Several days ago, a beast caught a beautiful princess and the princess was put in *. To rescue the princess, a prince who wanted to marry the princess set out immediately. Yet, the beast set a maze. Only if the prince find out the maze’s exit can he save the princess.

Now, here comes the problem. The maze is a dimensional plane. The beast is smart, and he hidden the princess snugly. He marked two coordinates of an equilateral triangle in the maze. The two marked coordinates are A(x₁,y₁) and B(x₂,y₂). The third coordinate C(x₃,y₃) is the maze’s exit. If the prince can find out the exit, he can save the princess. After the prince comes into the maze, he finds out the A(x₁,y₁) and B(x₂,y₂), but he doesn’t know where the C(x₃,y₃) is. The prince need your help. Can you calculate the C(x₃,y₃) and tell him?

The first line is an integer T(1 <= T <= 100) which is the number of test cases. T test cases follow. Each test case contains two coordinates A(x₁,y₁) and B(x₂,y₂), described by four floating-point numbers x₁, y₁, x₂, y₂ ( |x₁|, |y₁|, |x₂|, |y₂| <= 1000.0).
    Please notice that A(x₁,y₁) and B(x₂,y₂) and C(x₃,y₃) are in an anticlockwise direction from the equilateral triangle. And coordinates A(x₁,y₁) and B(x₂,y₂) are given by anticlockwise.

    For each test case, you should output the coordinate of C(x₃,y₃), the result should be rounded to 2 decimal places in a line.

4

-100.00 0.00 0.00 0.00

0.00 0.00 0.00 100.00

0.00 0.00 100.00 100.00

1.00 0.00 1.866 0.50

(-50.00,86.60)

(-86.60,50.00)

(-36.60,136.60)

(1.00,1.00)

 

2013年山东省第四届ACM大学生程序设计竞赛

分两种：一种是A点B点的x坐标相同，二种：X坐标不相同。

#include<stdio.h>

#include<math.h>

int main()

{

    double b[2],c[3],a,x[3],y[3],tx[2],ty[2],k,bb,edglen;

    int t;

    scanf("%d",&t);

    while(t--)

    {

        scanf("%lf%lf%lf%lf",&x[0],&y[0],&x[1],&y[1]);

        if(x[0]==x[1])

        {

            edglen=sqrt(pow(x[0]-x[1],2.0)+pow(y[0]-y[1],2.0));

            tx[0]=x[0]+sqrt(3.0)/2*edglen;

            tx[1]=x[0]-sqrt(3.0)/2*edglen;

            if(y[1]>y[0])

            {

                ty[0]=y[0]+edglen/2;

                printf("(%.2lf,%.2lf)\n",tx[1],ty[0]);

            }

            else

            {

                ty[0]=y[1]+edglen/2;

                printf("(%.2lf,%.2lf)\n",tx[0],ty[0]);

            }

            continue;

        }

        c[0]=(x[0]*x[0]-x[1]*x[1]+y[0]*y[0]-y[1]*y[1])/(2*x[0]-2*x[1]);

        b[0]=-(y[0]-y[1])/(x[0]-x[1]);

        a=b[0]*b[0]+1;  b[1]=2*(c[0]*b[0]-x[1]*b[0]-y[1]);

        c[1]=x[0]*x[0]-2*x[0]*x[1]+y[0]*y[0]-2*y[0]*y[1];

        c[2]=c[0]*c[0]-2*c[0]*x[1]-c[1];

        k=-b[0]; bb=y[0]-k*x[0];

        ty[0]=(-b[1]+sqrt(b[1]*b[1]-4*a*c[2]))/(2*a);tx[0]=c[0]+b[0]*ty[0];

        ty[1]=(-b[1]-sqrt(b[1]*b[1]-4*a*c[2]))/(2*a);tx[1]=c[0]+b[0]*ty[1];

        if(x[0]<x[1])

        {

            if(ty[0]-k*tx[0]-bb>0)

            {

                x[2]=tx[0];y[2]=ty[0];

            }

            else

            {

                 x[2]=tx[1];y[2]=ty[1];

            }

        }

        else if(x[0]>x[1])

        {

             if(ty[0]-k*tx[0]-bb<0)

            {

                x[2]=tx[0];y[2]=ty[0];

            }

            else

            {

                 x[2]=tx[1];y[2]=ty[1];

            }

        }

        printf("(%.2lf,%.2lf)\n",x[2],y[2]);

    }

}