CodeForces 1C Ancient Berland Circus

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<stack>
#include<string>
#include<map>
#include<iostream>
using namespace std;
#define eps 1e-2
#define pi acos(-1.0)
int n,m;
double s,r,S,p;
double x[4],y[4],ang[4],L[5];
double dis(double x1,double y1,double x2,double y2){
return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}
double gcd(double a,double b)
{
if(a<eps) return b;
if(b<eps) return a;
return gcd(b,fmod(a,b));
}
int main()
{
int i,j,k;
for(i=0;i<3;i++){
scanf("%lf%lf",&x[i],&y[i]);
}
for(i=0;i<3;i++){
L[i]=dis(x[i],y[i],x[(i+1)%3],y[(i+1)%3]); //求距离
}
p=(L[0]+L[1]+L[2])/2;
s=sqrt(p*(p-L[0])*(p-L[1])*(p-L[2])); //海伦凯勒公式求面积
r=L[0]*L[1]*L[2]/(4*s); //外接圆半径
for(i=0;i<3;i++) ang[i]=acos(1-L[i]*L[i]/(2*r*r));
ang[2]=2*pi-ang[1]-ang[0]; //三个圆心角，余弦定理
double unit=0;
for(i=0;i<3;i++) unit=gcd(unit,ang[i]); //圆心角的最大公约数
printf("%.6f\n",pi*r*r*sin(unit)/unit); //2*pi/unit为个数，S为每个三角形面积
return 0;
}