The Circumference of the Circle
題目描述
To calculate the circumference of a circle seems to be an easy task - provided you know its diameter. But what if you don’t?
You are given the cartesian coordinates of three non-collinear points in the plane.
Your job is to calculate the circumference of the unique circle that intersects all three points.
Iutput
For each test case, print one line containing one real number telling the circumference of the circle determined by the three points. The circumference is to be printed accurately rounded to two decimals. The value of pi is approximately 3.141592653589793.
Output
For each test case, print one line containing one real number telling the circumference of the circle determined by the three points. The circumference is to be printed accurately rounded to two decimals. The value of pi is approximately 3.141592653589793.
Sample Input
0.0 -0.5 0.5 0.0 0.0 0.5
0.0 0.0 0.0 1.0 1.0 1.0
5.0 5.0 5.0 7.0 4.0 6.0
0.0 0.0 -1.0 7.0 7.0 7.0
50.0 50.0 50.0 70.0 40.0 60.0
0.0 0.0 10.0 0.0 20.0 1.0
0.0 -500000.0 500000.0 0.0 0.0 500000.0
Sample Output
3.14
4.44
6.28
31.42
62.83
632.24
3141592.65
思路：
用海倫公式來求面積，然後根據面積和周長之間的關係求外接圓的半徑；
代碼
#include <cstdio>
#include <cmath>
using namespace std;
#define pi 3.141592653589793
double func(double x1,double y1,double x2,double y2)
{
double temp=abs(x1-x2)*abs(x1-x2)+abs(y1-y2)*abs(y1-y2);
return sqrt(temp);
}
int main()
{
double x1,y1,x2,y2,x3,y3;
while(scanf("%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3)!=EOF){
double l1=func(x1,y1,x2,y2);
double l2=func(x2,y2,x3,y3);
double l3=func(x1,y1,x3,y3);
double l=(l1+l2+l3)/2;
double temp=l*(l-l1)*(l-l2)*(l-l3);
double r=l1*l2*l3/(4*sqrt(temp));
printf("%.2f\n",2*r*pi);
}
return 0;
}