解题报告

http://blog.csdn.net/juncoder/article/details/38102263

n和m跟木棍相交,问一人取一交点(必须是交点。且取完后去掉交点的两根木棍),最后谁赢

思路:

取最大正方形,以对角线上的交点个数推断输赢。

#include <iostream>
#include <cstdio> using namespace std; int main()
{
int m,n;
while(cin>>n>>m)
{
if(n>m)
{
if(m%2==0)
printf("Malvika\n");
else
{
printf("Akshat\n");
}
}
else
{
if(n%2==0)
printf("Malvika\n");
else
{
printf("Akshat\n");
}
}
}
return 0;
}
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.


Codeforces Round #258 (Div. 2/A)/Codeforces451A_Game With Sticks的更多相关文章

  1. Codeforces Round #258 (Div. 2) A. Game With Sticks 水题

    A. Game With Sticks 题目连接: http://codeforces.com/contest/451/problem/A Description After winning gold ...

  2. Codeforces Round #258 (Div. 2)[ABCD]

    Codeforces Round #258 (Div. 2)[ABCD] ACM 题目地址:Codeforces Round #258 (Div. 2) A - Game With Sticks 题意 ...

  3. Codeforces Round #258 (Div. 2) 小结

    A. Game With Sticks (451A) 水题一道,事实上无论你选取哪一个交叉点,结果都是行数列数都减一,那如今就是谁先减到行.列有一个为0,那么谁就赢了.因为Akshat先选,因此假设行 ...

  4. Codeforces Round #297 (Div. 2)C. Ilya and Sticks 贪心

    Codeforces Round #297 (Div. 2)C. Ilya and Sticks Time Limit: 2 Sec  Memory Limit: 256 MBSubmit: xxx  ...

  5. 贪心 Codeforces Round #297 (Div. 2) C. Ilya and Sticks

    题目传送门 /* 题意:给n个棍子,组成的矩形面积和最大,每根棍子可以-1 贪心:排序后,相邻的进行比较,若可以读入x[p++],然后两两相乘相加就可以了 */ #include <cstdio ...

  6. Codeforces Round #258 (Div. 2) B. Sort the Array

    题目链接:http://codeforces.com/contest/451/problem/B 思路:首先找下降段的个数,假设下降段是大于等于2的,那么就直接输出no,假设下降段的个数为1,那么就把 ...

  7. Codeforces Round #258 (Div. 2) E. Devu and Flowers 容斥

    E. Devu and Flowers 题目连接: http://codeforces.com/contest/451/problem/E Description Devu wants to deco ...

  8. Codeforces Round #258 (Div. 2) D. Count Good Substrings 水题

    D. Count Good Substrings 题目连接: http://codeforces.com/contest/451/problem/D Description We call a str ...

  9. Codeforces Round #258 (Div. 2) C. Predict Outcome of the Game 水题

    C. Predict Outcome of the Game 题目连接: http://codeforces.com/contest/451/problem/C Description There a ...

随机推荐

  1. c++ heap学习

    heap并不属于STL容器组件,它分为 max heap 和min heap,在缺省情况下,max-heap是优先队列(priority queue)的底层实现机制. 而这个实现机制中的max-hea ...

  2. Mac版有道云笔记不能自动同步

    删除本地资源文件夹 /Users/xxxx/Library/Containers/com.youdao.note.YoudaoNoteMac 直接删除整个文件夹,之后重新登录账号.

  3. 【转载】linux之sed用法

    linux之sed用法 原文地址:http://www.cnblogs.com/dong008259/archive/2011/12/07/2279897.html   sed是一个很好的文件处理工具 ...

  4. [uiautomator篇] 基类

      package com.softwinner.performance.benchmark; /** * UiAssistant public class * @author liuzhipeng ...

  5. 【Android】android:ellipsize的使用以及一个点解决方法

    EidtText和textview中内容过长的话自动换行,使用android:ellipsize与android:singleine可以解决,使只有一行. EditText不支持marquee 用法如 ...

  6. jQuery中文文档

    http://www.jquery123.com/ http://www.shifone.cc/

  7. [BZOJ1582] [Usaco2009 Hol]Holiday Painting 节日画画(线段树)

    传送门 线段树区间修改傻题 #include <cstdio> #include <cstring> #include <iostream> #define N 5 ...

  8. 【容斥】HDU 4135 Co-prime

    acm.hdu.edu.cn/showproblem.php?pid=4135 [题意] 询问[a,b]中与n互质的数有多少个 [思路] 考虑[1,m]中与n互质的数有多少个,答案就是query(b) ...

  9. JSP表单提交中文乱码

    简要笔记:由于jsp默认表单提交编码方式是:ISO-8859-1,而我们需要的是utf-8或者是gbk码,故需要转化. 具体方法是:在表单处理文件中,将获取到的变量进行转换. String userN ...

  10. Python入门--3--操作符

    一.算数操作符 有:+.-.*././/.%.**(幂) a= 3; a = 3+1; #等同于a += 1  这相当与a加一 同样 也可以-.*././/          需要注意的是//是直接舍 ...