Uva 10250 The Other Two Trees

题目链接：https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1191

题意：有一个平行四边形然后分别以四条边为边长向外作正方形，四棵树在四个正方形中心的位置。其实四棵树构成正方形，相当于给出正方形的对角线的两个点，求另外两个。

证明正方形如下：来源：http://m.blog.csdn.net/blog/u014491224/29647971

下面证△FOE≌△HGO

 

设CD=2a，BC=2b

 

∴EF=OH=a,

    OF=HG=b

 

易知∠DFO=∠OHB

 

∴∠EFO=∠OHG=∠DFO+90°

∴△FOE≌△HGO（边角边）

 

∴∠E=∠GOH

∠EOG=∠EOF+∠DFO+∠E=90°      (因为再加上∠DFE就等于内角和180°了)

 

综上，OE=OG且OE⊥OG

 

由于对称性，四棵树就构成了个正方形。

需要注意pi

 代码：

#include<cstdio>
#include<cmath>
#include<iostream>
using namespace std;
const double pi=acos(-1);
const double eps=1e-10;
double cmp(double x)
{
    if(fabs(x)<eps) return 0;
    else return x;
}
struct Point
{
    double x,y;
    Point(double x,double y):x(x),y(y){};
};
typedef Point Vector;
Vector Rotate(Vector a,double rad)
{
    return Vector(a.x*cos(rad)-a.y*sin(rad),a.x*sin(rad)+a.y*cos(rad));
}
Vector reduce(Vector a,double time)
{
    return Vector(a.x*time,a.y*time);
}
int main()
{
    double x1,x2,y1,y2;
    while(cin>>x1>>y1>>x2>>y2)
    {
        Vector hypo(x2-x1,y2-y1);
        Vector v=reduce(hypo,sqrt(2)/2);
        Vector v1=Rotate(v,pi/4);
        Vector v2=Rotate(v,-pi/4);
        //cout<<v1.x<<" "<<v1.y<<" "<<v2.x<<" "<<v2.y<<endl;
        Vector ans1(cmp(v1.x+x1),cmp(v1.y+y1));
        Vector ans2(cmp(v2.x+x1),cmp(v2.y+y1));
        printf("%.10lf %.10lf %.10lf %.10lf\n",ans1.x,ans1.y,ans2.x,ans2.y);
        //cout<<ans1.x<<" "<<ans1.y<<" "<<ans2.x<<" "<<ans2.y<<endl;
    }
    return 0;
}