第十二周任务四

分手后的思念是犯贱 2022-06-12 06:06 186阅读 0赞

/* (程序头部注释开始)
    * 程序的版权和版本声明部分
    * Copyright (c) 2012, 烟台大学计算机学院学生 
    * All rights reserved.
    * 文件名称：  点与圆
    
    * 作 者：     薛广晨
    * 完成日期：  2012 年 5 月 1 日
    * 版 本 号：  x1.0
    * 对任务及求解方法的描述部分
    * 输入描述：  
    
    * 问题描述：  
    *【任务4】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
    
    * 程序输出： ......
    
    * 程序头部的注释结束
    */
    
    #include <iostream>
    
    #define PI 3.1415926
    
    #include<Cmath>
    
    using namespace std;
    
    class Point //定义坐标点类
    {
    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 & input, Point & c); 
    protected:
    	double x, y;   //点的横坐标和纵坐标
    };  
    
    ostream &operator << (ostream & output, Point & c)
    {
    	output << "Point:(" << c.x << ", " << c.y << ")";
    	return output;
    }
    
    
    class Circle : public Point
    {
    public:
    	Circle(){r = 0;}
    	Circle(double x0, double y0, double r);     //构造函数   
        ~Circle(){};  
        double getr(){return r;} 
        friend ostream &operator << (ostream & out, Circle & c);  
    	friend double locate(Point &p, Circle &c); 
        friend void crossover_point1(Point &p1,Circle &c1,Point &p4,Point &p5);    
        bool operator > (Circle &t);      
        bool operator < (Circle &t);      
        bool operator >= (Circle &t);      
        bool operator <= (Circle &t);      
        bool operator == (Circle &t);      
        bool operator != (Circle &t);
    	double area0();
    private:
    	double r;
    };
    
    Circle :: Circle(double x0, double y0, double r1) : Point(x0, y0), r(r1){}    //构造函数 
    
    ostream &operator << (ostream & output, Circle & c)  
    {  
        output << "以" << "(" << c.getx() << "," << c.gety() << ")" << "为圆心， "  << "以" << c.r << "为半径的圆" << endl;  
      
        return output;  
    }  
    
    double Circle :: area0()
    {
    	return PI * r * r;
    }
    
    bool Circle ::operator > (Circle &t) 
    {
    	if(area0() > t.area0()) return true;
    	else return false;
    }
    
    bool Circle ::operator < (Circle &t)
    {
    	if(area0() < t.area0()) return true;
    	else return false;
    }
    
    bool Circle ::operator >= (Circle &t)
    {
    	if(area0() < t.area0()) return false;
    	else return true;
    }
    
    bool Circle ::operator <= (Circle &t) 
    {
    	if(area0() > t.area0()) return false;
    	else return true;
    }
    bool Circle ::operator == (Circle &t)
    {
    	if(area0() >= t.area0() && area0() <= t.area0()) return true;
    	else return false;
    }
    
    bool Circle ::operator != (Circle &t) 
    {
    	if(area0() > t.area0() || area0() < t.area0()) return true;
    	else return false;
    }
    
    double locate(Point &p, Circle &c)
    {
    	double d;
    	d = (p.getx() - c.getx()) * (p.getx() - c.getx()) + (p.gety() - c.gety()) * (p.gety() - c.gety());
    	return (d - c.getr() * c.getr());
    }
    
    void crossover_point1(Point & p1, Circle & c1, Point & p4, Point & p5)  
    {  
    	 double d;
        double x0 = (c1.getx() - p1.getx()) * c1.r / sqrt((c1.getx() - p1.getx()) * (c1.getx() - p1.getx()) + (c1.gety() - p1.gety()) * (c1.gety() - p1.gety()));  
        double y0 = (c1.gety() - p1.gety()) * c1.r / sqrt((c1.getx() - p1.getx()) * (c1.getx() - p1.getx()) + (c1.gety() - p1.gety()) * (c1.gety() - p1.gety()));  
      
        d = c1.getx() + x0; 
    	p4.setx(d);
        d = c1.gety() + y0;
    	p4.sety(d);
      
        d = c1.getx() - x0; 
    	p5.setx(d);
        d = c1.gety() - y0;
    	p5.sety(d);
    }  
    //方法二
    /*void crossover_point1(Point &p1,Circle &c1,Point &p4,Point &p5)
    {
    	double d;
    	d = c1.getx() + (p1.getx() - c1.getx()) * c1.getr() / sqrt((p1.getx() - c1.getx()) * (p1.getx() - c1.getx()) + (p1.gety() - c1.gety()) * (p1.gety() - c1.gety()));
    	p4.setx(d);
    
    	d = c1.getx() - (p1.getx() - c1.getx()) * c1.getr() / sqrt((p1.getx() - c1.getx()) * (p1.getx() - c1.getx()) + (p1.gety() - c1.gety()) * (p1.gety() - c1.gety()));
    	p5.setx(d);
    
    	d = c1.gety() + (p1.gety() - c1.gety()) * c1.getr() / sqrt((p1.getx() - c1.getx()) * (p1.getx() - c1.getx()) + (p1.gety() - c1.gety()) * (p1.gety() - c1.gety()));
    	p4.sety(d);
    
    	d = c1.gety() - (p1.gety() - c1.gety()) * c1.getr() / sqrt((p1.getx() - c1.getx()) * (p1.getx() - c1.getx()) + (p1.gety() - c1.gety()) * (p1.gety() - c1.gety()));
    	p5.sety(d);
    }*/
    
    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_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;
    }
    
    
     上机感言：通过这个题我进一步的了解了类，类内成员的调用方式也更加熟悉

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