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Acdream 1234 Two Cylinders(自适应辛普森积分法)

来源:程序员人生   发布时间:2016-08-17 10:32:45 阅读次数:1391次

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Two Cylinders
Special Judge Time Limit: 2000/1000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others)
Submit Statistic Next 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 10⑷.

Sample Input

1 1
Sample Output

5.3333
Source

Andrew Stankevich Contest 3
Manager

mathlover

题目大意:
给了两个圆柱的半径 r1,r2,他们的轴线垂直相交,让你求的是相交部份的体积。

解题思路:
这个题目首先我们对其积分:得到1个公式:

r0r12x2r22x2dy

我们现在用正常的积分公式积不出来,我们就能够采取辛普森积分法进行积分,可以当作模板用
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; }

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