Ancient Berland Circus
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 0.0000001.000000 1.0000000.000000 1.000000
Output
1.00000000 题意： 给出三个点，以三个点为顶点的正多边形的最小面积思路： 三个点能围成一个愿，那么在圆内，边数越小面积越小，所以求边数最小的， 先求出边对应的圆心角，再求三个边圆心角的最大公约数知识相关：知道三边求面积，知道3边及面积求外接圆半径。 AC代码：
# include <iostream>
# include <cmath>
# include <cstdio>
using namespace std;
#define eqs 0.01
const double PI = acos(-1.0);
struct Point
{
double x;
double y;
};
Point operator-(Point v1, Point v2) // 向量相减
{
v1.x -= v2.x;
v1.y -= v2.y;
return v1;
}
double dis(Point v) // 取模
{
return sqrt(v.x * v.x + v.y * v.y);
}
double fgcd(double a, double b)
{
return a < eqs ? b : fgcd(fmod(b,a), a);
}
int main()
{
Point a, b, c;
while(cin >> a.x >> a.y >> b.x >> b.y >> c.x >> c.y)
{
double A = dis(a - b);
double B = dis(b - c);
double C = dis(a - c);
// 求面积
double p = (A + B + C) / 2.0 ;
double s = sqrt(p * (p - A) * (p - B) * (p - C)) ;
// 求半径
double r = A * B * C / ( 4 * s );
if(A > C)
swap(A, C);
if(B > C)
swap(B, C);
double angA = 2 * asin(A / (2 * r)) ;
double angB = 2 * asin(B / (2 * r)) ;
double angC = 2 * PI - angA - angB ;
double g = fgcd(angC, fgcd(angA, angB));
// 求内接正多边形面积
double result = PI*r*r*sin(g)/g;
printf("%.6lf\n", result);
}
return 0;
}