感谢楼上的那位哥们儿提供的代码，把我从wa的深渊里面解救了出来。。
我也附上我的代码。。。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
#define maxn 1005
#define sqr(x) ((x) * (x))
int n;
const double eps = 1e-5;
int sgn(double x)
{
	return fabs(x) < eps ? 0 : x < -eps ? -1: 1;
}
struct point
{
	double x,y;
	point(){}
	point(double a,double b):x(a),y(b){}
	point operator - (point p){return point(x - p.x,y - p.y);}
	point operator + (point p){return point(x + p.x,y + p.y);}
	double norm(){return sqrt(sqr(x) + sqr(y));}
	double operator * (point p){return x * p.x + y * p.y;}
	point operator * (double a){return point(x * a,y * a);}
	double operator ^ (point p){return x * p.y - y * p.x;}
}p[maxn];
bool line(point a,point b,point c){return sgn((a - b) ^ (a - c)) == 0;}
struct circle
{
	point o;
	double r;
	circle(){}
	circle(point _o,double _r):o(_o),r(_r){}
	circle(point a,point b)
	{
		o = (a + b) * 0.5;
		r = (a - b).norm() * 0.5;
	}
	circle(point a,point b,point c)
	{
		if(sgn((a - b) * (c - b)) == 0)
		{
			o = point((a + c) * 0.5);
			r = (a - c).norm() * 0.5;
		}
		else if(sgn((c - a) * (b - a)) == 0)
		{
			o = point((c + b) * 0.5);
			r = (c - b).norm() * 0.5;
		}
		else if(sgn((a - c) * (b - c)) == 0)
		{
			o = point((a + b) * 0.5);
			r = (a - b).norm() * 0.5;
		}
		else if(sgn((b.x - a.x) * (b.x - c.x)))
		{
			o.y = (((sqr(b.y) - sqr(c.y)) * (b.x - a.x) - (sqr(b.y) - sqr(a.y)) * (b.x - c.x)) + (b.x - c.x) * (b.x - a.x) * (c.x - a.x)) / (2.0 * ((b - a) ^ (b - c)));
			o.x = ((a.y + b.y) * 0.5 - o.y) * (b.y - a.y) / (b.x - a.x) + (a.x + b.x) * 0.5;
			r = (b - o).norm();
		}
		else
		{
			o.x = (((sqr(b.x) - sqr(a.x)) * (b.y - c.y) - (sqr(b.x) - sqr(c.x)) * (b.y - a.y)) + (a.y - c.y) * (b.y - a.y) * (b.y - c.y)) / (2.0 * ((b - a) ^ (b - c)));
			o.y = ((a.x + b.x) * 0.5 - o.x) * (b.x - a.x) / (b.y - a.y) + (a.y + b.y) * 0.5;
			r = (b - o).norm();
		}
	}
	bool in(point x)
	{
		return sgn((x - o).norm() - r) <= 0;
	}
};
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) == 6)
	{
		circle ans(p[0],p[1],p[2]);
		double h = ans.o.x,k = ans.o.y,r = ans.r,c = 2.0 * h,d = 2.0 * k,e = sqr(h) + sqr(k) - sqr(r);
		printf("(x %c %.3lf)^2 + (y %c %.3lf)^2 = %.3lf^2\n",sgn(h)<=0?'+':'-',fabs(h),sgn(k)<=0?'+':'-',fabs(k),r);
		printf("x^2 + y^2 %c %.3lfx %c %.3lfy %c %.3lf = 0\n",sgn(c)<=0?'+':'-',fabs(c),sgn(d)<=0?'+':'-',fabs(d),sgn(e)>=0?'+':'-',fabs(e));
		puts("");
	}
}

• Re:坑爹题，0.000是照样输出的，只不过符号是正罢了。。。 (110) 455462586 2013-10-10 02:39:06