Rescue The Princess （SDIBT 3230）

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?

Input

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.

Output

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.

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)

题意：

给定等边三角形两点坐标，求在这两点逆时针方向上的第三点坐标。

这个题有一个公式可以求得第三点坐标：

点（x1，y1）绕点（x2，y2）逆时针旋转a角度后的新坐标（x，y）

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x = (x2-x1)*cos(a)-(y2-y1)*sin(a)+x1;

y = (x2-x1)*sin(a)+(y2-y1)*cos(a)+y1;

注意：此处a为弧度，该题为 π/3。

推断过程有些复杂，有兴趣的可以去百度一下，这里就不再说了^_^

代码如下：

#include <stdio.h>
#include <math.h>
#define PI 3.1415926
int main()
{
    int t;
    double l,x1,x2,x3,y1,y2,y3;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
        x3 = (x2-x1)*cos(PI/3.0) - (y2-y1)*sin(PI/3.0)+x1;  
        y3 = (x2-x1)*sin(PI/3.0) + (y2 - y1)*cos(PI/3.0)+y1;
        printf("(%.2lf,%.2lf)\n",x3,y3);

    }
    return 0;
}

Rescue The Princess （SDIBT 3230）

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