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
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：
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)