[nowcoder5667K]Keyboard Free

不妨设$r1le r2le r3$，令$f(alpha)=E(S_{Delta}ABC)$，其中AB坐标分别为$(r_{1},0)$和$(r_{2}cos alpha,r_{2}sin alpha)$，C在原点为圆心、$r_{3}$为半径的圆上，那么有答案$ans=lim_{n oinfty}limitsfrac{sum_{i=1}^{n}f(frac{2pi i}{n})}{n}$，而由于答案误差可以较大，因此n取1000左右即可

考虑求$f(alpha)$，可以求出$l=|AB|=sqrt{(r_{2}cosalpha-r_{1})^{2}+(r_{2}sin alpha)^{2}}$，作$CHperp AB$交AB于点H，那么有$f(alpha)=frac{lcdot E(CH)}{2}$

考虑求$E(CH)$，延长AB交最大的圆于点D，作$OEverb|//|AB$交最大的圆于点E交CH延长线于点H'，那么根据这些可以求出$eta=angle DOE=angle ADO=arcsin frac{r_{1}r_{2}sinalpha}{r_{3}l}$，$h=HH'=r_{3}sineta$，然后对$gamma=angle COA-angle DOA$分类讨论：

1.$etale gamma< pi+eta$，那么$E(CH)=egin{equation*}int_{0}^{pi} h+r_{3}sin(gamma-eta)  m d(gamma-eta)end{equation*}=pi h+2r_{3}$

2.$pi+2eta le gamma<2pi$，那么$E(CH)=egin{equation*}int_{eta}^{pi-eta} (r_{3}sin (gamma-eta-pi)-h)  m d(gamma-eta-pi)end{equation*}=2r_{3}coseta-(pi-2eta)h$

3.$0le gamma<eta$或$pi+etale gamma<pi+2eta$，那么$E(CH)=2egin{equation*}int_{0}^{eta} (h-r_{3}sin (eta-gamma))  m d(eta-gamma)end{equation*}=2eta h+2r_{3}(coseta-1)$

综上，$E(CH)=frac{(pi h+2r_{3})+(2r_{3}coseta-(pi-2eta )h)+(2eta h+2r_{3}(coseta-1))}{2pi}=frac{2eta h+2r_{3}coseta}{pi}$，代入即可得到$f(alpha)=frac{l(eta h+r_{3}coseta)}{pi}$

#include<bits/stdc++.h>
using namespace std;
#define pi acos(-1.0) 
int t,r1,r2,r3;
double sqr(double a){
    return a*a;
}
double f(double a){
    double l=sqrt(sqr(r2*cos(a)-r1)+sqr(r2*sin(a)));
    double h=r1*r2*sin(a)/l;
    double b=asin(h/r3);
    return l*(b*h+cos(b)*r3)/pi;
}
int main(){
    scanf("%d",&t);
    while (t--){
        scanf("%d%d%d",&r1,&r2,&r3);
        if (r1>r2)swap(r1,r2);
        if (r1>r3)swap(r1,r3);
        if (r2>r3)swap(r2,r3);
        double ans=0;
        for(int i=1;i<=1000;i++)ans+=f(2*i*pi/1000.0)/1000.0;
        printf("%.1f
",ans);
    }
}