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. Day05基本运算符,if判断和while循环

    day05 1.常量 变量名全大写 2.基本运算符 ①算术运算 10/3除法 10//3取整 10*3乘法 10**3幂 ②赋值运算 增量赋值 age += 1#age = age + 1 age * ...

  2. linux 下常见命令

    ===============安装和登陆命令============================================================= Mount: 挂载命令.把存储介 ...

  3. UVa 11987 并查集 Almost Union-Find

    原文戳这 与以往的并查集不同,这次需要一个删除操作.如果是叶子节点还好,直接修改父亲指针就好. 但是如果要是移动根节点,指向它的所有子节点也会跟着变化. 所以要增加一个永远不会被修改的虚拟根节点,这样 ...

  4. 15,re正则表达式

    判断手机号是否合法. phone_number = input('请输入手机号:') if re.match('^(13|14|15|18)[0-9]{9}$',phone_number): prin ...

  5. android 之 TabHost

    TabHost的实现有两种方式,第一种继承TabActivity,从TabActivity中用getTabHost()方法获取TabHost.各个Tab中的内容在布局文件中定义就行了. mainAct ...

  6. c#笔记2018-12-26

    using System; /*C#学习笔记2018-12-26 * 1.@逐字字符串 * 2.数据类型转换 * 3.变量声明和占位符使用 * 4.接收用户输入值 * 5.const 关键字 * 6. ...

  7. Sequence Models

    Sequence Models This is the fifth and final course of the deep learning specialization at Coursera w ...

  8. 【bzoj4200】[Noi2015]小园丁与老司机 STL-map+dp+有上下界最小流

    题目描述 小园丁 Mr. S 负责看管一片田野,田野可以看作一个二维平面.田野上有 nn 棵许愿树,编号 1,2,3,…,n1,2,3,…,n,每棵树可以看作平面上的一个点,其中第 ii 棵树 (1≤ ...

  9. 学习系列 - 马拉车&扩展KMP

    Manacher(马拉车)是一种求最长回文串的线性算法,复杂度O(n).网上对其介绍的资料已经挺多了的,请善用搜索引擎. 而扩展KMP说白了就是是求模式串和主串的每一个后缀的最长公共前缀[KMP更像是 ...

  10. SPOJ GSS4 Can you answer these queries IV ——树状数组 并查集

    [题目分析] 区间开方+区间求和. 由于区间开方次数较少,直接并查集维护下一个不是1的数的位置,然后暴力修改,树状数组求和即可. 这不是BZOJ上上帝造题7分钟嘛 [代码] #include < ...