坐标旋转公式

x1=cos(angle)*x-sin(angle)*y;

y1=cos(angle)*y+sin(angle)*x;

其中x，y表示物体相对于旋转点旋转angle的角度之前的坐标，x1，y1表示物体旋转angle后相对于旋转点的坐标

从数学上来说，此公式可以用来计算某个点绕另外一点旋转一定角度后的坐标，例如：A（x，y）绕B（a，b)旋转β度后的位置为C（c，d），则x，y，a，b，β，c，d有如下关系式：

1。设A点旋转前的角度为δ，则旋转(逆时针)到C点后角度为δ+β

2。求A，B两点的距离：dist1=|AB|=y/sin(δ)=x/cos(δ)

3。求C，B两点的距离：dist2=|CB|=d/sin(δ+β)=c/cos(δ+β)

4。显然dist1=dist2，设dist1=r所以：

　　r=x/cos(δ)=y/sin(δ)=d/sin(δ+β)=c/cos(δ+β)

5。由三角函数两角和差公式知：

　　sin(δ+β)=sin(δ)cos(β)+cos(δ)sin(β)

　　cos(δ+β)=cos(δ)cos(β)-sin(δ)sin(β)

　　所以得出：

　　c=r*cos(δ+β)=r*cos(δ)cos(β)-r*sin(δ)sin(β)=xcos(β)-ysin(β)

　　d=r*sin(δ+β)=r*sin(δ)cos(β)+r*cos(δ)sin(β)=ycos(β)+xsin(β)

即旋转后的坐标c，d只与旋转前的坐标x，y及旋转的角度β有关

从图中可以很容易理解出A点旋转后的C点总是在圆周上运动，圆周的半径为|AB|，利用这点就可以使物体绕圆周运动，即旋转物体。

上面公式是相对于B点坐标来的，也就是假如B点位（0,0）可以这么做。现在给出可以适合任意情况的公式：

x0 = dx * cos(β) - dy * sin(β)

y0 = dy * cos(β) + dx * sin(β)

参数解释：

x0,y0是旋转后相对于中心点的坐标，也就是原点的坐标，但不是之前点旋转后的实际坐标，还要计算一步，

β旋转角度，可以是顺时针或者逆时针。

dx是旋转前的x坐标-旋转后的x坐标

dy是旋转前的y坐标-旋转后的y坐标

x1=a+x0;

y1=b+y0;

上面才是旋转后的实际坐标，其中a，b是原点坐标

下面是上面图的公式解答：

x0=(x-a)*cos(β)-(y-b)*sin(β);

y0=(y-b)*cos(β)+(x-a)*sin(β);

x1=x0+a;

y1=y0+b;

Rescue The Princess

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

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 0.00
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

2013年山东省第四届ACM大学生程序设计竞赛

       cin>>a>>b>>x>>y ;
       X=(x-a)*0.5-(y-b)*sqrt(3.0)*0.5 +a;
       Y=(y-b)*0.5+(x-a)*sqrt(3.0)*0.5 +b;
       printf("(%.2lf,%.2lf)
",X,Y) ;