C. Ancient Berland Circus
time limit per test
2 seconds
memory limit per test
64 megabytes
input
standard input
output
standard output

Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different.

In Ancient Berland arenas in circuses were shaped as a regular (equiangular) polygon, the size and the number of angles could vary from one circus to another. In each corner of the arena there was a special pillar, and the rope strung between the pillars marked the arena edges.

Recently the scientists from Berland have discovered the remains of the ancient circus arena. They found only three pillars, the others were destroyed by the time.

You are given the coordinates of these three pillars. Find out what is the smallest area that the arena could have.

Input

The input file consists of three lines, each of them contains a pair of numbers –– coordinates of the pillar. Any coordinate doesn't exceed 1000 by absolute value, and is given with at most six digits after decimal point.

Output

Output the smallest possible area of the ancient arena. This number should be accurate to at least 6 digits after the decimal point. It's guaranteed that the number of angles in the optimal polygon is not larger than 100.

Examples
Input 
0.000000 0.000000
1.000000 1.000000
0.000000 1.000000
Output 
1.00000000

【题意】

            给你一个正凸多边形的三个点，然后求出这个正凸多边形的面积的最小值。

【解题方法】

            简单计算几何，以这三个点做一个三角形，求出这个三角形的外心（外接圆的圆心），这个点也就是外接多边形的中心。然后找出内角所对应的边数的GCD。当然，内角所对应的边数不一定是整数，我们需要用double的GCD和LCM进行计算，求出最小边数。特别坑的是，要注意eps的设置。

【AC 代码】

#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-4;
const double PI = acos(-1.0);
struct point{
    double x,y;
    void read()
    {
        scanf("%lf%lf",&x,&y);
    }
}p[3];
double gcd(double x,double y)
{
    return y>eps?gcd(y,x-floor(x/y)*y):x;
}
double getdis(point a,point b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
double solve(double a,double b,double c)
{
    return acos((a*a+b*b-c*c)/(2.0*a*b));
}
int main()
{
    for(int i=0; i<3; i++) p[i].read();
    double a=getdis(p[0],p[1]);
    double b=getdis(p[0],p[2]);
    double c=getdis(p[1],p[2]);
    //printf("%.5f %.5f %.5f\n",a,b,c);
    double p=(a+b+c)/2.0;
    double s=sqrt(p*(p-a)*(p-b)*(p-c));
    double R=(a*b*c)/(4.0*s);
    double A=solve(b,c,a);
    double B=solve(a,c,b);
    double C=solve(a,b,c);
    double n=PI/gcd(A,gcd(B,C));
    printf("%.10f\n",R*R*sin(2*PI/n)*n/2.0);
    return 0;
}