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  2 * sqrt(sqr(fb) * (1 - sqr(x) / sqr(fa)));
}

double simpson(double a, double b) {
    double mid = a + (b - a) / 2;
    return (func(a) + 4 * func(mid) + func(b)) * (b - a) / 6;
}

double asr(double a, double b, double eps, double A) {
    double mid = a + (b - a) / 2;
    double l = simpson(a, mid), r = simpson(mid, b);
    if(fabs(l + r - A) <= 15 * eps) return l + r + (l + r - A) / 15;
    return asr(a, mid, eps / 2, l) + asr(mid, b, eps / 2, 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-5));
    }
}