HDU 1724 ——————自适应辛普森法

桃扇骨 2022-05-12 14:34 158阅读 0赞

# Ellipse #

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)  
Total Submission(s): 2916 Accepted Submission(s): 1334

### 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:

![在这里插入图片描述][C61-1005-1.JPG]

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![在这里插入图片描述][C61-1005-2.JPG] , 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 ###

2  
2 1 -2 2  
2 1 0 2

### Sample Output ###

6.283  
3.142

--------------------

#include<bits/stdc++.h>
    using namespace std;
    const double PI = 3.1415926;
    int a,b,l,r;
    
    double F(double x)
    {
        
            return sqrt(b*b*(a*a-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)/15.0;
            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()
    {
        
            int t;
            scanf("%d",&t);
            while(t--)
            {
        
                    scanf("%d %d %d %d",&a,&b,&l,&r);
                    printf("%.3lf\n",asr(l,r,(1e-6))*2);
            }
            return 0;
    }

--------------------

以下转自 [数值问题专题小结：自适应辛普森算法求定积分][Link 1]：

## 三点辛普森公式 ##

![Center][]

该公式要求f(x)必须是一个全局函数，用它可以近似的来求解一个定积分，但精度不够高。因此衍生出一个重要的“变种”，称为“自适应辛普森法”。

## 自适应辛普森法 ##

（1）概述：自适应辛普森法(Adaptive Simpson's Rule)是一种数值积分方法，适用于无法求出原函数时的定积分。比直接用辛普森公式的精度更高，而且效率也可观。

（2）原理：该算法还是基于三点辛普森公式进行计算，不过需要设置一个精度eps，然后可以根据情况递归的划分区间：容易近似的地方少划分，不容易近似的地方多划分。近似程度利用如下公式来判断：

![Center 1][]

其中的三个S值是在对应的区间中利用“三点辛普森”公式计算出来的值。c是区间\[a,b\]的中点，ε就是上述的eps。如果满足该不等式，就直接返回结果，这里的结果指的是S(a,c)+S(c,b)+ΔS（ΔS就是上述不等式中小于号之前的部分），否则递归调用，即再次划分区间。递归调用时精度也要相应地减小一半。

double F(double x)
    {
        
    	//Simpson公式用到的函数
    }
    double simpson(double a, double b)//三点Simpson法，这里要求F是一个全局函数
    {
        
    	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)//自适应Simpson公式（递归过程）。已知整个区间[a,b]上的三点Simpson值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) / 15.0;
    	return asr(a, c, eps / 2, L) + asr(c, b, eps / 2, R);
    }
    double asr(double a, double b, double eps)//自适应Simpson公式（主过程）
    {
        
    	return asr(a, b, eps, simpson(a, b));
    }

[C61-1005-1.JPG]: /images/20220504/fd3786d399624a30a32622d566e60e3c.png
[C61-1005-2.JPG]: /images/20220504/c818ca43d8b34fc69e60e1ae226cab91.png
[Link 1]: https://blog.csdn.net/u014800748/article/details/44343955
[Center]: /images/20220504/71d118670d3b44e49a0be2b01456c116.png
[Center 1]: /images/20220504/1b252768eb6b44eab48d111049de8ed8.png