GZHU18级寒假训练：Gemini's Trial C

最新推荐文章于 2020-11-25 19:40:19 发布

原创最新推荐文章于 2020-11-25 19:40:19 发布 · 162 阅读

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本文介绍了一个基于网格的游戏，其中两名玩家轮流移除交点处的所有棍子，直到无法再进行移动。通过分析游戏规则和玩家策略，我们提供了一个简单的算法来预测最终赢家。输入为网格的水平和垂直棍子数量，输出则是预测的胜者名称。

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After winning gold and silver in IOI 2014, Akshat and Malvika want to have some fun. Now they are playing a game on a grid made of n horizontal and m vertical sticks.
An intersection point is any point on the grid which is formed by the intersection of one horizontal stick and one vertical stick.
In the grid shown below, n = 3 and m = 3. There are n + m = 6 sticks in total (horizontal sticks are shown in red and vertical sticks are shown in green). There are n·m = 9 intersection points, numbered from 1 to 9.
The rules of the game are very simple. The players move in turns. Akshat won gold, so he makes the first move. During his/her move, a player must choose any remaining intersection point and remove from the grid all sticks which pass through this point. A player will lose the game if he/she cannot make a move (i.e. there are no intersection points remaining on the grid at his/her move).
Assume that both players play optimally. Who will win the game?
Input
The first line of input contains two space-separated integers, n and m (1 ≤ n, m ≤ 100).
Output
Print a single line containing “Akshat” or “Malvika” (without the quotes), depending on the winner of the game.
Examples
Input
2 2
Output
Malvika
Input
2 3
Output
Malvika
Input
3 3
Output
Akshat

#include <iostream>
using namespace std;
int main()
{   
	int n,m;
	while(cin>>n>>m)    
	{        
		cout<<(((n<m?n:m)%2==0)?"Malvika":"Akshat")<<endl;    
	}    
	return 0;
}