acdream 1234 Two Cylinders
Special JudgeTime Limit: 10000/5000MS (Java/Others)Memory
Limit: 128000/64000KB (Java/Others)
SubmitStatisticNext
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
題解及代碼：
這道題的意思很簡單，就是求兩個垂直相交的圓柱的重合體積。推導的思想可以見：點擊打開鏈接
根據上面的方法我們可以推出截面公式是sqrt(R*R-x*x)*sqrt(r*r-x*x)，然後積分的上下限是0--r。
雖然這公式推出來了，但是積分很困難（沒推出來= =！），下面的代碼用的是辛普森積分法，學習了一下，挺簡單的。
主要就是這個：
vc/QobXEyrG68qOszvOy7rK7tPOjrLWrysfK/bHIvc+087XEyrG68qOszvOy7r7Nuty088HLo6zL+dLUvMbL47n9s8zW0NKqyrnTw7b+t9bAtLDRx/i85Lz10KGjrLz10KHO87LuoaM8L3A+CjxwPjxicj4KPC9wPgo8cD48cHJlIGNsYXNzPQ=="brush:java;">#include
#include
#include
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)
