【自适应Simpson积分】hdu 1724 Ellipse

Ellipse

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1013    Accepted Submission(s): 363

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

22 1 -2 22 1 0 2

 

Sample Output

6.2833.142

 

题目大意：

给定四个实数a,b,l,r，求椭圆x^2/a^2+y^2/b^2=1与半平面x≥l和x≤r相交部分的面积
做法：
这一道题需要使用自适应Simpson积分法来求解函数的定积分
我们发现椭圆上下对称，所以题目所求的面积就是函数f(x)=sqrt(b^2(1−x^2/a^2))在区间[l,r]上的积分的2倍

#include<bits/stdc++.h>#include <ctime>using namespace std;typedef long long ll;double a, b;double f(double x)           ///被积函数{    return b * sqrt(1.0 - x * x / (a * a));}double simpson(double a, double b){    double c = a + (b - a) / 2;    return (f(a) + 4 * f(c) + f(b)) * (b - a) / 6;}double asr(double a, double b, double epss, double A){    double c = a + (b - a) / 2;    double L = simpson(a, c) , R = simpson(c, b);    if (fabs(L + R - A) <= 15 * epss)    {        return L + R + (L + R - A) / 15;    }    return asr(a, c, epss / 2, L) + asr(c, b, epss / 2, R);}double solve(double l, double r, double epss){    return asr(l, r, epss, simpson(l, r));}int main(){    std::ios::sync_with_stdio(false);    int t;    cin >> t;    while (t--)    {        double l , r;        cin >> a >> b >> l >> r;        printf("%0.3f\n", solve(l, r, 1e-6));    }    return 0;}