SDUTRescue The Princess（数学问题）

题目描述

Several days ago, a beast caught a beautiful princess and the princess was put in prison. 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?

输入

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.

输出

    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.

示例输入

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

示例输出

(-50.00,86.60)
(-86.60,50.00)
(-36.60,136.60)
(1.00,1.00)

提示

 

来源

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]);
    }

}