Acdream 1234 Two Cylinders（自适应辛普森积分法）

Two Cylinders
Special Judge Time Limit: 2000/1000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others)
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Problem Description

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.

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 10⑷.

Sample Input

1 1
Sample Output

5.3333
Source

Andrew Stankevich Contest 3
Manager

mathlover

r0r12x2r22x2dy

My Code:

/** 求两个交叉圆柱的体积 **/ #include <iostream> #include <cstdio> #include <algorithm> #include <cmath> const double eps = 1e⑼;///精度 using namespace std; double r1,r2; double f(double x)///积分函数 { return 8.0*(sqrt(r1*r1-x*x))*(sqrt(r2*r2-x*x)); } double simpson(double x,double y) { return (y-x)*(f(x)+f(y)+4*f((x+y)/2))/6.0; } double jifen(double x,double y)///积分区间[x,y] { double ans = simpson(x,y);///2分区间积分减小误差 double mid = (x+y)/2.0; double left = simpson(x,mid); double right = simpson(mid,y); if(fabs(ans-(left+right)) < eps) return ans; return jifen(x,mid)+jifen(mid,y); } int main() { while(~scanf("%lf%lf",&r1,&r2)) { if(r1<r2) swap(r1,r2); printf("%.6lf\n",jifen(0.0,r2)); } return 0; } ------分隔线----------------------------

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