Gym - 100814L Candy Jars (博弈)

最新推荐文章于 2020-09-13 18:24:08 发布

原创最新推荐文章于 2020-09-13 18:24:08 发布 · 382 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#17暑假训练

博弈相关算法专栏收录该内容

13 篇文章

订阅专栏

本文介绍了一个简单的糖果罐博弈游戏：两名玩家轮流操作，选择一个罐子扔掉其他罐子的所有糖果，并将选中的罐子里的糖果均匀分配到剩余的罐子中。游戏的目标是迫使对手无法进行有效操作而输掉游戏。文章提供了判断先手玩家是否能赢得比赛的算法。

Alice and Bob love to play the following game, they have N jars and the rules are as follows:

Each jar has a strictly positive number of candies.
Alice plays first and the two players alternate.
In his/her turn, the player selects any jar X, throws away all candies in all the other jars and redistributes the candies in X among all jars in anyway he/she wishes, as long as in the end there is at least one candy in each jar (including X).
The player who cannot make a move in his/her turn loses the game.

Assuming both players play optimally, you are asked the following question who wins the game? Playing optimally means that both players will have insight into all possible next moves and will play in such a way to maximize their chance of winning without making a mistake. However, if all moves lead to the other player winning, then this player still has to play, and any move would be equivalent.

Input

The first line contains the number of test cases T. Each of the next T lines contains an integer (2 ≤ N ≤ 1, 000) the number of jars, andN integers (1 ≤ ai ≤ 1000) where ai is the number of candies in jar i.

Output

Output T lines, one for each test case, containing "Alice" if Alice wins the game, or "Bob" otherwise.

Examples

input
3
4 1 2 3 3
4 1 2 3 4
2 1 3

output
Bob
Alice
Bob

123

博弈, 选择一个罐子,扔掉其他, 将选择的罐子里的数目分配给所有的罐子,(每个罐子至少要分到一个) 如果有罐子分不到, 则输;

可以发现如果所有的罐子里的数目都<n 则肯定输,

并且当所有罐子中存在几个罐子里的数目为为 n - n-1 那么无论是选择n 还是选择 n-1 先手一定赢;

否则先手一定输;

所以: 先手赢得条件为存在罐子里的数目 x % (n*(n-1)) =n || =0 即先手输为 1<=x % (n*(n-1)) <n

#include <stdio.h>
#include <iostream>
using namespace std;

int main()
{
	int sum=0;
	int T;
	cin>>T;
	int n;
	while(T--)
	{
		sum=0;
		scanf("%d",&n);
		int x;
		int flag=0;
		for(int i=1;i<=n;i++)
		{
			scanf("%d",&x);
			
			if(flag) continue;
			 
			int t= x%((n-1)*n);
			 
			if(t<n&&t>=1) 
			 	continue;
			flag=1;
		}
		if(!flag)
			printf("Bob\n");
		else
			printf("Alice\n");
	}
}