959A--Mahmoud and Ehab and the even-odd game

最新推荐文章于 2021-08-01 01:32:11 发布

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最新推荐文章于 2021-08-01 01:32:11 发布

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分类专栏： ACM

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本文介绍了一种基于博弈论的游戏，名为Even-Odd Game，由Mahmoud和Ehab两人参与。游戏规则简单，但胜负关键在于理解奇偶数的选择策略。文章详细解释了游戏流程及胜负条件，并提供了解题思路和AC代码实现。

题目

Mahmoud and Ehab play a game called the even-odd game. Ehab chooses his favorite integer n and then they take turns, starting from Mahmoud. In each player’s turn, he has to choose an integer a and subtract it from n such that:

1,1 ≤ a ≤ n.
2,If it’s Mahmoud’s turn, a has to be even, but if it’s Ehab’s turn, a has to be odd.

If the current player can’t choose any number satisfying the conditions, he loses. Can you determine the winner if they both play optimally?

Input

The only line contains an integer n (1 ≤ n ≤ 109), the number at the beginning of the game.

Output

Output “Mahmoud” (without quotes) if Mahmoud wins and “Ehab” (without quotes) otherwise.

Examples

input

output

Ehab

input

output

Mahmoud

Note

In the first sample, Mahmoud can’t choose any integer a initially because there is no positive even integer less than or equal to 1 so Ehab wins.

In the second sample, Mahmoud has to choose a = 2 and subtract it from n. It’s Ehab’s turn and n = 0. There is no positive odd integer less than or equal to 0 so Mahmoud wins.

题意：

Mahmoud 选择一个正整数n, Mahmoud和Ehab轮流选择一个数a（1 ≤ a ≤ n），由Ehab开始，Ehab每次必须选择一个偶数，Mahmoud每次选择一个奇数，每次选择后n=n-a；直到对方无法选择，则胜利。

解题思路：

巴什博奕思路，当n为偶数是Mahmoud胜利，当n为奇数时，Ehab胜利。

AC-Code

import java.util.Scanner;

public class ACM959A {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		if(n%2==1)
		{
			System.out.println("Ehab");
		}
		else
		{
			System.out.println("Mahmoud");
		}
		sc.close();

	}

}