joj1184 The Circumference of the Circle

 1184: The Circumference of the Circle

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Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	1s	8192K	558	338	Standard

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.

Input Specification

The input file will contain one or more test cases. Each test case consists of one line containing six real numbers x₁,y₁, x₂,y₂,x₃,y₃, representing the coordinates of the three points. The diameter of the circle determined by the three points will never exceed a million. Input is terminated by end of file.

Output Specification

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

 

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This problem is used for contest: 70  167  197

 

这个题很值得吐槽...记得《那些年》里说以后会连log都不知道是什么...

其实，这题给出三个点坐标算外接圆半径什么的我果真一下子没想起来.....高中的知识.....

其实就是一个小公式，r = (a*b*c) / (4*s)。可以用正弦定理推导。面积用到行列式，这个还是记得的。

#include <stdio.h>
#include <math.h>

#define PI 3.141592653589793

double distance(double x1, double y1, double x2, double y2)
{
    return sqrt((x1-x2)*(x1-x2) + (y1-y2)*(y1-y2));
}
double mianji(double x1, double y1, double x2, double y2, double x3, double y3)
{
    return 0.5*(x1*y2+x2*y3+x3*y1-x3*y2-x2*y1-x1*y3);
}

int main()
{
    //freopen("in.txt", "r", stdin);

    double x1, y1, x2, y2, x3, y3;
    double a, b, c;
    double r;
    double s;
    double C;

    while (scanf("%lf%lf%lf%lf%lf%lf", &x1, &y1, &x2, &y2, &x3, &y3) != EOF)
    {
        a = distance(x1, y1, x2, y2);
        b = distance(x2, y2, x3, y3);
        c = distance(x3, y3, x1, y1);

        s = mianji(x1, y1, x2, y2, x3, y3);
        r = (a*b*c) / (4*s);

        C = 2 * PI * r;
        printf("%.2lf\n", fabs(C));
    }

    return 0;
}