Two Cylinders （辛普森公式处理积分）

Two Cylinders （辛普森公式处理积分）：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=120113#problem/I   传送门：nefu

题面描述:

I - Two Cylinders

Time Limit:1000MS     Memory Limit:64000KB     64bit IO Format:%lld & %llu

Submit Status

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 R ₁ and R ₂ (1 <= R ₁,R ₂ <= 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

题目大意：

此题要计算两个圆柱体垂直相交之后形成的体积，且两个圆柱体的半径不一样。

题目分析：

可以先依据两个半径相等的圆柱体相交所得物体（称为：牟合方盖）的体积计算方法来推导半径不相等的物体的体积，计算牟合方盖的体积时，我们都是利用定积分，先计算在第一卦限中的体积，最后乘以8就行了，具体过程如下。

求解两柱体x2+y2=R2和x2+z2=R2相交部分的体积。

而此题所要求的是柱体x2+y2=R22和x2+z2=R12相交的体积。

我们同样也是只算第一卦限的体积，最后乘以8即可。

代码实现如下：

#include <iostream>#include <cstdio>#include <cmath>using namespace std;const double esp=1e-10;double r1,r2;double f(double x){    return sqrt((r1*r1-x*x)*(r2*r2-x*x));}double Simpson(double a,double b)  {    return (b-a)*(f(a)+f(b)+4*f((a+b)/2.0))/6.0;}double Midd(double a,double b) ///迭代求解数值积分{    double mid=(a+b)/2.0;    double res=Simpson(a,b);    if(fabs(res-Simpson(a,mid)-Simpson(mid,b))<esp)    {        return res;    }    else return Midd(a,mid)+Midd(mid,b);}int main(){    while(scanf("%lf%lf",&r1,&r2)!=EOF)    {        if(r1==r2)            printf("%.4lf\n",16.0/3*r1*r1*r1);        else        {            if(r1>r2)            {                double t=r1;                r1=r2;                r2=t;            }            printf("%.4lf\n",8.0*Midd(0,r1));        }    }    return 0;}