POJ 1329

模板题，注意一下输出就可以。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>

using namespace std;
struct point {
	double x,y;
};
struct triangle{
	point t[3];
}t;
struct Circle{
	double x,y;
	double r;
}tmp;

double dist(point p1, point p2){
	return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
}

double triangleArea(){
	return fabs(t.t[0].x*t.t[1].y+t.t[1].x*t.t[2].y+t.t[2].x*t.t[0].y-
	t.t[1].x*t.t[0].y-t.t[2].x*t.t[1].y-t.t[0].x*t.t[2].y)/2;
}

int main(){
	double x1,y1,x2,y2,x3,y3;
	while(scanf("%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3)!=EOF){
		t.t[0].x=x1; t.t[0].y=y1;
		t.t[1].x=x2; t.t[1].y=y2;
		t.t[2].x=x3; t.t[2].y=y3;
		double a,b,c;
		a=dist(t.t[0],t.t[1]);
		b=dist(t.t[1],t.t[2]);
		c=dist(t.t[2],t.t[0]);
		tmp.r=(a*b*c)/triangleArea()/4;
		double c1=(x1*x1+y1*y1-x2*x2-y2*y2)/2;
		double c2=(x1*x1+y1*y1-x3*x3-y3*y3)/2;
		tmp.x=(c1*(y1-y3)-c2*(y1-y2))/((x1-x2)*(y1-y3)-(x1-x3)*(y1-y2));
		tmp.y=(c1*(x1-x3)-c2*(x1-x2))/((y1-y2)*(x1-x3)-(y1-y3)*(x1-x2));
		printf("(x");
		if(tmp.x>=0) printf(" - %0.3lf)^2 + (y",tmp.x);
		else printf(" + %0.3lf)^2 + (y",-tmp.x);
		if(tmp.y>=0) printf(" - %0.3lf)^2 =",tmp.y);
		else printf(" + %0.3lf)^2 =",-tmp.y);
		printf(" %0.3lf^2\n",tmp.r);
		printf("x^2 + y^2");
		if(tmp.x>0)
		printf(" - %0.3lfx",2*tmp.x);
		else printf(" + %0.3lfx",-2*tmp.x);
		if(tmp.y>0)
		printf(" - %0.3lfy",2*tmp.y);
		else printf(" + %0.3lfy",-2*tmp.y);
		double ans=tmp.x*tmp.x+tmp.y*tmp.y-tmp.r*tmp.r;
		if(ans<0){
			printf(" - %.3lf = 0\n",-ans);
		}
		else printf(" + %.3lf = 0\n",ans);
		printf("\n");
		}
	return 0;
}