Rescue The Princess (数学水题)

Rescue The Princess

 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?

  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 0.000.00 0.00 0.00 100.000.00 0.00 100.00 100.001.00 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)

就是给出a，b的坐标求等边三角形c点的坐标；

代码“：

#include <stdio.h>#include<math.h>#define l sqrt(3)int main(){    double xa,ya,xb,yb,x0,y0;    int n,i;    while(~scanf("%d",&n))    {        for(i=0;i<n;i++)        {            scanf("%lf%lf%lf%lf",&xa,&ya,&xb,&yb);            x0=(xa+xb)/2;            y0=(ya+yb)/2;            if(xa==xb)                printf("(%.2lf,%.2lf)\n",xa+l*(ya-yb)/2,y0);            else if(ya==yb)                printf("(%.2lf,%.2lf)\n",x0,ya+l*(xb-xa)/2);            else            {                double kb,k0,xf,yf;                k0=-(xb-xa)/(yb-ya);                kb=(k0+l/3)/(1-k0*l/3);                xf=(yb-y0+k0*x0-kb*xb)/(k0-kb);                yf=((xa-xb)/2+yb/kb-y0/k0)/(1/kb-1/k0);                printf("(%.2lf,%.2lf)\n",xf,yf);            }        }    }    return 0;}