zoukankan      html  css  js  c++  java
  • Toy Army

    Description

    The hero of our story, Valera, and his best friend Arcady are still in school, and therefore they spend all the free time playing turn-based strategy "GAGA: Go And Go Again". The gameplay is as follows.

    There are two armies on the playing field each of which consists of n men (n is always even). The current player specifies for each of her soldiers an enemy's soldier he will shoot (a target) and then all the player's soldiers shot simultaneously. This is a game world, and so each soldier shoots perfectly, that is he absolutely always hits the specified target. If an enemy soldier is hit, he will surely die. It may happen that several soldiers had been indicated the same target. Killed soldiers do not participate in the game anymore.

    The game "GAGA" consists of three steps: first Valera makes a move, then Arcady, then Valera again and the game ends.

    You are asked to calculate the maximum total number of soldiers that may be killed during the game.

    Input

    The input data consist of a single integer n (2 ≤ n ≤ 108, n is even). Please note that before the game starts there are 2n soldiers on the fields.

    Output

    Print a single number — a maximum total number of soldiers that could be killed in the course of the game in three turns.

    Sample Input

    Input
    2
    
    Output
    3
    
    Input
    4
    
    Output
    6
    

    Sample Output

    Hint

    The first sample test:

    1) Valera's soldiers 1 and 2 shoot at Arcady's soldier 1.

    2) Arcady's soldier 2 shoots at Valera's soldier 1.

    3) Valera's soldier 1 shoots at Arcady's soldier 2.

    There are 3 soldiers killed in total: Valera's soldier 1 and Arcady's soldiers 1 and 2.

    题意    有两对士兵  人数为偶数n    进行三次射击   一队先射击,二队再射击,一队再射击  每次可以一人射击一人也可以多人射击一人     问三次射击之后死亡的最多士兵人数 

    思路    为使伤亡人数最大其中一队要全部死亡,另一方尽可能有最大的伤亡数   最大伤亡数位n/2     最终两队伤亡数最大为n*3/2

    #include<cstdio>
    int main()
    {
    	int n;
    	scanf("%d",&n);
    	printf("%d",n*3/2);
    	return 0;
    	
    }


     

       

  • 相关阅读:
    C++头文件相互引用,最好一个#include,另一个class C;
    Git 安装配置
    loadrunner字符串赋值
    loadrunner 调用外部dll
    redis启动、清缓存命令
    solr-6.4.1 学习安装与配置 和 Elasticsearch(1.5.2)学习文档
    分页
    JS原型理解
    angular2 依赖注入新坑。
    javascript数组传值与地址。
  • 原文地址:https://www.cnblogs.com/kingjordan/p/12027236.html
Copyright © 2011-2022 走看看