zoukankan      html  css  js  c++  java
  • Codeforces Round #199 (Div. 2) C. Cupboard and Balloons

    C. Cupboard and Balloons
    time limit per test
    2 seconds
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    A girl named Xenia has a cupboard that looks like an arc from ahead. The arc is made of a semicircle with radius r (the cupboard's top) and two walls of height h (the cupboard's sides). The cupboard's depth is r, that is, it looks like a rectangle with base r and height h + rfrom the sides. The figure below shows what the cupboard looks like (the front view is on the left, the side view is on the right).

     

     

    Xenia got lots of balloons for her birthday. The girl hates the mess, so she wants to store the balloons in the cupboard. Luckily, each balloon is a sphere with radius . Help Xenia calculate the maximum number of balloons she can put in her cupboard.

    You can say that a balloon is in the cupboard if you can't see any part of the balloon on the left or right view. The balloons in the cupboard can touch each other. It is not allowed to squeeze the balloons or deform them in any way. You can assume that the cupboard's walls are negligibly thin.

    Input

    The single line contains two integers r, h (1 ≤ r, h ≤ 107).

    Output

    Print a single integer — the maximum number of balloons Xenia can put in the cupboard.

    Sample test(s)
    input
    1 1
    
    output
    3
    
    input
    1 2
    
    output
    5
    
    input
    2 1
    
    output
    2
    几何题,有点坑,有个地方没想清楚,一直wrong,我们可以先把长方形的那部分尽可能的多放,然后,我们可以,找到剩下的空间,刚好放两个球,和刚好放三个球的情况,这样,我们就可以得出公式了!
    #include <iostream>
    #include <stdio.h>
    #include <string.h>
    #include <math.h>
    using namespace std;
    
    int main()
    {
        double h,r;double temp=sqrt(3.0)/2.0;
        while(scanf("%lf%lf",&r,&h)!=EOF){
            int ans=2*(int)(h/r);
            double t=h-ans/2.0*1.0*r;
            //printf("%d %.6f %.6f
    ",ans,t,temp*r);
            if(t<r/2.0)ans++;
            else if(t<temp*r)ans+=2;
            else ans+=3;
            printf("%d
    ",ans);
        }
        return 0;
    }
    


  • 相关阅读:
    lucene建立索引时候的用到的一些文档和目录操作
    MYSQL Cast函数 类型转换
    windows和linux下执行java程序+获取本机IP
    apache lucene 的核心类
    apache lucene 一个最简单的实例
    lucene 索引非txt文档 (pdf word rtf html xml)
    MyISAM & InnoDB
    软件开发者面试百问
    比较lucene各种英文分析器Analyzer
    浅谈Base64编码
  • 原文地址:https://www.cnblogs.com/riskyer/p/3318076.html
Copyright © 2011-2022 走看看