/*
*程序的版权和版本声明部分:
*Copyright(c)2014,烟台大学计算机学院学生
*All rights reserved.
*文件名称:
*作者:田成琳
*完成日期:2014 年 5 月 19 日
*版本号:v1.0
*对任务及求解方法的描述部分:
*输入描述: -
*问题描述:(1)先建立一个Point(点)类,包含数据成员x,y(坐标点);
(2)以Point为基类,派生出一个Circle(圆)类,增加数据成员(半径),基类的成员表示圆心;
(3)编写上述两类中的构造、析构函数及必要运算符重载函数(本项目主要是输入输出);
(6)给定一点p,求出该点与圆心相连成的直线与圆的两个交点并输出
*程序输出:点与圆的关系,圆与圆的大小关系
*问题分析:
*算法设计:
*/
#include<iostream.h>
//using namespace std;
#include<cmath>
const double pi=3.14;
class Point
{
public:
Point():x(0),y(0){}
Point(double X,double Y):x(X),y(Y){}
~Point(){}
double getX()
{
return x;
}
double getY()
{
return y;
}
void setX(double X)
{
x=X;
}
void setY(double Y)
{
y=Y;
}
friend ostream & operator << (ostream &out,Point &p);
protected:
double x,y;
};
class Circle:public Point
{
public:
Circle(double a,double b,double R):Point(a,b),r(R){}
~Circle(){}
double getR()
{
return r;
}
friend ostream & operator << (ostream &out,Circle &c);
private:
double r;
};
ostream & operator << (ostream &out,Point &p)
{
out<<"("<<p.getX()<<","<<p.getY()<<")"<<endl;
return out;
}
ostream & operator << (ostream &out,Circle &c)
{
out<<"("<<c.getX()<<","<<c.getY()<<")"<<" "<<"r="<<c.getR()<<endl;
return out;
}
void crossoverPoint(Point &p,Circle &c,Point &p1,Point &p2)
{//copy来至贺老师去年发布的这道题答案。。。
p1.setX (c.getX() + sqrt(c.getR()*c.getR()/(1+((c.getY()-p.getY())/(c.getX()-p.getX()))*((c.getY()-p.getY())/(c.getX()-p.getX())))));
p2.setX (c.getX() - sqrt(c.getR()*c.getR()/(1+((c.getY()-p.getY())/(c.getX()-p.getX()))*((c.getY()-p.getY())/(c.getX()-p.getX())))));
p1.setY (p.getY() + (p1.getX() -p.getX())*(c.getY()-p.getY())/(c.getX()-p.getX()));
p2.setY (p.getY() + (p2.getX() -p.getX())*(c.getY()-p.getY())/(c.getX()-p.getX()));
}
int main( )
{
Circle c1(3,2,4);
Point p1(1,1),p2,p3;
cout<<"点p1: "<<p1;
cout<<"与圆c1: "<<c1;
crossoverPoint(p1,c1,p2,p3);
cout<<"的圆心相连,与圆交于两点,分别是:"<<endl;
cout<<"交点: "<<p2;
cout<<"交点: "<<p3;
return 0;
}
运行结果:
心得体会:不会做啊不会做