poj 1329(已知三点求外接圆方程.)

Circle Through Three Points

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 3766		Accepted: 1570

Description

Your team is to write a program that, given the Cartesian coordinates of three points on a plane, will find the equation of the circle through them all. The three points will not be on a straight line.
The solution is to be printed as an equation of the form

	(x - h)^2 + (y - k)^2 = r^2				(1)

and an equation of the form

	x^2 + y^2 + cx + dy - e = 0				(2)

Input

Each line of input to your program will contain the x and y coordinates of three points, in the order Ax, Ay, Bx, By, Cx, Cy. These coordinates will be real numbers separated from each other by one or more spaces.

Output

Your program must print the required equations on two lines using the format given in the sample below. Your computed values for h, k, r, c, d, and e in Equations 1 and 2 above are to be printed with three digits after the decimal point. Plus and minus signs in the equations should be changed as needed to avoid multiple signs before a number. Plus, minus, and equal signs must be separated from the adjacent characters by a single space on each side. No other spaces are to appear in the equations. Print a single blank line after each equation pair.

Sample Input

7.0 -5.0 -1.0 1.0 0.0 -6.0
1.0 7.0 8.0 6.0 7.0 -2.0

Sample Output

(x - 3.000)^2 + (y + 2.000)^2 = 5.000^2
x^2 + y^2 - 6.000x + 4.000y - 12.000 = 0

(x - 3.921)^2 + (y - 2.447)^2 = 5.409^2
x^2 + y^2 - 7.842x - 4.895y - 7.895 = 0

Source

恶心的输出..看了discuss才知道0.000要原样输出。。

#include<stdio.h>
#include<iostream>
#include<string.h>
#include <stdlib.h>
#include<math.h>
#include<algorithm>
using namespace std;
const double pi =  3.141592653589793;
const double eps = 1e-8;
struct Point
{
    double x,y;
} p[3];
double dis(Point a,Point b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
///外接圆圆心坐标
Point waixin(Point a,Point b,Point c)
{
    Point p;
    double a1 = b.x - a.x, b1 = b.y - a.y, c1 = (a1*a1 + b1*b1)/2;
    double a2 = c.x - a.x, b2 = c.y - a.y, c2 = (a2*a2 + b2*b2)/2;
    double d = a1*b2 - a2*b1;
    p.x = a.x + (c1*b2 - c2*b1)/d, p.y=a.y + (a1*c2 -a2*c1)/d;
    return p;
}
char check(double x)
{
    if(x<-eps) return '+';
    return '-';
}
char check2(double x)
{
    if(x<-eps) return '-';
    return '+';
}
int main()
{

    while(scanf("%lf%lf%lf%lf%lf%lf",&p[0].x,&p[0].y,&p[1].x,&p[1].y,&p[2].x,&p[2].y)!=EOF)
    {
        double a = dis(p[0],p[1]);
        double b = dis(p[1],p[2]);
        double c = dis(p[0],p[2]);
        double r = a*b*c/sqrt((a+b+c)*(-a+b+c)*(a-b+c)*(a+b-c));
        Point center;
        center = waixin(p[0],p[1],p[2]);
        if(fabs(center.x)<eps) printf("x^2 + ");
        else printf("(x %c %.3lf)^2 + ",check(center.x),fabs(center.x));
        if(fabs(center.y)<eps) printf("y^2");
        else printf("(y %c %.3lf)^2",check(center.y),fabs(center.y));
        printf(" = %.3lf^2
",r);

        printf("x^2 + y^2");
        double c1 = 2*center.x,d1=2*center.y;
        double r1 = center.x*center.x+center.y*center.y-r*r;
        printf(" %c %.3lfx %c %.3lfy %c %.3lf = 0

",check(c1),fabs(c1),check(d1),fabs(d1),check2(r1),fabs(r1));
    }
    return 0;
}