codeforces Round #258(div2) A解题报告

A. Game With Sticks

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

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 nhorizontal 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.

Sample test(s)

input

2 2

output

Malvika

input

2 3

output

Malvika

input

3 3

output

Akshat

Note

Explanation of the first sample:

The grid has four intersection points, numbered from 1 to 4.



If Akshat chooses intersection point 1, then he will remove two sticks (1 - 2 and 1 - 3).
The resulting grid will look like this.



Now there is only one remaining intersection point (i.e. 4). Malvika must choose it and remove both remaining sticks. After her move the grid will be empty.

In the empty grid, Akshat cannot make any move, hence he will lose.

Since all 4 intersection points of the grid are equivalent, Akshat will lose no matter which one he picks.

题目大意：

两个人轮流移动，移动到某个点，就将该点的横竖两根棍子拿走，求最终胜利者是谁。

解法：

如若横竖两个都有棍子，则必然有点可以继续走。求出n和m的最小值，就知道有多少步可以走，然后就可以得知最终的胜利者是谁了。

代码：

#include <cstdio>

int n, m, min;

int main() {
scanf("%d%d", &n, &m);

if (n < m)
min = n;
else
min = m;

if (min%2)
printf("Akshat\n");
else
printf("Malvika\n");
}