Rescue The Princess

Description
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)

涉及到向量旋转角度的问题；

结论：

对于任意两个不同点A和B，A绕B旋转θ角度后的坐标为：

(Δx*cosθ- Δy * sinθ+ xB, Δy*cosθ + Δx * sinθ+ yB )

注：xB、yB为B点坐标。

具体证明过程：http://www.tuicool.com/m/articles/FnEZJb

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>#include<math.h>using namespace std;int main(){    int t;    scanf("%d",&t);    while (t--)    {        double x1,x2,y1,y2,x3,y3;        scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);    double x,y;    x=x2-x1;<span style="white-space:pre"></span>//以第一个点为原点，顺时针旋转60度    y=y2-y1;    x3=x*(1.0/2.0)-y*(sqrt(3.0)/2.0)+x1;    y3=y*(1.0/2.0)+x*(sqrt(3.0)/2.0)+y1;    printf("(%.2lf,%.2lf)\n",x3,y3);    }    return 0;}