【题目】点类与圆类
(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
下面给出用于测试的main()函数,涉及到的类请自行定义
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: "<<c1; cout<<endl; Point p4,p5; crossover_point(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; }
【简析】本题首先要求弄清各个类的继承关系:Point为基类,Circle类为派生类。
Point类的成员应该包括x,y两个座标,除了main()函数中已经体现的对<<运算符的重载外,为了在Circle类中判断点与圆的位置关系,求两点间的距离是必要的,可以将之定义为Point类的成员函数(当然也可以是其他设计)。
对于Circle类,需要重载<<运算符(友元函数);根据上面main()函数中08、10、12行的调用形式,显然locate()函数应该为返回int型的友元函数;判断两圆大小关系的6个比较运算符处理为成员函数;crossover_point()函数的设计是个难点,从main()函数中可以看出,crossover_point()应该是一友元函数或一般函数,其返回值为void型,要求位置关系的点和圆分别是第1个参数p1和第2个参数c1,而后两个参数p4和p5则用于返回值(回想函数部分的相关知识,p4和p5应该为引用,这很重要。)
crossover_point()求解仅用基本的解析几何知识就可解决。学编程的孩子,千万不能被数学式子吓倒(本文后附了求解公式的推导),现在学的高数、后面的线性代数、概率统计,都是顶有用的。
【参考解答1】
#include <iostream> #include<Cmath> using namespace std; class Point { public: Point(double x=0,double y=0); //构造函数 double distance(const Point &p) const; //求距离 double getx() const {return x;} double gety() const {return y;} void setx(double x1){x=x1;} void sety(double y1){y=y1;} friend ostream & operator<<(ostream &,const Point &);//重载运算符“<<” protected: //受保护成员 double x,y; }; //Point的构造函数 Point::Point(double a,double b):x(a),y(b){} double Point::distance(const Point &p) const //求距离 { double dx = x-p.x; double dy = y-p.y; return sqrt(dx*dx+dy*dy); } ostream & operator<<(ostream &output,const Point &p) { output<<"["<<p.x<<","<<p.y<<"]"<<endl; return output; } class Circle:public Point //circle是Point类的公用派生类 { public: Circle(double x=0,double y=0,double r=0); //构造函数 double area ( ) const; //计算圆面积 friend ostream &operator<<(ostream &,const Circle &);//重载运算符“<<” friend int locate(const Point &p, const Circle &c); //判断点p在圆上、圆内或圆外,返回值:<0圆内,==0圆上,>0 圆外 //重载关系运算符(种)运算符,使之能够按圆的面积比较两个圆的大小; bool operator>(const Circle &); bool operator<(const Circle &); bool operator>=(const Circle &); bool operator<=(const Circle &); bool operator==(const Circle &); bool operator!=(const Circle &); //给定一点p,求出该点与圆c的圆心相连成的直线与圆的两个交点p1和p2(为了带回计算结果,p1和p2需要声明为引用) friend void crossover_point(Point &p,Circle &c, Point &p1,Point &p2 ) ; protected: double radius; }; //定义构造函数,对圆心坐标和半径初始化 Circle::Circle(double a,double b,double r):Point(a,b),radius(r){ } //计算圆面积 double Circle::area( ) const { return 3.14159*radius*radius; } //重载运算符“<<”,使之按规定的形式输出圆的信息 ostream &operator<<(ostream &output,const Circle &c) { output<<"Center=["<<c.x<<", "<<c.y<<"], r="<<c.radius<<endl; return output; } //判断点p在圆内、圆c内或圆c外 int locate(const Point &p, const Circle &c) { const Point cp(c.x,c.y); //圆心 double d = cp.distance(p); if (abs(d - c.radius) < 1e-7) return 0; //相等 else if (d < c.radius) return -1; //圆内 else return 1; //圆外 } //重载关系运算符(种)运算符,使之能够按圆的面积比较两个圆的大小; bool Circle::operator>(const Circle &c) { return (this->radius - c.radius) > 1e-7; } bool Circle::operator<(const Circle &c) { return (c.radius - this->radius) > 1e-7; } bool Circle::operator>=(const Circle &c) { return !(*this < c); } bool Circle::operator<=(const Circle &c) { return !(*this > c); } bool Circle::operator==(const Circle &c) { return abs(this->radius - c.radius) < 1e-7; } bool Circle::operator!=(const Circle &c) { return abs(this->radius - c.radius) > 1e-7; } //给定一点p,求出该点与圆c的圆心相连成的直线与圆的两个交点p1和p2(为了带回计算结果,p1和p2需要声明为引用) void crossover_point(Point &p, Circle &c, Point &p1,Point &p2 ) { p1.setx (c.getx() + sqrt(c.radius*c.radius/(1+((c.gety()-p.gety())/(c.getx()-p.getx()))*((c.gety()-p.gety())/(c.getx()-p.getx()))))); p2.setx (c.getx() - sqrt(c.radius*c.radius/(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),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_point(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; }【改进想法】
crossover_point()有4个参数的处理可行,但是否还能更直观些?该操作涉及圆与另外一点,作为圆的一个成员函数显然合理。这样,将另外一点作参数就好了。在返回值的处理上,利用Point类的两个引用返回值是可行的。直观的做法是,求两个交点,去接将两个交点(Point类的对象)返回不就行了?但函数返回值只有一个,我们用一个技巧,由两点组合成一个结构体Two_points即可解决(用成class或者是对象数组也可以)。
【参考解答2】
#include <iostream> #include<Cmath> using namespace std; class Point { public: Point(double x=0,double y=0); //构造函数 double distance(const Point &p) const; //求距离 double getx() const{return x;} double gety() const{return y;} void setx(double x1){x=x1;} void sety(double y1){y=y1;} friend ostream & operator<<(ostream &,const Point &);//重载运算符“<<” protected: //受保护成员 double x,y; }; //Point的构造函数 Point::Point(double a,double b):x(a),y(b){} double Point::distance(const Point &p) const //求距离 { double dx = x-p.x; double dy = y-p.y; return sqrt(dx*dx+dy*dy); } ostream & operator<<(ostream &output,const Point &p) { output<<"["<<p.x<<","<<p.y<<"]"<<endl; return output; } struct Two_points //专为crossover_point()函数返回值定义的结构体 { Point p1; Point p2; }; class Circle:public Point //circle是Point类的公用派生类 { public: Circle(double x=0,double y=0,double r=0); //构造函数 double area ( ) const; //计算圆面积 friend ostream &operator<<(ostream &,const Circle &);//重载运算符“<<” friend int locate(const Point &p, const Circle &c); //判断点p在圆上、圆内或圆外,返回值:<0圆内,==0圆上,>0 圆外 //重载关系运算符(种)运算符,使之能够按圆的面积比较两个圆的大小; bool operator>(const Circle &); bool operator<(const Circle &); bool operator>=(const Circle &); bool operator<=(const Circle &); bool operator==(const Circle &); bool operator!=(const Circle &); //再给一种解法:给定一点p,求出该点与圆c的圆心相连成的直线与圆的两个交点(返回Two_points型值) Two_points crossover_point(Point &p); protected: double radius; }; //定义构造函数,对圆心坐标和半径初始化 Circle::Circle(double a,double b,double r):Point(a,b),radius(r){ } //计算圆面积 double Circle::area( ) const { return 3.14159*radius*radius; } //重载运算符“<<”,使之按规定的形式输出圆的信息 ostream &operator<<(ostream &output,const Circle &c) { output<<"Center=["<<c.x<<", "<<c.y<<"], r="<<c.radius<<endl; return output; } //判断点p在圆内、圆c内或圆c外 int locate(const Point &p, const Circle &c) { const Point cp(c.x,c.y); //圆心 double d = cp.distance(p); if (abs(d - c.radius) < 1e-7) return 0; //相等 else if (d < c.radius) return -1; //圆内 else return 1; //圆外 } //重载关系运算符(种)运算符,使之能够按圆的面积比较两个圆的大小; bool Circle::operator>(const Circle &c) { return (this->radius - c.radius) > 1e-7; } bool Circle::operator<(const Circle &c) { return (c.radius - this->radius) > 1e-7; } bool Circle::operator>=(const Circle &c) { return !(*this < c); } bool Circle::operator<=(const Circle &c) { return !(*this > c); } bool Circle::operator==(const Circle &c) { return abs(this->radius - c.radius) < 1e-7; } bool Circle::operator!=(const Circle &c) { return abs(this->radius - c.radius) > 1e-7; } //再给一种解法:给定一点p,求出该点与圆的圆心相连成的直线与圆的两个交点(返回Two_points型值) Two_points Circle::crossover_point(Point &p) { Two_points pp; pp.p1.setx ( x + sqrt(radius*radius/(1+((y-p.gety())/(x-p.getx()))*((y-p.gety())/(x-p.getx()))))); pp.p2.setx ( x - sqrt(radius*radius/(1+((y-p.gety())/(x-p.getx()))*((y-p.gety())/(x-p.getx()))))); pp.p1.sety ( p.gety() + (pp.p1.getx() -p.getx())*(y-p.gety())/(x-p.getx())); pp.p2.sety ( p.gety() + (pp.p2.getx() -p.getx())*(y-p.gety())/(x-p.getx())); return pp; } 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; cout<<"用另外一种方法求交点:"<<endl; Two_points twoPoint = c1.crossover_point(p1); 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: "<<twoPoint.p1; cout<<"交点: "<<twoPoint.p2; system("pause"); return 0; }
【附】crossover_point()中求解公式的简单推导