HDOJ 1724 Ellipse（自适应simpson积分）

最新推荐文章于 2019-01-27 14:31:43 发布

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本文介绍了一种使用自适应辛普森积分方法快速计算椭圆与直线交集面积的算法实现。通过输入椭圆参数及直线坐标，利用C++编程语言实现了精确到小数点后三位的面积计算。

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Ellipse

TimeLimit: 1000/1000 MS (Java/Others) Memory Limit:32768/32768 K (Java/Others)
Total Submission(s): 2025 Accepted Submission(s): 869

Problem Description

Math isimportant!! Many students failed in 2+2’s mathematical test, so let's AC thisproblem 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 verticalthrough the X-axis. The problem is calculating the blue intersection area. Butcalculating the intersection area is dull, so I have turn to you, a talent ofprogrammer. Your task is tell me the result of calculations.(definedPI=3.14159265 , The area of an ellipse A=PI*a*b )

Input

Inputmay 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 willconsist 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

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

Sample Input

2 1 -2 2

2 1 0 2

Sample Output

6.283

3.142

此题用自适应simpson积分能很快解决

自适应simpson积分：给定一个函数，以及左右区间，求出该函数在这个区间中的图像与x轴围成的面积

#include <bits/stdc++.h>
using namespace std;
#define mst(a,b) memset((a),(b),sizeof(a))
#define f(i,a,b) for(int i=(a);i<(b);++i)
typedef long long ll;
const int maxn= 105;
const int mod = 475;
const ll INF = 0x3f3f3f3f;
const double eps = 1e-6;
#define rush() int T;scanf("%d",&T);while(T--)

double a,b,l,r;

double F(double x)
{
    return sqrt(b*b*(1-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 eps ,double A)
{
    double c = a+ (b-a)/2;
    double L = simpson(a,c) ,R = simpson(c,b);
    if(fabs(L+R-A)<=15*eps) return L+R+(L+R-A)/15;
    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()
{
    rush()
    {
        scanf("%lf%lf%lf%lf",&a,&b,&l,&r);
        double sum=asr(l,r,eps);
        sum*=2;
        printf("%.3lf\n",sum);
    }
    return 0;
}