清北学堂 站军姿

 

/*2bc*cosA=b^2+c^2-a^2 模拟计算 50分*/
#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
const double t=3.1415926535898;
int n;
double a,b,c,x,y,z,x2,y2,z2,s,k,m,w,p;
int main ()
{
    freopen ("standing.in","r",stdin);
    freopen ("standing.out","w",stdout);
    scanf ("%d",&n);
    for (int i=1;i<=n;i++)
    {
        scanf ("%lf%lf%lf%lf%lf%lf",&x,&y,&z,&x2,&y2,&z2);
        if (sqrt(abs(x-x2)*abs(x-x2)+abs(y-y2)*abs(y-y2))>z+z2)
            printf ("%.3lf\n",z*z*t+z2*z2*t);
        else if ((x+z>x2+z2&&x-z<x2-z2&&y+z>y2+z2&&y-z<y2-z2)||(x+z<x2+z2&&x-z>x2-z2&&y+z<y2+z2&&x-z>y2-z2))
            printf ("%.3lf\n",max(z*z*t,z2*z2*t));
        else 
        {
            s=z*z*t+z2*z2*t;
            k=sqrt(abs(x-x2)*abs(x-x2)+abs(y-y2)*abs(y-y2));
            a=(z*z-z2*z2+k*k)/(2*k);
            b=k-a;
            
            c=2*sqrt(z*z-a*a);
            w=(2*z*z-c*c)/(2*z*z);
            p=acos(w);
            m=(z+z+c)/2;
            s-=((p*z*z)/2-sqrt(m*(m-c)*(m-z)*(m-z)));
            
            c=2*sqrt(z2*z2-b*b);
            w=(2*z2*z2-c*c)/(2*z2*z2);
            p=acos(w);
            m=(z2+z2+c)/2;
            s-=((p*z2*z2)/2-sqrt(m*(m-c)*(m-z2)*(m-z2)));
            printf ("%.3lf\n",s);
        }
    }
    return 0;
}