Codeforces-935C：Fifa and Fafa

C. Fifa and Fafa

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Fifa and Fafa are sharing a flat. Fifa loves video games and wants to download a new soccer game. Unfortunately, Fafa heavily uses the internet which consumes the quota. Fifa can access the internet through his Wi-Fi access point. This access point can be accessed within a range of r meters (this range can be chosen by Fifa) from its position. Fifa must put the access point inside the flat which has a circular shape of radius R. Fifa wants to minimize the area that is not covered by the access point inside the flat without letting Fafa or anyone outside the flat to get access to the internet.

The world is represented as an infinite 2D plane. The flat is centered at (x1, y1) and has radius R and Fafa's laptop is located at (x2, y2), not necessarily inside the flat. Find the position and the radius chosen by Fifa for his access point which minimizes the uncovered area.

Input

The single line of the input contains 5 space-separated integers R, x1, y1, x2, y2 (1 ≤ R ≤ 105, |x1|, |y1|, |x2|, |y2| ≤ 105).

Output

Print three space-separated numbers xap, yap, r where (xap, yap) is the position which Fifa chose for the access point and r is the radius of its range.

Your answer will be considered correct if the radius does not differ from optimal more than 10 - 6 absolutely or relatively, and also the radius you printed can be changed by no more than 10 - 6 (absolutely or relatively) in such a way that all points outside the flat and Fafa's laptop position are outside circle of the access point range.

Examples

input

Copy

5 3 3 1 1

output

3.7677669529663684 3.7677669529663684 3.914213562373095

input

Copy

10 5 5 5 15

output

5.0 5.0 10.0

思路：主要是列方程解出直线(x1,y1) 和(x2,y2)与圆的2个交点。不过利用向量更简单。

#include
using namespace std;
const int MAX=1e6;
const int MOD=1e9+7;
typedef __int64 ll;
double dis(double x1,double y1,double x2,double y2){return (x1-x2)*(x1-x2)+(y1-y2)*(y1-y2);}
int main()
{
    ll R,x1,y1,x2,y2;
    cin>>R>>x1>>y1>>x2>>y2;
    if(dis(x1,y1,x2,y2)>=R*R)cout<fabs(y1-R-y2))printf("%.12lf %.12lf %.12lf\n",1.0*x1,1.0*(y1+R+y2)/2,1.0*fabs(y1+R-y2)/2);
            else printf("%.12lf %.12lf %.12lf\n",1.0*x1,1.0*(y1-R+y2)/2,1.0*fabs(y1-R-y2)/2);
        }
        else
        {
            double k=1.0*(y2-y1)/(x2-x1);
            double A=k*k+1;
            double B=-2*x1-2*k*k*x2+2*k*y2-2*k*y1;
            double C=k*k*x2*x2-2*k*x2*y2+y2*y2+2*k*y1*x2-2*y1*y2+x1*x1+y1*y1-R*R;
            double D=B*B-4*A*C;
            double x3=(-B+sqrt(D))/(2*A),y3=k*(x3-x2)+y2;
            double x4=(-B-sqrt(D))/(2*A),y4=k*(x4-x2)+y2;
            if(dis(x3,y3,x2,y2)>dis(x4,y4,x2,y2))printf("%.12lf %.12lf %.12lf\n",(x3+x2)/2,(y3+y2)/2,sqrt(dis(x3,y3,x2,y2))/2);
            else printf("%.12lf %.12lf %.12lf\n",(x4+x2)/2,(y4+y2)/2,sqrt(dis(x4,y4,x2,y2))/2);
        }
    }
    return 0;
}