8.2.6 John

John

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)

Total Submission(s): 31 Accepted Submission(s): 23

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

anti-nim

sg函數的題 給我寫過的一篇sg函數的連接

http://www.cnblogs.com/cssystem/p/3204826.html

 1 #include <cmath>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstring>
 5 #include <string>
 6 #include <cstdlib>
 7 using namespace std;
 8 
 9 const int maxn=210;
10 int n,ans,T,x;
11 bool flag;
12 
13 void close()
14 {
15 exit(0);
16 }
17 
18 
19 void init()
20 {
21     scanf("%d",&T);
22     while (T--)
23     {
24         scanf("%d",&n);
25         flag=false;
26         scanf("%d",&ans);
27         if (ans>1) flag=true;
28         for (int i=2;i<=n;i++)
29         {
30             scanf("%d",&x);
31             if (x>1) flag=true;
32             ans=ans ^ x;
33         }
34         if (not flag) //所有是1
35         {
36             if (n % 2==0)
37                 printf("John\n");
38             else
39                 printf("Brother\n");
40         }
41         else
42         {        
43             if (ans!=0)
44                 printf("John\n");
45             else
46                 printf("Brother\n");
47         }
48     }
49 }
50 
51 int main ()
52 {
53     init();
54     close();
55     return 0;
56 }