zoj 1090 三角形外接圆周长

#include <iostream>
#include <iomanip>
#include <math.h>

using namespace std;

/*此题为数学问题，根据三角形外接圆的性质和正余弦定理，可计算出圆的直径 
 * ((a*a+b*b-c*c)/2ab)*((a*a+b*b-c*c)/2ab) + c*c/d*d = 1
 */

int main() {
	const double pi = 3.141592653589793;
	double x1 = 0 , y1 = 0 , x2 = 0, y2 = 0, x3 = 0, y3 = 0;
	while(cin >> x1 >> y1 >> x2 >> y2 >> x3 >> y3) {
		double ap = 0, bp = 0, cp = 0, d = 0, temp = 0;
		ap = pow(x1-x3, 2) + pow(y1-y3, 2);
		bp = pow(x1-x2, 2) + pow(y1-y2, 2);
		cp = pow(x2-x3, 2) + pow(y2-y3, 2);
		temp = pow(ap+bp-cp, 2) / (4*ap*bp);
		d = sqrt(cp/(1-temp));
		double peri = pi * d;
		
		cout << fixed << setprecision(2) << peri << endl;		//格式化输出浮点数 
	}	
	return 0;
}