第12周 项目4 点、圆的关系

问题及代码：

/*。*Copyright(c)2014,烟台大学计算机学院*All right reserved,*文件名：test.cpp*作者：liu_feng_zi_*完成日期：2015年5月31日*版本号：v1.0*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）；（4）定义友元函数double locate，判断点p与圆的位置关系（返回值<0圆内，==0圆上，>0 圆外）；*输入描述：*程序输出：*/#include <iostream>#include <cmath>using namespace std;class Point{protected:    int x,y;public:    Point(int a=0,int b=0):x(a),y(b) {}    friend ostream&operator<<(ostream&output,Point &p);    int getx()    {        return x;    }    int gety()    {        return y;    }};ostream&operator<<(ostream&output,Point &p){    output<<"("<<p.x<<","<<p.y<<")"<<endl;    return output;}class Circle:public Point{protected:    int r;public:    Circle(int a=0,int b=0,int c=0):        Point(a,b),r(c) {}    ~Circle() {}    friend ostream&operator<<(ostream&output,Circle &c);    friend double locate(Point p,Circle c);};ostream&operator<<(ostream&output,Circle &c){    output<<"[("<<c.x<<","<<c.y<<")"<<","<<c.r<<"]"<<endl;    return output;}double locate(Point p,Circle c){    int dx=p.getx()-c.x;    int dy=p.gety()-c.y;    double d=sqrt(dx*dx+dy*dy);    double t=d-c.r;    return t;}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;    return 0;}

运行结果：