hdu 3668 Volume 求两个相交的圆柱体的总体积

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

 

 

Recommend

lcy

#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;
}