hdu-1907-反nim博弈

John

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 5793    Accepted Submission(s): 3358

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

Southeastern Europe 2007

 

　　　　 对于所有不同颜色M&M的个数都为1的情况,显然:

　　     if(sg==0)  win       -->状态A

　　　 if(sg!=0)   lose       -->状态B

　　 对于对立情况，

　　　　当数量不为1的种类只有一种的时候(状态C)，显然先手总可以将其转化到状态B这个败态，所以C是胜态。

　　　　当数量不为1的种类大于1的时候,分为sg=0(状态D)和sg!=0(状态E)两种情况，D只能转化到E,E能转化到D或者

C,由于C是败态，所以E尽可能向D转化，但随着石子的不断减少导致E只能转化到C，所以E是败态，D是胜态。

　　

　　

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int main(){
 4     int t,n,m;
 5     cin>>t;
 6     while(t--){
 7         int a,sg=0,tot=0;
 8         cin>>n;
 9         for(int i=1;i<=n;++i){
10             cin>>a;
11             sg^=a;
12             if(a==1) tot++;
13         }
14         if( (tot==n&&sg==0) || (tot!=n&&sg!=0)) puts("John");
15         else puts("Brother");
16     }
17     return 0;
18 }