首页系统综合问题假设有若干 (如10)个Circle,Rectangle和Square形状,计算总面积

假设有若干 (如10)个Circle,Rectangle和Square形状,计算总面积

时间2023-03-26 04:59:18发布分享专员分类系统综合问题浏览185

今天小编给各位分享rectangle的知识,文中也会对其通过假设有若干 (如10)个Circle,Rectangle和Square形状,计算总面积和java编程等多篇文章进行知识讲解,如果文章内容对您有帮助,别忘了关注本站,现在进入正文!

内容导航:

  • 假设有若干 (如10)个Circle,Rectangle和Square形状,计算总面积
  • java编程
  • java类求N个图形的面积和
  • 应用纯虚函数的使用,定义几个派生类如三角形,矩形,圆形等,求出所有形状的总面积。
  • 一、假设有若干 (如10)个Circle,Rectangle和Square形状,计算总面积

    package OctoberTwentyfour;

    abstract class Shape{

    protected String type;

    protected double area, perim;

    public abstract void computeArea();

    public String toString()

    { return type + ":Area = " + area + "Perimeter = " + perim;

    }

    }//父类

    package OctoberTwentyfour;

    class Rectangle extends Shape{

    protected double width, height;

    public Rectangle(double _width, double _height)

    { type = "Rectangle";

    width = _width;

    height = _height;

    }

    public void computeArea()

    { area = width * height; }

    public void computePerimeter()

    { perim = 2*width + 2*height; }

    public String toString()

    { return super.toString() + "Width = " + width + "Height = " + height;

    }

    }//矩形子类

    package OctoberTwentyfour;

    class Circle extends Shape

    { protected double radius;

    public Circle(double _radius)

    { type = "Circle";

    radius = _radius;

    }

    public void computeArea()

    { area = Math.PI * radius * radius; }

    public void computePerimeter()

    { perim = 2.0 * Math.PI * radius; }

    public String toString()

    { return super.toString() +"Radius = " + radius;

    }

    }

    //圆子类

    package OctoberTwentyfour;

    class Square extends Rectangle

    { public Square(double _side)

    { super(_side, _side);//重写父类方法super

    type = "Square";

    }

    }

    //继承于矩形的正方形子类

    package OctoberTwentyfour;

    public class ShapeTest

    { public static void main(String args[])

    { Shape shapes[] = new Shape[10];

    double total_area=0.0;

    int k=0;

    for (int i=0; i<shapes.length; i++ ){

    k=(int)Math.ceil(Math.random()*3);

    System.out.println("k is :"+k);

    switch (k) {

    case 1: shapes[i]=new Circle(Math.random()*10);

    break;

    case 2: shapes[i]=new Rectangle(Math.random()*10, Math.random()*10);

    break;

    case 3: shapes[i]=new Square(Math.random()*10);

    break;

    }

    }

    for (int i = 0; i < shapes.length; i++) {

    shapes[i].computeArea();

    total_area+=shapes[i].area;

    }

    System.out.println("Total area is :"+total_area);

    }

    }

    //测试

    运行结果图

    一、java编程

    //亲手为你所写

    abstract class Shape{
    public abstract double area();
    }

    class Circle extends Shape{
    int r;
    final double pi = 3.14;

    public Circle(int r){
    this.r= r;

    }
    public double area(){
    return pi*r*r;
    }
    }

    class Rectang extends Shape{
    int w;
    int h;

    public Rectang(int w,int h){
    this.w = w;
    this.h = h;
    }

    public double area(){
    return w*h;
    }
    }

    public class Test{
    public static void main(String[] args){
    double area_total=0;
    int i;

    Circle c1 = new Circle(2);
    Circle c2 = new Circle(3);

    Rectang rg1 = new Rectang(2,3);
    Rectang rg2 = new Rectang(3,4);

    Shape sshape[] = {c1,c2,rg1,rg2};
    for (i=0; i *pvec )
    {
    // ...
    for ( ; it != end_it; ++it )
    {
    Query *pq = *it;
    // ...
    delete pq;
    }
    }
    为了使函数能够正确执行,在应用delete 表达式时,必须调用pq 指向的动态类型的析构函数。为此,必须把Query 类析构函数声明为虚拟的。

    关于rectangle的问题,通过《java类求N个图形的面积和》、《应用纯虚函数的使用,定义几个派生类如三角形,矩形,圆形等,求出所有形状的总面积。》等文章的解答希望已经帮助到您了!如您想了解更多关于rectangle的相关信息,请到本站进行查找!

    爱资源吧版权声明:以上文中内容来自网络,如有侵权请联系删除,谢谢。

    rectangle
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