ACM HDU 3668Volume

Volume

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 531    Accepted Submission(s): 159

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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数学题，积分求体积

#include<stdio.h>
#include<iostream>
#include<cmath>
using namespace std;

#define PI acos(-1.0)

int main()
{
    double r,h;
    double v;
    while(cin>>r>>h)
    {
        if(h>=2*r)
        v=2*PI*r*r*h-16*r*r*r/3;
        else
        v=2*PI*r*r*h-8*r*r*(r-sqrt(r*r-h*h/4))+8*(r*r*r-sqrt((r*r-h*h/4)*(r*r-h*h/4)*(r*r-h*h/4)))/3-8*h*h*sqrt(r*r-h*h/4)/4;
        printf("%.4lf\n",v);
    }
    return 0;    
}