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;}