自适应辛普森积分：

最新推荐文章于 2022-10-16 16:44:23 发布

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最新推荐文章于 2022-10-16 16:44:23 发布

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分类专栏：数论基本算法

本文链接：https://blog.youkuaiyun.com/zhaoxinxin1234/article/details/100543640

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该博客介绍了如何使用自适应辛普森积分法解决数学问题，提供了三个具体的例子，分别涉及椭圆区域的面积计算和两个不同类型的积分计算。通过示例详细解释了积分的计算过程，并强调在特定条件下积分的收敛性和发散性的判断。

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例①：Ellipse HDU - 1724：

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

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=PIab )
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 , 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

将 x 轴上方的 y 化为 x 的函数，积分出来之后乘以2即可。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<set>
#include<deque>
#include<map>
#include<vector>
#include<cmath>
#define ll long long
#define llu unsigned ll
using namespace std;
const double eps = 1e-5;
const ll lnf=0x3f3f3f3f3f3f3f3f;
const int inf = 0x3f3f3f3f;
const int maxn=100010;
const int mod=1e9+7;
double a,b,l,r;

double f(double x)
{
   
    return b*sqrt(1.0-(x*x)/(a*a));
}

double s(double a,double b)
{
   
    double c=(a+b)/2;
    return (b-a)*(f(a)+f(b)+f(c)

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