计算几何初步-向量的旋转 Rescue The Princess

题目地址：http://acm.upc.edu.cn/problem.php?id=2217

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 theequilateral 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 0.000.00 0.00 0.00 100.000.00 0.00 100.00 100.001.00 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)

告诉你是一个等边三角形并且给你两个点a(x1,y1)和b(x2,y2)，让你求第三个点，按照逆时针顺序求。

我这里是用的纯三角函数做的。

#include <stdio.h>#include <math.h>int n;double xx1,xx2,yy1,yy2;int main() {double l, atana, pi = 3.14159265358;freopen("in.txt", "r", stdin);scanf("%d", &n);while(n--){scanf("%lf %lf %lf %lf", &xx1,&yy1, &xx2, &yy2);double tmp = (double)(xx2-xx1) * (xx2-xx1) + (yy2-yy1)*(yy2-yy1);l = sqrt(tmp);atana = atan2((yy2-yy1) , (xx2-xx1)) + pi/3;printf("(%.2lf,%.2lf)\n", cos(atana)*l + xx1, sin(atana)*l + yy1);}return 0;}

这里有推导的公式：

http://www.cnblogs.com/E-star/archive/2013/06/11/3131563.html

我们可以这样认为：对于任意点A(x,y)，A非原点，绕原点旋转θ角后点的坐标为：(x*cosθ- y * sinθ, y*cosθ + x * sinθ)

因此又有此解法：

1. #include<stdio.h>  
2. #include<math.h>  
3. int main()  
4. {  
5.     int t;  
6.     scanf("%d",&t);  
7.     while(t--)  
8.     {  
9.         double x1,x2,y1,y2,x3,y3,x,y;  
10.         scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);  
11.         x=x2-x1;  
12.         y=y2-y1;  
13.         x3=x/2-y*sqrt(3)/2+x1;  
14.         y3=y/2+x*sqrt(3)/2+y1;  
15.         printf("(%.2f,%.2f)\n",x3,y3);  
16.     }  
17.     return 0;  
18. }  

http://www.cnblogs.com/E-star/archive/2013/06/11/3131563.html