HDU 1724 Ellipse（数值积分の辛普森公式）

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.

题目大意：给椭圆的a、b参数，求区间[l, r]的椭圆积分的和。

思路：直接套用辛普森公式，贴个模板。

代码（93MS）：

 #include <cmath>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <iostream>
 using namespace std;

 double fa, fb, fl, fr;
 int n;

 double sqr(double x) {
     return x * x;
 }

 double func(double x) {
     return   * sqrt(sqr(fb) * ( - sqr(x) / sqr(fa)));
 }

 double simpson(double a, double b) {
     double mid = a + (b - a) / ;
     return (func(a) +  * func(mid) + func(b)) * (b - a) / ;
 }

 double asr(double a, double b, double eps, double A) {
     double mid = a + (b - a) / ;
     double l = simpson(a, mid), r = simpson(mid, b);
     if(fabs(l + r - A) <=  * eps) return l + r + (l + r - A) / ;
     return asr(a, mid, eps / , l) + asr(mid, b, eps / , r);
 }

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

 int main() {
     scanf("%d", &n);
     while(n--) {
         scanf("%lf%lf%lf%lf", &fa, &fb, &fl, &fr);
         printf("%.3f\n", asr(fl, fr, 1e-));
     }
 }