sdut 2603 向量的旋转变换与旋转角的关系（山东第4届省赛）

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

    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.

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

2013年山东省第四届ACM大学生程序设计竞赛

本题在二维坐标中给出等边三角形其中的两个点（逆时针顺序），求第三个点的坐标

用到了向量的旋转变换：



向量p1(x1,y1)逆时针旋转

到p(x,y)，二者的坐标关系为





将向量AB平移到坐标原点求出旋转后的点，然后再平移回去即可

#include<iostream>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<vector>
#include<stack>
#include<queue>
#include<algorithm>
#include<sstream>
#define inf 0x3f3f3f3f
#define ll long long
using namespace std;
const double pi=3.141592;
int main()
{
int T;
cin>>T;
while(T--)
{
double x1,x2,y1,y2;
cin>>x1>>y1>>x2>>y2;
double d=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
double dx=x2-x1;
double dy=y2-y1;
double xx=dx*0.5-dy*sqrt(3)/2+x1;
double yy=dx*sqrt(3)/2+dy*0.5+y1;
printf("(%0.2f,%0.2f)\n",xx,yy);
}
return 0;
}