莫名其妙就AC了……

圆的反演……

神马是反演？

快去恶补奥数……

#include<iostream>
#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
const double pi=acos(-1.0);
const double eps=1e-9;
int dcmp(double x){return fabs(x)<eps?0:x<0?-1:1;}
struct dot
{
    double x,y;
    dot(){}
    dot(double a,double b){x=a;y=b;}
    dot operator +(dot a){return dot(x+a.x,y+a.y);}
    dot operator -(dot a){return dot(x-a.x,y-a.y);}
    dot operator *(double a){return dot(x*a,y*a);}
    double operator *(dot a){return x*a.y-y*a.x;}
    dot operator /(double a){return dot(x/a,y/a);}
    double operator /(dot a){return x*a.x+y*a.y;}
    bool operator ==(dot a){return x==a.x&&y==a.y;}
    void in(){scanf("%lf%lf",&x,&y);}
    void out(){printf("%f %f\n",x,y);}
    dot norv(){return dot(-y,x);}
    dot univ(){double a=mod();return dot(x/a,y/a);}
    dot ro(double a){return dot(x*cos(a)-y*sin(a),x*sin(a)+y*cos(a));}
    double mod(){return sqrt(x*x+y*y);}
    double dis(dot a){return sqrt(pow(x-a.x,2)+pow(y-a.y,2));}
};
struct cir
{
    dot o;
    double r;
    cir(){}
    cir(dot a,double b){o=a;r=b;}
    void in(){o.in();scanf("%lf",&r);}
};
struct seg
{
    dot s,e;
    seg(){}
    seg(dot a,dot b){s=a;e=b;}
};
cir sivs(dot a,dot b,dot c)
{
    dot dir,a1,b1;
    double t,d,w;
    t=fabs((b-a)*(c-a));
    d=a.dis(b);
    t/=d;
    w=0.5/t;
    dir=(b-a).norv();
    a1=c+dir*(w/d);
    b1=c-dir*(w/d);
    if(fabs((b-a)*(a1-a))<fabs((b-a)*(b1-a)))
        return cir(a1,w);
    else
        return cir(b1,w);
}
cir civs(cir a,dot b)
{
    cir c;
    double t,x,y,s;
    t=a.o.dis(b);
    x=1.0/(t-a.r);
    y=1.0/(t+a.r);
    c.r=(x-y)/2.0;
    s=(x+y)/2.0;
    c.o=b+(a.o-b)*(s/t);
    return c;
}
seg se[2];
void comseg(dot a,double r1,dot b,double r2)
{
    double ang;
    ang=acos((r1-r2)/a.dis(b));
    se[0].s=a+(b-a).ro(ang).univ()*r1;
    se[1].s=a+(b-a).ro(-ang).univ()*r1;
    ang=pi-ang;
    se[0].e=b+(a-b).ro(-ang).univ()*r2;
    se[1].e=b+(a-b).ro(ang).univ()*r2;
}
int main()
{
    int T,cnt,i;
    cir a,b,a1,b1,ans[2];
    dot c;
    scanf("%d",&T);
    while(T--)
    {
        a.in();
        b.in();
        c.in();
        a1=civs(a,c);
        b1=civs(b,c);
        comseg(a1.o,a1.r,b1.o,b1.r);
        cnt=0;
        for(i=0;i<2;i++)
            if(dcmp((a1.o-se[i].s)*(se[i].e-se[i].s))==dcmp((c-se[i].s)*(se[i].e-se[i].s)))
                if(dcmp((b1.o-se[i].s)*(se[i].e-se[i].s))==dcmp((c-se[i].s)*(se[i].e-se[i].s)))
					ans[cnt++]=sivs(se[i].s,se[i].e,c);
        printf("%d\n",cnt);
        for(i=0;i<cnt;i++)
            printf("%.8f %.8f %.8f\n",ans[i].o.x,ans[i].o.y,ans[i].r);
    }
}

Problem of Apollonius

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Special Judge

Problem Description

　　Apollonius of Perga (ca. 262 BC - ca. 190 BC) was a Greek geometer and astronomer. In his noted work Epaphai, he posed and solved such a problem: constructing circles that are tangent to three given circles in a plane. Two tangent
circles can be internally or externally tangent to each other, thus Apollonius‘s problem generically have eight solutions.

　　Now considering a simplified case of Apollonius‘s problem: constructing circles that are externally tangent to two given circles, and touches a given point(the given point must be on the circle which you find, can‘t be inside the circle). In addition, two
given circles have no common points, and neither of them are contained by the other, and the given point is also located strictly outside the given circles. You should be thankful that modern mathematics provides you with plenty of useful tools other than
euclidean geometry that help you a lot in this problem.

Input

　　The first line of input contains an integer T (T ≤ 200), indicating the number of cases.

　　Each ease has eight positive integers x1, y1, r1, x2, y2, r2, x3, y3 in a single line, stating two circles whose centres are (x1, y1), (x2, y2) and radius are r1 and r2 respectively, and a point located at (x3, y3). All integers are no larger than one hundred.

Output

　　For each case, firstly output an integer S, indicating the number of solutions.

　　Then output S lines, each line contains three float numbers x, y and r, meaning that a circle, whose center is (x, y) and radius is r, is a solution to this case. If there are multiple solutions (S > 1), outputing them in any order is OK. Your answer
will be accepted if your absolute error for each number is no more than 10^-4.

Sample Input

1
12 10 1 8 10 1 10 10

Sample Output

2
10.00000000 8.50000000 1.50000000
10.00000000 11.50000000 1.50000000

Hint

This problem is special judged.


 

Source

2013 Asia Hangzhou Regional Contest