acdream 1234 Two Cylinders

Two Cylinders

Special JudgeTime Limit: 10000/5000MS (Java/Others)Memory
Limit: 128000/64000KB (Java/Others)

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Problem

Problem Description

In this problem your task is very simple.

Consider two infinite cylinders in three-dimensional space, of radii R1 and R2 respectively,
located in such a way that their axes intersect and are perpendicular.

Your task is to find the volume of their intersection.

Input

Input file contains two real numbers R1 and R2 (1
<= R1,R2 <= 100).

Output

Output the volume of the intersection of the cylinders. Your answer must be accurate up to 10-4.

Sample Input

1 1

Sample Output

5.3333

Source

Andrew Stankevich Contest 3

Manager

mathlover

题解及代码：

这道题的意思很简单，就是求两个垂直相交的圆柱的重合体积。推导的思想可以见：点击打开链接

根据上面的方法我们可以推出截面公式是sqrt(R*R-x*x)*sqrt(r*r-x*x)，然后积分的上下限是0--r。

虽然这公式推出来了，但是积分很困难（没推出来= =！），下面的代码用的是辛普森积分法，学习了一下，挺简单的。

主要就是这个：

这个公式在数比较小的时候，误差不大，但是数比较大的时候，误差就很大了，所以计算过程中要使用二分来把区间减小，减小误差。

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
const double eps=1e-7;

double R,r;

double f(double n)
{
    return 8*sqrt(R*R-n*n)*sqrt(r*r-n*n);
}

double simpson(double a,double b)
{
    return (b-a)/6.0*(f(a)+4*f((a+b)/2.0)+f(b));
}

double cal(double a,double b)
{
    double sum=simpson(a,b),mid=(a+b)/2.0;
    double t=simpson(a,mid)+simpson(mid,b);

    if(fabs(t-sum)<eps) return sum;
    return cal(a,mid)+cal(mid,b);
}
int main()
{
    scanf("%lf%lf",&R,&r);
    if(R<r) swap(R,r);

    printf("%.5lf\n",cal(0,r));
    return 0;
}