给三点，求外切圆周长
代码：
#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;struct Point {double x, y;Point() {}Point(double x, double y) {this->x = x;this->y = y;}void read() {scanf("%lf%lf", &x, &y);}};typedef Point Vector;Vector operator + (Vector A, Vector B) {return Vector(A.x + B.x, A.y + B.y);}Vector operator – (Vector A, Vector B) {return Vector(A.x – B.x, A.y – B.y);}Vector operator * (Vector A, double p) {return Vector(A.x * p, A.y * p);}Vector operator / (Vector A, double p) {return Vector(A.x / p, A.y / p);}bool operator < (const Point& a, const Point& b) {return a.x < b.x || (a.x == b.x && a.y < b.y);}const double eps = 1e-8;const double PI = acos(-1.0);int dcmp(double x) {if (fabs(x) < eps) return 0;else return x < 0 ? -1 : 1;}bool operator == (const Point& a, const Point& b) {return dcmp(a.x – b.x) == 0 && dcmp(a.y – b.y) == 0;}double Dot(Vector A, Vector B) {return A.x * B.x + A.y * B.y;} //点积double Length(Vector A) {return sqrt(Dot(A, A));} //向量的模double Angle(Vector A, Vector B) {return acos(Dot(A, B) / Length(A) / Length(B));} //向量夹角double Cross(Vector A, Vector B) {return A.x * B.y – A.y * B.x;} //叉积double Area2(Point A, Point B, Point C) {return Cross(B – A, C – A);} //有向面积double angle(Vector v) {return atan2(v.y, v.x);}struct Circle {Point c;double r;Circle(Point c, double r) {this->c = c;this->r = r;}Point point(double a) {return Point(c.x + cos(a) * r, c.y + sin(a) * r);}};Circle CircumscribedCircle(Point p1, Point p2, Point p3) {double Bx = p2.x – p1.x, By = p2.y – p1.y;double Cx = p3.x – p1.x, Cy = p3.y – p1.y;double D = 2 * (Bx * Cy – By * Cx);double cx = (Cy * (Bx * Bx + By * By) – By * (Cx * Cx + Cy * Cy)) / D + p1.x;double cy = (Bx * (Cx * Cx + Cy * Cy) – Cx * (Bx * Bx + By * By)) / D + p1.y;Point p = Point(cx, cy);return Circle(p, Length(p1 – p));}Point p[3];int main() {while (~scanf("%lf%lf", &p[0].x, &p[0].y)) {p[1].read(); p[2].read();Circle ans = CircumscribedCircle(p[0], p[1], p[2]);printf("%.2f\n", ans.r * 2 * PI);}return 0;}
，此刻睡觉的口水将变成明天流下的泪水。
