自适应辛普森法通过(Simpson)公式，用二次函数来拟合，实现时用二分递归来自动控制区间分割的大小，既保证精度，又保证速度
(Simpson)公式推导
[int_a^bf(x)dx
]
[approxint_a^bAx^2+Bx+C
]
[=frac{A}{3}(b^3-a^3)+frac{B}{2}(b^2-a^2)+C(b-a)
]
[=frac{(b-a)}{6}(2Ab^2+2Aab+2Aa^2+3Bb+3Ba+6C)
]
[=frac{(b-a)}{6}(Aa^2+Ba+C+Ab^2+Bb+C+4A(frac{a+b}{2})^2+4B(frac{a+b}{2})+4C)
]
[=frac{(b-a)}{6}(f(a)+f(b)+4f(frac{a+b}{2}))
]
求(displaystyle{int_l^rfrac{cx+d}{ax+b}mathrm{d}x})
(code:)
double f(double x)
{
return (c*x+d)/(a*x+b);
}
double simpson(double l,double r)
{
double mid=(l+r)/2;
return (r-l)*(f(l)+f(r)+4*f(mid))/6;
}
double solve(double l,double r,double eps)
{
double mid=(l+r)/2;
double a=simpson(l,mid),b=simpson(mid,r),m=simpson(l,r);
if(fabs(a+b-m)<=15*eps) return a+b+(a+b-m)/15;
return solve(l,mid,eps/2)+solve(mid,r,eps/2);
}
月下柠檬树：求一个图形的面积
(code:)
double f(double x)
{
double y=0;
for(int i=1;i<n;++i)
if(x>=h[i]-r[i]&&x<=h[i]+r[i])
y=max(y,sqrt(r[i]*r[i]-(h[i]-x)*(h[i]-x)));
for(int i=1;i<n;++i)
if(x>=p[i].l&&x<=p[i].r)
y=max(y,p[i].k*x+p[i].b);
return y;
}
double simpson(double l,double r)
{
double mid=(l+r)/2;
return (r-l)*(f(l)+f(r)+4*f(mid))/6;
}
double solve(double l,double r,double eps)
{
double mid=(l+r)/2;
double a=simpson(l,mid),b=simpson(mid,r),m=simpson(l,r);
if(fabs(a+b-m)<=15*eps) return a+b+(a+b-m)/15;
return solve(l,mid,eps/2)+solve(mid,r,eps/2);
}