hdu 1907 John (尼姆博弈)

John

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 6000    Accepted Submission(s): 3486

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

C/C++：

 1 #include <map>
 2 #include <queue>
 3 #include <cmath>
 4 #include <vector>
 5 #include <string>
 6 #include <cstdio>
 7 #include <cstring>
 8 #include <climits>
 9 #include <iostream>
10 #include <algorithm>
11 #define INF 0x3f3f3f3f
12 #define LL long long
13 #define wzf ((1 + sqrt(5.0)) / 2.0)
14 using namespace std;
15 const int MAX = 100;
16 
17 int n, t, num[MAX], temp;
18 
19 int main()
20 {
21     scanf("%d", &t);
22     while (t --)
23     {
24         scanf("%d", &n);
25         for (int i = 0; i < n; ++ i)
26             scanf("%d", &num[i]);
27         sort(num, num + n);
28         if (num[n - 1] == 1)
29         {
30             if (n & 1) printf("Brother\n");
31             else printf("John\n");
32             continue;
33         }
34         temp = num[0] ^ num[1];
35         for (int i = 2; i < n; ++ i)
36             temp ^= num[i];
37         if (temp == 0)
38             printf("Brother\n");
39         else
40             printf("John\n");
41     }
42     return 0;
43 }