/* (程序头部注释开始)
* 程序的版权和版本声明部分
* Copyright (c) 2011, 烟台大学计算机学院学生
* All rights reserved.
* 文件名称: 第十二周任务四类的组合与继承
* 作 者:石丽君
* 完成日期: 2012 年 5 月 9 日
* 版 本 号:
* 对任务及求解方法的描述部分
* 输入描述:类的组合与继承
(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、析构函数及必要的输入输出函数
(4)定义友元函数int locate,判断点p在圆c上、圆c内或圆c外,返回值<0圆内,==0圆上,>0 圆外;
(5)重载关系运算符(6种)运算符,使之能够按圆的面积比较两个圆的大小;
(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
* 问题描述:
* 程序输出:
* 程序头部的注释结束
*/
#include <iostream>
#include<cmath>
using namespace std;
#define pi 3.1415926
class Point
{
protected:
double x,y;
public:
Point () {x=0;y=0;}
Point (double x0,double y0)
{
x=x0;
y=y0;
}
~Point();
double getx() {return x;}
double gety() {return y;}
void setx(double n) {x=n;}
void sety(double n) {y=n;}
friend ostream& operator<<(ostream &output,Point c);
};
class Circle:public Point
{
protected:
double r;
public:
Circle(double x0,double y0,double r0):Point(x0,y0), r(r0) {}
friend ostream&operator<< (ostream& output,Circle c);
bool operator >(Circle &c1);
bool operator <(Circle &c1);
bool operator >=(Circle &c1);
bool operator <=(Circle &c1);
bool operator ==(Circle &c1);
bool operator !=(Circle &c1);
friend void crossover_point1(Point &p1,Circle &c1, Point &p4, Point &p5);
friend double locate(Point &p,Circle &c);
double getr() {return r;}
~Circle();
};
ostream& operator<<(ostream &output,Point c)
{
output<<'('<<c.getx()<<','<<c.gety()<<')'<<endl;
return output;
}
ostream& operator<< (ostream& output,Circle c)
{
output<<"圆心坐标为:"<<"("<<c.getx()<<','<<c.gety()<<")"<<"半径为:"<<c.getr()<<endl;
return output;
}
Point::~Point()
{
}
Circle::~Circle()
{
}
bool Circle::operator >(Circle &c1)
{
if(r>c1.r)
return true;
return false;
}
bool Circle::operator <(Circle &c1)
{
if(r>c1.r)
return false;
return true;
}
bool Circle::operator >=(Circle &c1)
{
if(r>=c1.r)
return true;
return false;
}
bool Circle::operator <=(Circle &c1)
{
if(r>=c1.r)
return false;
return true;
}
bool Circle::operator ==(Circle &c1)
{
if(r==c1.r)
return true;
return false;
}
bool Circle::operator !=(Circle &c1)
{
if(r!=c1.r)
return true;
return false;
}
void crossover_point1(Point &p1,Circle &c1, Point &p4, Point &p5)
{
double n,m;
n=(c1.getx()-sqrt(((c1.r)*(c1.r))/((c1.gety()-p1.gety())/(c1.getx()-p1.getx()))*((c1.gety()-p1.gety())/(c1.getx()-p1.getx()))));
p4.setx(n);
n=(c1.getx()+sqrt(((c1.r)*(c1.r))/((c1.gety()-p1.gety())/(c1.getx()-p1.getx()))*((c1.gety()-p1.gety())/(c1.getx()-p1.getx()))));
p5.setx(n);
m=(c1.gety()-p1.gety())/(c1.getx()-p1.getx())*(p4.getx()-p1.getx())+p1.gety();
p4.sety(m);
m=(c1.gety()-p1.gety())/(c1.getx()-p1.getx())*(p5.getx()-p1.getx())+p1.gety();
p5.sety(m);
}
double locate(Point &p,Circle &c)
{
double d,d1;
d=sqrt((p.getx()-c.getx())*(p.getx()-c.getx())+(p.gety()-c.gety())*(p.gety()-c.gety()));
d1=d-c.r;
return d1;
}
int main( )
{
Circle c1(3,2,4),c2(4,5,5); //c2应该大于c1
Point p1(1,1),p2(3,-2),p3(7,3); //分别位于c1内、上、外
cout<<"圆c1: "<<c1;
cout<<"点p1: "<<p1;
cout<<"点p1在圆c1之"<<((locate(p1, c1)>0)?"外":((locate(p1, c1)<0)?"内":"上"))<<endl;
cout<<"点p2: "<<p2;
cout<<"点p2在圆c1之"<<((locate(p2, c1)>0)?"外":((locate(p2, c1)<0)?"内":"上"))<<endl;
cout<<"点p3: "<<p3;
cout<<"点p3在圆c1之"<<((locate(p3, c1)>0)?"外":((locate(p3, c1)<0)?"内":"上"))<<endl;
cout<<endl;
cout<<"圆c1: "<<c1;
if(c1>c2) cout<<"大于"<<endl;
if(c1<c2) cout<<"小于"<<endl;
if(c1>=c2) cout<<"大于等于"<<endl;
if(c1<=c2) cout<<"小于等于"<<endl;
if(c1==c2) cout<<"等于"<<endl;
if(c1!=c2) cout<<"不等于"<<endl;
cout<<"圆c2: "<<c2;
cout<<endl;
Point p4,p5;
crossover_point1(p1,c1, p4, p5);
cout<<"点p1: "<<p1;
cout<<"与圆c1: "<<c1;
cout<<"的圆心相连,与圆交于两点,分别是:"<<endl;
cout<<"交点: "<<p4;
cout<<"交点: "<<p5;
cout<<endl;
system("pause");
return 0;
}
经验积累:在使用void setx()函数时,其无返回值,在使用void crossover_point1(Point &p1,Circle &c1, Point &p4, Point &p5)时,应为个double getx()