HDU-1907 John

http://acm.hdu.edu.cn/showproblem.php?pid=1907

                       John

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others) Total Submission(s): 2385    Accepted Submission(s): 1289

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
 1 #include<iostream>
 2 #include<algorithm>
 3 using namespace std;
 4 int main()
 5 {
 6     int t,n,i,a[50];
 7     int temp,ans;
 8     scanf("%d",&t);
 9     while(t--)
10     {
11         temp=0;
12         ans=-1;
13         scanf("%d",&n);
14         for(i=0;i<n;i++)
15         {
16             scanf("%d",&a[i]);
17             if(i==0)
18                 ans=a[i];
19             else
20                 ans=ans^a[i];
21             if(a[i]>1)
22                  temp=1;
23         }
24         if(temp==0)
25         {
26             if(n%2==1)
27               printf("Brother
");
28             else
29               printf("John
");
30         }
31         else
32         {
33             if(ans==0)
34               printf("Brother
");
35             else
36               printf("John
");
37             
38         }
39     }
40     return 0;
41 }
 
原文地址:https://www.cnblogs.com/cancangood/p/3512783.html