《C++第十二周实验报告4-1》---类的组合与继承
2012-05-07 22:43
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/* 【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: vehicle.cpp * 作 者: 计114-3 王兴锋 * 完成日期: 2012 年 5 月 7 日 * 版 本 号: V 1.0 */ #include <iostream> #include <cmath> using namespace std; #define PI 3.1415926 //自行定义类 class Point //定义坐标点类 { public: double x, y;//点的横坐标和纵坐标 Point(){x = 0; y = 0;} ~Point(){} Point(double x0, double y0) {x = x0; y = y0;} void setXY(double x0, double y0){x = x0; y = y0;} double getX(){return x;} double getY(){return y;} friend ostream& operator << (ostream&, Point&); friend istream& operator >> (istream&, Point&); }; ostream& operator << (ostream& output, Point& p) { output << "(" << p.x << "," << p.y << ")" << endl; return output; } istream& operator >> (istream& input, Point& p) { cout << "输入点的横纵坐标:(x y)"; input >> p.x >> p.y; return input; } class Circle : public Point { protected: double r; public: Circle(){} Circle(double x, double y, double r = 0):Point(x, y){this->r = r;} ~Circle(){} void setR(double r){this->r = r;} double getR(){return r;} double getL(){return 2*PI*r;} double getAre(){return PI*r*r/2;} friend ostream& operator << (ostream&, Circle&); friend istream& operator >> (istream&, Circle&); friend double locate(Point p, Circle c); bool operator > (Circle &c); bool operator < (Circle &c); bool operator >= (Circle &c); bool operator <= (Circle &c); bool operator == (Circle &c); bool operator != (Circle &c); void crossover_point1(Point p, Circle c, Point &p1, Point &p2); }; void crossover_point1(Point p, Circle c, Point &p1, Point &p2) { double len = c.getR()/(locate(p, c) + c.getR()); double x = (p.x-c.x)*len; double y = (p.y-c.y)*len; p1.x = c.x + x; p1.y = c.y + y; p2.x = c.x - x; p2.y = c.y - y; } bool Circle::operator > (Circle &c) { if (getAre() > c.getAre()) return true; else return false; } bool Circle::operator < (Circle &c) { if (getAre() < c.getAre()) return true; else return false; } bool Circle::operator >= (Circle &c) { if (getAre() >= c.getAre()) return true; else return false; } bool Circle::operator <= (Circle &c) { if (getAre() <= c.getAre()) return true; else return false; } bool Circle::operator == (Circle &c) { if (getAre() == c.getAre()) return true; else return false; } bool Circle::operator != (Circle &c) { if (getAre() != c.getAre()) return true; else return false; } double locate(Point p, Circle c) { return (sqrt((c.x - p.x)*(c.x - p.x) + (c.y - p.y)*(c.y - p.y)) - c.r); } ostream& operator << (ostream& out, Circle& c) { out << "圆心:" << "(" << c.x << "," << c.y << ")" << endl; out << "半径:" << c.r << endl; return out; } istream& operator >> (istream& in, Circle& c) { cout << "输入圆心坐标:(x y)"; in >> c.x >> c.y; cout << "请输入圆的半径:(r)" ; in >> c.r; return in; } //用下面的main()函数测试 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; } /* 圆c1: 圆心:(3,2) 半径:4 点p1: (1,1) 点p1在圆c1之内 点p2: (3,-2) 点p2在圆c1之上 点p3: (7,3) 点p3在圆c1之外 圆c1: 圆心:(3,2) 半径:4 小于 小于等于 不等于 圆c2: 圆心:(3,2) 半径:4 点p1: (1,1) 与圆c1: 圆心:(3,2) 半径:4 的圆心相连,与圆交于两点,分别是: 交点: (-0.577709,0.211146) 交点: (6.57771,3.78885) 请按任意键继续. . . */
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