John(POJ-3480)
题面
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.
输入
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
输出
Output T lines each of them containing information about game winner. Print “John” if John will win the game or “Brother” in other case.For each test case, write a single line with an integer indicating the number of winning moves from the given Nim position.
样例输入
12
23
33 5 1
41
51
样例输出
1John
2Brother
提示
无
思路
Anti-Nim博弈,属于Anti-SG游戏的一种。
Anti-SG游戏定义:
决策集合为空的操作者胜。
其余规则与SG游戏一致。
SJ定理:
对于任意一个Anti-SG游戏,如果定义所有子游戏的SG值为0时游戏结束,先手必胜的条件:
- 游戏的SG值为0且所有子游戏SG值均不超过1。
- 游戏的SG值不为0且至少一个子游戏SG值超过1。
代码
1using namespace std;
2
3int main()
4{
5 int T; scanf("%d", &T);
6 while(T--)
7 {
8 int n; scanf("%d", &n);
9
10 int nim=0, anti=0;
11 for(int i=0; i<n; i++){
12 int x; scanf("%d", &x);
13 if(x>1) anti = 1;
14 nim ^=x;
15 }
16 if((!nim&&!anti) || (nim&&anti)){
17 printf("John\n");
18 }else{
19 printf("Brother\n");
20 }
21 }
22 return 0;
23}