hdu 3668 Volume 求两个相交的圆柱体的总体积

最新推荐文章于 2025-08-09 13:45:03 发布

kongming_acm

最新推荐文章于 2025-08-09 13:45:03 发布

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分类专栏： acm_数学问题文章标签： orthogonal output less numbers each object

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本文介绍了一种特殊的几何体——两个正交圆柱体相交的体积计算方法。该几何体由两个圆柱体组成，它们在中间位置相交。文章提供了一个C++程序示例，用于计算不同半径和高度的此类几何体体积。

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Volume

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 214 Accepted Submission(s): 68

Problem Description

This time your job is to calculate the volume of a special object. The object consists of two orthogonal cylinders. The two cylinders intersect each other in the middle place. One example is shown in Fig. 1. The radiuses of the bottom disk of both cylinders are R, and the heights of both cylinders are H.

Input

We test the problem in many cases. Each case includes two integers, the first one is R and the second one is H. All the numbers given are positive integers and are less than 100.

Output

The output consists of the volumes. The results must be round to 4 decimal numbers. Remember that R may be less than half of H.

Sample Input

  
   10 30
10 40

Sample Output

  
   13516.2226
19799.4079

Source

2010 Asia Regional Harbin

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#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
const double pi=acos(-1.0);
int main()
{
    int r,h;
    while(scanf("%d%d",&r,&h)==2)
    {
        double v=pi*r*r*h;
        double t;
        if(r*2<h) t=16.0/3*r*r*r;
        else
        {
            double k=1.0*h/2;
            double x0=sqrt(0.0+r*r-k*k);
            double s1,s2;
           // cout<<"r="<<r<<"...."<<"k="<<k<<"..."<<"x0="<<x0<<endl;
            s1=(r*r*r-r*r*x0-1.0/3*r*r*r+1.0/3*x0*x0*x0);
            //s2=k*(r*r*1.0/2*asin(x0*1.0/r)+1.0/2*x0*sqrt(r*r-x0*x0+0.0));
            s2=k*k*x0;
            t=(s1+s2)*8;
            //cout<<"s1="<<s1<<"..."<<"s2="<<s2<<endl;
        }
        double cnt=2*v-t;
        //cout<<"v="<<v<<"...."<<2*v<<"..."<<"t="<<t<<endl;
        printf("%.4lf/n",cnt);
    }
    return 0;
}