hdu1724 Ellipse（数值积分，自适应辛普森）

题意：

解法：

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$b^2{x}^2+a^2{y}^2=a^2{b}^2$

$y^2=\frac{a^2{b}^2-b^2{x}^2}{a^2}$

$y=\sqrt{\frac{a^2{b}^2-b^2{x}^2}{a^2}}$

$y=\frac{\sqrt{a^2{b}^2-b^2{x}^2}}{a}$

$y=\frac{b\ast \sqrt{a^2-x^2}}{a}$

$答案为y在[l,r]上的积分\ast 2,$

$乘2是因为有上下两部分.$

$用自适应辛普森来计算积分的值.$

code：

#include<bits/stdc++.h>
using namespace std;
const double eps=1e-6;
double a,b,l,r;
double f(double x){
    
    //被积函数求值
    return b*sqrt(a*a-x*x)/a;
}
double simpson(double l,double r){
    
    
    double mid=(l+r)/2;
    return (r-l)*(f(l)+4*f(mid)+f(r))/6;
}
double ask(double l,double r,double eps,double ans){
    
    
    double mid=(l+r)/2;
    double fl=simpson(l,mid),fr=simpson(mid,r);
    if(fabs(fl+fr-ans)<=15*eps)return fl+fr+(fl+fr-ans)/15;
    return ask(l,mid,eps/2,fl)+ask(mid,r,eps/2,fr);
}
signed main(){
    
    
    int T;scanf("%d",&T);
    while(T--){
    
    
        scanf("%lf%lf%lf%lf",&a,&b,&l,&r);
        double ans=2*ask(l,r,eps,simpson(l,r));
        printf("%.3f\n",ans);
    }
    return 0;
}

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