挑战程序设计 CGL_1+B:Reflection

#include<bits/stdc++.h>
#define eps (1e-10)
using namespace std;
class Point
{
	public:
	double x,y;
	Point(double _x=0,double _y=0):x(_x),y(_y){}
	Point operator + (Point p){ return Point(x+p.x,y+p.y);} 
	Point operator - (Point p){ return Point(x-p.x,y-p.y);}
	Point operator * (double a){return Point(a*x,a*y);} 
	Point operator / (double a){return Point(x/a,y/a);}
	double norm(){return x*x+y*y;}
	double ABS(){return sqrt(norm());} 
};
struct Segment
{
	Point p1,p2;	
};
//判断是否正交 向量内积：a*b=|a||b|*cos(Y) 当cos为0时正交(90,-90垂直) 
//a*b=a.x*b.x+a.y*b.y;
bool solve1(Point a,Point b)//是否正交 
{
	double f=a.x*b.x+a.y*b.y; 
	if(fabs(f-0.0)<eps) return true;
	else return false;
}
//判断是否平行  向量外积：|a*b|=|a||b|sin(Y) 当sin为0时平行(180,0平行)
//|a*b|=|a|*|b|*sin(Y) 
bool solve2(Point a,Point b)//是否平行 
{
	double f=a.x*b.y-a.y*b.x;
	if(fabs(f-0.0)<eps) return true;
	else return false;
}
double dot(Point a,Point b)
{
	return a.x*b.x+a.y*b.y;
}
/*
求垂足x：对于给定的三点p1,p2,p从点p向通过p1,p2的直线引一条垂线
base=p2-p1;
hypo=p-p1;
 x=s.p1+base*(hypo*base/|base|^2); hypo*base 可以用向量内积求 
*/
Point solve3(Segment s,Point p)//求投影点 
{
	Point base=s.p2-s.p1;
	double r=dot(p-s.p1,base)/base.norm();
	return s.p1+base*r; 
}
/*
求投影点x：对于给定的三点p1,p2,p从点p向通过p1,p2的直线为对称轴与点p
成线对称点为x 
通过求solve3的垂足延长一倍就可以求x 
*/
Point solve4(Segment s,Point p)//求映象 
{
	return p+(solve3(s,p)-p)*2.0;
}
int main()
{
	double x1,y1,x2,y2;
	scanf("%lf %lf %lf %lf",&x1,&y1,&x2,&y2);
	Point p1(x1,y1),p2(x2,y2);
	Segment s;
	s.p1=p1,s.p2=p2;
	int T;scanf("%d",&T);
	while(T--){
		double x,y;scanf("%lf %lf",&x,&y);
		Point p(x,y);
		Point w=solve4(s,p);
		printf("%.10f %.10f\n",w.x,w.y);
	}
	return 0;
}