CF_1C_AncientBerlandCircus

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

题目意思给一个正多边形的三个顶点组成的三角形

然后问这个正多边形的最小面积，即边数最少的正多边形

这个题目要求的有两个要点1多边形的外接圆半径

2多边形边数

因为多边形的外接圆与所给三角形外接圆一致

因此可以靠求面积（海伦公式，这里貌似用正弦定理过不了数据）然后正弦定理得到r

得到边数的方法比较奇特

首先要掌握一个新技术 小数的gcd

因为三角形的每边的圆心角一定是所对应正多边形每边圆心角的整数倍

因此这个gcd得出的就是多边形的圆心角

至于三角形边的圆心角由两个r和一边的三角形的余弦定理得出。

虽然结果保留6位

但是过程中gcd的精度却是十分敏感的，这里要注意

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;

const double PI=acos(-1.0);    //π的大小
const double eps=1e-2;         //允许误差
int dcmp(double x)              //误差修正
{
if(fabs(x)<eps)return 0;
if(x>0)        return 1;
return -1;
}
double dgcd(double a,double b)  //小数的不精确最大公约数
{
if(!dcmp(a))
return b;
if(!dcmp(b))
return a;
return dgcd(b,fmod(a,b));
//fmod返回x-n*y，符号同y。n=[x/y](向离开零的方向取整)
}
inline double sqr(double x)    //平方
{
return (x*x);
}

struct Point                                              //点类，向量类
{
double x,y;
Point(){}
Point(double a,double b):x(a),y(b){}
void input()
{
scanf("%lf%lf",&x,&y);
}
friend Point operator +(const Point &a,const Point &b)      //点之和
{
return Point(a.x+b.x,a.y+b.y);
}
friend Point operator -(const Point &a,const Point &b)      //点之差
{
return Point(a.x-b.x,a.y-b.y);
}
double norm()                                  //向量的模长
{
return sqrt(sqr(x)+sqr(y));
}
};
typedef Point Vector;                              //向量是点的别名

double dist(const Point &a,const Point &b)         //点a与点b间距离
{
return (a-b).norm();
}
double cos_a(double a,double b,double c)
{
return acos((sqr(b)+sqr(c)-sqr(a))/(2*b*c));
}

int main()
{
//freopen("1.in","r",stdin);
Point a,b,c;
a.input();b.input();c.input();
double AB=dist(a,b),AC=dist(a,c),BC=dist(b,c);
double p=(AB+AC+BC)/2;
double s=sqrt(p*(p-AB)*(p-AC)*(p-BC));  //三角形面积
double r=AB*AC*BC/4.0/s; //外接圆半径
double A=cos_a(BC,r,r),B=cos_a(AC,r,r);
double C=2*PI-A-B;
double O=dgcd(A,dgcd(B,C));
//cout<<A<<" "<<B<<" "<<C<<" "<<O<<endl;
printf("%.6lf\n",r*r*sin(O)*PI/O);
return 0;
}