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函数的题 给我写过的一篇sg函数的链接
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) //全部是1 35 { 36 if (n % 2==0) 37 printf("John "); 38 else 39 printf("Brother "); 40 } 41 else 42 { 43 if (ans!=0) 44 printf("John "); 45 else 46 printf("Brother "); 47 } 48 } 49 } 50 51 int main () 52 { 53 init(); 54 close(); 55 return 0; 56 }