向量旋转 UPC 2217

这道题目是13山东省省赛的签到题，题目大意是给等边三角形的两个定点，让求逆时针旋转之后的第三个点的坐标，原来不会向量的旋转，在网上找了找，找到一篇挺好的，直接贴过来。

向量的旋转

实际做题中我们可能会遇到很多有关及计算几何的问题，其中有一类问题就是向量的旋转问题，下面我们来具体探讨一下有关旋转的问题。

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如图：A' 就是A点绕B点旋转θ角度后得到的点，问题是我们要如何才能得到A' 点的坐标。（向逆时针方向旋转角度正，反之为负）研究一个点绕另一个点旋转的问题，我们可以先简化为一个点绕原点旋转的问题，这样比较方便我们的研究。之后我们可以将结论推广到一般的形式上。令B是原点，我们先以A点向逆时针旋转为例，我们过A' 做AB的垂线，交AB于C，过C做x轴的平行线交过A' 做x轴的垂线于D。过点C做x轴的垂线交x轴于点E。

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" alt="" name="角度" width="226" height="215" align="left" border="0" hspace="12" />A的坐标(x,y)，A' 坐标(x1,y1)，B的坐标(0,0)。我们可以轻松的获取AB的长度，而且显而易见A' B长度等于AB。假设我们已知θ角的大小那么我们可以很快求出BC和A' C的长度。BC=A' B x cosθ，A' C=A' B x sinθ。

因为∠A' CB和∠DCE为直角（显然的结论），则∠A' CD +∠DCB =∠ECD +∠DCB=90度。

则∠A' CD=∠ECD，∠A' DC=∠CEB=90度，因此可以推断⊿CA' D ∽⊿CBE。由此可以退出的结论有：

BC/BE=A' C/A' D和BC/CE=A' C/CD

当然了DC和A' D都是未知量，需要我们求解，但是我们却可以通过求出C点坐标和E点坐标间接获得A' C和CD的长度。我们应该利用相似的知识求解C点坐标。

C点横坐标等于：((|AB| x cosθ) / |AB|) * x = x*cosθ

C点纵坐标等于：((|AB| x cosθ) / |AB|) * y = y*cosθ

则CE和BE的的长度都可以确定。

我们可以通过相⊿CA' D ∽⊿CBE得出：

AD = x * sinθ   DC = y * sinθ

那么接下来很容易就可以得出：

x1 = x*cosθ- y * sinθ y1 = y*cosθ + x * sinθ

则A' 的坐标为(x*cosθ- y * sinθ, y*cosθ + x * sinθ)

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)

接下来我们对这个结论进行一下简单的推广，对于任意两个不同的点A和B（对于求点绕另一个点旋转后的坐标时，A B重合显然没有太大意义），求A点绕B点旋转θ角度后的坐标，我们都可以将B点看做原点，对A和B进行平移变换，计算出的点坐标后，在其横纵坐标上分别加上原B点的横纵坐标，这个坐标就是A' 的坐标。

推广结论：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 )//结论

注：xB、yB为B点坐标。

结论的进一步推广：对于任意非零向量AB（零向量研究意义不大），对于点C进行旋转，我们只需求出点A和B对于点C旋转一定角度的坐标即可求出旋转后的向量A' B' ，因为向量旋转后仍然是一条有向线段。同理，对于任意二维平面上的多边形旋转也是如此。

以上是比较简单的，最后把结论记好就行了。上面是绕B点进行旋转的，所以Δx = x_a - x_b, Δy = y_a - y_{b，如果绕A反过来就行了。}

题目代码：

/*************************************************************************
    > File Name:            upc_2217.cpp
    > Author:               Howe_Young
    > Mail:                 1013410795@qq.com
    > Created Time:         2015年04月29日 星期三 20时03分21秒
 ************************************************************************/

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>

using namespace std;

int main()
{
    int t;
    double x1, y1, x2, y2;
    scanf("%d", &t);
    while (t--)
    {
        scanf("%lf %lf %lf %lf", &x1,  &y1, &x2, &y2);
        double x3, y3;
        x3 = (x2 - x1) * 0.5 - (y2 - y1) * sqrt(3.0) / 2.0 + x1;
        y3 = (y2 - y1) * 0.5 + (x2 - x1) * sqrt(3.0) / 2.0 + y1;
        printf("(%.2f,%.2f)\n", x3, y3);
    }
    return ;
}