Sdut 2603:rescue the Princess__others

Rescue the Princess Time limit:1000ms Memory limit:65536k have questions. Dot here ^_^ Topic Description

Several days ago, a beast caught a beautiful princess and the princess were 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 to 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? Enter

The ' The ' is ' an integer t (1 <= t <=) 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 so A (x1,y1) and B (X2,y2) and C (X3,y3) are in A anticlockwise direction the equilateral. and coordinates A (x1,y1) and B (X2,y2) are given by anticlockwise.     Output For each test case, you should output the coordinate of C (X3,Y3), the result should is rounded to 2 decimal places in a Li NE. sample Input

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

Sample Output

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

Tips Source2013 Shandong Four ACM College students Program Design Competition
The point: In a counter-clockwise order to give two points of coordinates, to find the 3rd coordinate. Make these three points constitute a equilateral.

#include <stdio.h>
#include <math.h>
#define PI acos ( -1.0)
int main ()
{
    int n;
    scanf ("%d", &n);
    while (n--)
    {
        double x1,x2,y1,y2;
        scanf ("%lf%lf%lf%lf", &x1,&y1,&x2,&y2);
        Double t=atan2 (y2-y1,x2-x1) +pi/3.0;
        Double zz=sqrt (POW (x1-x2,2) +pow (y1-y2,2));
        Double X=zz*cos (t) +x1,y=zz*sin (t) +y1;
        printf ("(%.2LF,%.2LF) \ n", x,y);
    }
    return 0;
}