A. Mahmoud and Ehab and the even-odd game

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本文介绍了一个简单的两玩家游戏——Even-Odd游戏，并通过分析游戏规则，给出了一种快速判断胜者的算法实现。游戏由两名玩家轮流进行，依据当前轮次玩家必须选择的数的奇偶性不同而有所区别。

A. Mahmoud and Ehab and the even-odd game
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
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 ≤ a ≤ n.
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
inputCopy
1
outputCopy
Ehab
inputCopy
2
outputCopy
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.

#include <stdio.h>
#include <stdlib.h>
int main()
{
    int n;
    scanf("%d",&n);
    if(n%2==0)
    {
        printf("Mahmoud\n");
    }
    else printf("Ehab\n");
    return 0;
}

代码很简单，就是思考如果是奇数的话，那么第一个人要选择一个偶数，选完之后，奇数减掉偶数还是奇数，那第二个人就会把这个数都选走，就剩下0了，那到第一个人选择的时候a得是大于1的，所以第一个人就输了；当是偶数的时候，第一个人把数直接拿走剩个0，第二个人就输了，所以，综上，就是奇数的时候第二个人赢，偶数的时候第一个人赢