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
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
#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]);
}
}