第十二周项目4 ： 点、圆的关系

问题及代码：
（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）；

（4）定义友元函数int locate，判断点p与圆的位置关系（返回值<0圆内，==0圆上，>0 圆外）；

int main( ){Circle c1(3,2,4),c2(4,5,5);      //c2应该大于c1Point 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;return 0;}

解决代码：

#include<iostream>#include <cmath>using namespace std;class Point{public:    Point(double a=0,double b=0):x(a),y(b){};    double distance(Point &p);    friend ostream &operator<<(ostream &,const Point &);protected:    double x;    double y;};ostream &operator<<(ostream &output,const Point &p){    output<<"("<<p.x<<","<<p.y<<")"<<endl;    return output;}double Point::distance(Point &p){    return sqrt((x-p.x)*(x-p.x)+(y-p.y)*(y-p.y));}class Circle:public Point{public:    Circle(double a,double b,double c):Point(a,b),r(c){};    friend ostream &operator<<(ostream &,const Circle &);    friend int locate(Point &p,const Circle &c);protected:    double r;};ostream &operator<<(ostream &output,const Circle &c){    output<<"center:("<<c.x<<","<<c.y<<"),r="<<c.r<<endl;    return output;}int locate(Point &p,const Circle &c){    Point cp(c.x,c.y);    double d=cp.distance(p);    if(abs(d-c.r)<1e-7)        return 0;    if(d<c.x)        return -1;    if(d>c.x)        return 1;}int main( ){Circle c1(3,2,4),c2(4,5,5);      //c2应该大于c1Point 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;return 0;}

运行结果：