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.
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.
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
using namespace std;
struct point
{
double x,y;
}a,b, mid;
double dist(point a, point b)
{
return (double) ( (a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y) );
}
int main()
{
int t;
scanf("%d", &t);
while(t--)
{
scanf("%lf %lf %lf %lf", &a.x ,&a.y, &b.x, &b.y);
if(a.x==0 && b.x==0 ) //dou zai y zhou
{
//斜率为0
mid.x=(a.x+b.x)/2.0;
mid.y=(a.y+b.y)/2.0;
double lon=fabs(a.y-b.y);
lon=lon*lon;
lon=lon-(mid.y-b.y)*(mid.y-b.y);
lon=sqrt(lon);
printf("(%.2lf,%.2lf)\n", -1*lon, mid.y );
}
else if(a.y==0 && b.y==0)
{// 斜率 不存在
mid.x=(a.x+b.x)/2.0;
mid.y=(a.y+b.y)/2.0;
double lon=fabs(a.x-b.x);
lon=lon*lon;
lon=lon-(mid.x-a.x)*(mid.x-a.x);
lon=sqrt(lon);
printf("(%.2lf,%.2lf)\n", mid.x, lon);
}
else
{
double k, x, y;
k=(a.y-b.y)/(a.x-b.x);
k=(-1.0)/k;
mid.x=(a.x+b.x)/2.0;
mid.y=(a.y+b.y)/2.0;
double lon=dist(a, b);
double dd=k*mid.x-mid.y+a.y;
x=((2*a.x+2*dd*k)+sqrt( (2*a.x+2*dd*k)*(2*a.x+2*dd*k) - 4*(k*k+1)*(a.x*a.x+dd*dd-lon) ) )/(2.0*(k*k+1));
y=k*x-k*mid.x+mid.y;
point A, B;
A.x = b.x-a.x;
A.y = b.y-a.y;
B.x = x-b.x;
B.y = y-b.y;
if( (A.x*B.y - A.y*B.x) > 0 ){
printf("(%.2lf,%.2lf)\n", x, y );
}
else
{
x=((2*a.x+2*dd*k)-sqrt( (2*a.x+2*dd*k)*(2*a.x+2*dd*k) - 4*(k*k+1)*(a.x*a.x+dd*dd-lon) ) )/(2.0*(k*k+1));
y=k*x-k*mid.x+mid.y;
printf("(%.2lf,%.2lf)\n", x, y );
}
}
}
return 0;
}
