【题解】POJ-3480 John

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

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2
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2
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3 5 1
1
1

样例输出

1
2
John
Brother

提示

思路

Anti-Nim博弈,属于Anti-SG游戏的一种。

Anti-SG游戏定义:

  1. 决策集合为空的操作者胜。

  2. 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:

  1. 游戏的SG值为0且所有子游戏SG值均不超过1。
  2. 游戏的SG值不为0且至少一个子游戏SG值超过1。

代码

查看代码
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using namespace std;

int main()
{
int T; scanf("%d", &T);
while(T--)
{
int n; scanf("%d", &n);

int nim=0, anti=0;
for(int i=0; i<n; i++){
int x; scanf("%d", &x);
if(x>1) anti = 1;
nim ^=x;
}
if((!nim&&!anti) || (nim&&anti)){
printf("John\n");
}else{
printf("Brother\n");
}
}
return 0;
}
_/_/_/_/_/ EOF _/_/_/_/_/