题意：三角形ABC的内切圆把它的三边分别划分成 m1：n1，m2：n2 和 m3：n3 的比例。另外已知内切圆的半径 r ，求三角形ABC 的面积。

#include
#include
#include
#include
#define sqr(a) (a)*(a)
#define eps 1e-12
#define min(a,b) ab a:b
#define pi asin(1.0)

using namespace std;

int sig(double a)
{
    return (a>eps)-(a<-eps);
}

int main()
{
    int t;
    double r,m1,n1,m2,n2,m3,n3,k1,k2,k3;
    double left,right,mid,thy,th1,th2,th3;
    cin>>t;
    while(t--)
    {
        cin>>r>>m1>>n1>>m2>>n2>>m3>>n3;
        k1=sqr(n1);
        k2=sqr(n2/m2)*k1;
        k3=sqr(m1);
        left=min(sqrt(3/k1)*r,sqrt(3/k2)*r);
        left=min(left,sqrt(3/k3)*r);
        right=max(sqrt(3/k1)*r,sqrt(3/k2)*r);
        right=max(right,sqrt(3/k3)*r);
        mid=(left+right)/2;
        while(sig(right-left)>0)
        {
            th1=r/sqrt(k1*sqr(mid)+sqr(r));
            th2=r/sqrt(k2*sqr(mid)+sqr(r));
            th3=r/sqrt(k3*sqr(mid)+sqr(r));
            thy=asin(th1)+asin(th2)+asin(th3);
            int f=sig(thy-pi);
            if(f==0) break;
            else if(f<0) right=mid;
            else left=mid;
            mid=(left+right)/2;
        }
        thy=2*asin(r/sqrt(k1*sqr(mid)+sqr(r)));
        double area=(n1+m1)*mid/2*(n2+m2)*n1/m2*mid*sin(thy);
        cout<