自适应Simpson

Ellipse

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1268    Accepted Submission(s): 493

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 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).

2
2 1 -2 2
2 1 0 2

6.283
3.142

/* ***********************************************
Author        :CKboss
Created Time  :2014年12月27日 星期六 22时46分38秒
File Name     :HDOJ1724.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

double a,b,l,r;

double F(double x)
{
	return b*sqrt((1-x/a)*(1+x/a));
}

/// 3-points simpson


double simpson(double a,double b)
{
	double c=(a+b)/2;
	return (F(a)+4*F(c)+F(b))*(b-a)/6.;
}

double asr(double a,double b,double eps,double A)
{
	double c=(a+b)/2;
	double L=simpson(a,c),R=simpson(c,b);
	if(fabs(L+R-A)<=15*eps) return L+R+(L+R-A)/15.0;
	return asr(a,c,eps/2,L)+asr(c,b,eps/2,R);
}

double ASR(double a,double b,double eps)
{
	return asr(a,b,eps,simpson(a,b));
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);

	int T_T;
	scanf("%d",&T_T);
	while(T_T--)
	{
		scanf("%lf%lf%lf%lf",&a,&b,&l,&r);
		printf("%.3lf\n",2.*ASR(l,r,1e-6));
	}
    
    return 0;
}