ZOJ Problem Set

 

ZOJ Problem Set - 1090

The Circumference of the Circle

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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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Source: University of Ulm Local Contest 1996

Submit    Status

#include<bits/stdc++.h>
#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)
    {
        //求内接三角形的每一个边 
        double a=len(x1,x2,y1,y2),b=len(x1,x3,y1,y3),c=len(x2,x3,y2,y3);
        //求2R=a/sinA 因为不知道sinA 所以先求cosA 那么就是cosA = b^2-c^2/2bc 然后1-cosA的平方=sinA的平方  
        double cosa=(b*b+c*c-a*a)/(2*b*c);
        double sina=sqrt(1-cosa*cosa);
        double d=a/sina;
        printf("%.2f
",pi*d);
    }
    return 0;
}