The Circumference of the Circle

最新推荐文章于 2018-08-29 16:45:28 发布

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最新推荐文章于 2018-08-29 16:45:28 发布

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本文介绍了一种通过三个非共线点的笛卡尔坐标计算唯一相交圆周长的方法。利用正余弦定理计算出直径后，再求得圆的周长。

The Circumference of the Circle

Describe

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
The input 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
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

利用正余弦定理推出直径 $d=2R=\frac a {sinA}=\frac b {sinB}=\frac c {sinC}$ ，计算即可。

#include<iostream>
#include<cmath>
#include<cstdio>
#define pi 3.141592653589793

using namespace std;

double len(double a,double b,double c,double d)
{
    return sqrt((a-b)*(a-b)+(c-d)*(c-d));
}

int main()
{
    double x1,x2,x3,y1,y2,y3;
    while(cin>>x1>>y1>>x2>>y2>>x3>>y3)
    {
        double a=len(x1,x2,y1,y2),b=len(x1,x3,y1,y3),c=len(x2,x3,y2,y3);
        double cosa=(b*b+c*c-a*a)/(2*b*c);
        double sina=sqrt(1-cosa*cosa);
        double d=a/sina;
        printf("%.2f\n",pi*d);
    }
    return 0;
}