题目传送门：http://acm.sdut.edu.cn/onlinejudge2/index.php/Home/Index/problemdetail/pid/2603.html
Rescue The Princess
Problem 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
Example 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
Example Output
(-50.00,86.60) (-86.60,50.00) (-36.60,136.60) (1.00,1.00)
Hint
Author
题意：已知一个等边三角形的两个顶点A(x1,y1)和B(x2,y2)，求第三个顶点，按照逆时针顺序求。
A(x1,y1) B(x2,y2)
设x=x2-x1,y=y2-y1，B绕A点旋转a度之后到C，则C的坐标为：
C( x*cos(a)-y*sin(a)+x1 , y*cos(a)+x*sin(a)+y1)。
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
int main()
{int T;double x1, y1, x2, y2, x, y, c1, c2;cin >> T;while(T--){cin >> x1 >> y1 >> x2 >> y2;x = x2 - x1;y = y2 - y1;c1 = x/2 - y*sqrt(3)/2 + x1;c2 = y/2 + x*sqrt(3)/2 + y1;printf("(%.2f,%.2f)\n", c1, c2);}return 0;
}