CF 1C

思路（根据题目特殊条件切入）：

1、看到边数<=100 ——> 可以枚举

2、看到正多边形 ——>研究正多边形顶点之间的角度（X，无明显规律）正多边形在圆上（√）——>三个点在圆上

于是就可以通过找圆心、算夹角、枚举边数、检验是否为最小角度倍数，来找到最小圆，从而得到面积

注意：

1、特判斜率

2、误差范围不能太大，玄学

#include<bits/stdc++.h>
using namespace std;
#define ll long long
ll n,m,a,ans,ans2;
double iptf1x,iptf2x,iptf3x,iptf1y,iptf2y,iptf3y;
int i,j,k;

bool equ(double a,double b)
{
//printf("%.3f %.3f %.3f\n",a,b,fabs(a-b));
   if(fabs(a-b)<0.0001)return true;
   return false;
}
int main()
{
   scanf("%lf%lf%lf%lf%lf%lf",&iptf1x,&iptf1y,&iptf2x,&iptf2y,&iptf3x,&iptf3y);

double mid12x=(iptf1x+iptf2x)/2,mid12y=(iptf1y+iptf2y)/2;
double K12;
if(equ(iptf1y-iptf2y,0))K12=999999999;
else K12=-(iptf1x-iptf2x)/(iptf1y-iptf2y);

double B12=mid12y-K12*mid12x;
//特判水平竖直
double mid23x=(iptf2x+iptf3x)/2,mid23y=(iptf2y+iptf3y)/2;
double K23;
if(equ(iptf2y-iptf3y,0))K23=999999999;
else K23=-(iptf2x-iptf3x)/(iptf2y-iptf3y);

double B23=mid23y-K23*mid23x;
double X=(B23-B12)/(K12-K23);
double Y=X*K12+B12;
//换系
iptf1x=iptf1x-X;
iptf1y=iptf1y-Y;
iptf2x=iptf2x-X;
iptf2y=iptf2y-Y;
iptf3x=iptf3x-X;
iptf3y=iptf3y-Y;

double r2=fabs(iptf1x*iptf1x+iptf1y*iptf1y);
double cos12=(iptf1x*iptf2x+iptf1y*iptf2y)/r2;
double ang12=acos(cos12);
double cos23=(iptf2x*iptf3x+iptf2y*iptf3y)/r2;
double ang23=acos(cos23);
double cos13=(iptf1x*iptf3x+iptf1y*iptf3y)/r2;
double ang13=acos(cos13);
//printf("%.3lf %.3lf %.3lf %.3lf %.3lf \n",X,Y,ang12/3.1415926*180,ang23/3.1415926*180,ang13/3.1415926*180);

//枚举边数
for(i=3;i<=100;i++)
{ //
//判断角大小
double bz=3.1415926535793*2/i;//printf("%d %.9lf\n",i,ang12- (bz* (double)((int)(ang12/bz))));

if(equ(ang12- (bz* (double)((int)(ang12/bz+0.00001))),0))
if(equ(ang23- (bz* (double)((int)(ang23/bz+0.00001))),0))
if(equ(ang13- (bz* (double)((int)(ang13/bz+0.00001))),0))
{
   printf("%.6lf",(double)i*fabs(sin(bz)*r2)/2);
   return 0;
}

}
}

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