>1) || (ans&&!num)) puts("Brother");
13         else puts("John");
14      } 
15     
16     return 0;
17 }
View Code

 

posted @ 2019-01-28 19:37  StarHai  阅读(175)  评论(0编辑  收藏  举报