HDU 1724 自适应Simpson积分 解题报告

Ellipse

Problem Description

Math is important!! Many students failed in 2+2’s mathematical test, so let’s AC this problem to mourn for our lost youth..
Look this sample picture:

A ellipses in the plane and center in point O. the L,R lines will be vertical through the X-axis. The problem is calculating the blue intersection area. But calculating the intersection area is dull, so I have turn to you, a talent of programmer. Your task is tell me the result of calculations.(defined PI=3.14159265 , The area of an ellipse A=PI*a*b )

Input

Input may contain multiple test cases. The first line is a positive integer N, denoting the number of test cases below. One case One line. The line will consist of a pair of integers a and b, denoting the ellipse equation , A pair of integers l and r, mean the L is (l, 0) and R is (r, 0). (-a <= l <= r <= a).

Output

For each case, output one line containing a float, the area of the intersection, accurate to three decimals after the decimal point.

Sample Input

2
2 1 -2 2
2 1 0 2

Sample Output

6.283
3.142

【解题报告】
板子题

代码如下：

#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespace std;#define eps 1e-8int t;  double a,b,l,r;double f(double x)  {      return b*sqrt(1.0-(x*x)/(a*a));  }  double calc(double len,double fL,double fM,double fR) {    return (fL+4*fM+fR)*len/6;}double Simpson(double L,double M,double R,double fL,double fM,double fR,double sqr) {    double M1=(L+M)/2,M2=(M+R)/2;    double fM1=f(M1),fM2=f(M2);    double g1=calc(M-L,fL,fM1,fM),g2=calc(R-M,fM,fM2,fR);    if(fabs(sqr-g1-g2)<=eps) return g1+g2;    return Simpson(L,M1,M,fL,fM1,fM,g1)+Simpson(M,M2,R,fM,fM2,fR,g2);}int main()  {        for(scanf("%d",&t);t;--t)      {          scanf("%lf%lf%lf%lf",&a,&b,&l,&r);        double m=(l+r)/2;        printf("%.3lf\n",2*Simpson(l,m,r,f(l),f(m),f(r),calc(r-l,f(l),f(m),f(r))));      }      return 0;  }