hdu 1907 John（anti nim）

Posted 2020-09-14 gongpixin

John

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 4704    Accepted Submission(s): 2720

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

 

 

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 int main()
 5 {
 6     int t;
 7     int n;
 8     int a;
 9     int i;
10     int sum, ones;
11 
12     scanf("%d", &t);
13 
14     while (t--) {
15         scanf("%d", &n);
16         sum = 0, ones = 0;
17         for (i = 0; i < n; ++i) {
18             scanf("%d", &a);
19             sum ^= a;
20             if (a == 1) {
21                 ++ones;
22             }
23         }
24 
25         //win : t0, s1, s2
26         //lose : s0, t2
27         if (sum == 0) {//t
28             if (n - ones == 0) {//t0
29                 printf("John\n");
30             } else {//t2
31                 printf("Brother\n");
32             }
33         } else {//s
34             if (n - ones == 0) {//s0
35                 printf("Brother\n");
36             } else {//s1, s2
37                 printf("John\n");
38             }
39         }
40     }
41 
42     return 0;
43 }