Ancient Berland Circus CodeForces - 1C

题意：给定一个正多边形的三个顶点，求这个正多边形的最小面积。

思路：首先，边数越小面积越小，所以只要确定出包含这三个顶点的边数最小的正多边形即可。这个三角形和正多边形外接同一个圆。所以先求出外接圆的半径，再求出三个圆心角，易得这个多边形的边所对应的圆心角可被这三个圆心角整除，所以三个圆心角的gcd就是多边形边所对的圆心角，然后2π除一下就得到是几边形，之后就可计算面积了

海伦公式: p=(a+b+c)/2，S=√p(p-a)(p-b)(p-c)（a,b,c为三角形的三边，S为三角形面积） 

求外接圆半径r=a*b*c/4S 

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<stack>
#include<cstdlib>
#include<queue>
#include<set>
#include<string.h>
#include<vector>
#include<deque>
#include<map>
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-4
#define bug printf("*********\n")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
typedef long long LL;
typedef long long ll;
const int MAXN = 1e6 + 5;
const int mod = 998244353;

struct node{
    double x,y;
};
double len(node a,node b) {
    double tmp = sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
    return tmp;
}
double gcd(double x,double y) {
    while(fabs(x) > eps && fabs(y) > eps) {
        if(x > y)
            x -= floor(x / y) * y;
        else
            y -= floor(y / x) * x;
    }
    return x + y;
}
int main()
{
    node a,b,c;
    cin >> a.x >> a.y >> b.x >> b.y >> c.x >> c.y;
    double lena = len(a,b);
    double lenb = len(b,c);
    double lenc = len(a,c);

    double p = (lena + lenb + lenc) / 2.0;
    double S = sqrt(p * (p - lena) * (p - lenb) * (p - lenc));
    double R = lena * lenb * lenc / (4.0 * S);
    double A = acos((lenb * lenb + lenc * lenc - lena * lena) / (2 * lenb * lenc));
    double B = acos((lena * lena + lenc * lenc - lenb * lenb) / (2 * lena * lenc));
    double C = acos((lena * lena + lenb * lenb - lenc * lenc) / (2 * lena * lenb));
    double PI = acos(-1.0);
    double n = PI / gcd(gcd(A,B),C);
    double ans = n / 2 * R * R * sin(2 * PI / n);
    printf("%.10f\n",ans);
}