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Problem D: 从点到面

2024-09-08 14:15:03   215人阅读

Description

一个矩形可以由左上角和右下角的顶点而唯一确定。现在请定义两个类：Point和Rectangle。

其中Point类有x和y两个属性（均为int类型），表示二维空间内一个点的横纵坐标，并具有相应的构造函数、析构函数和拷贝构造函数。此外，还有getX()和getY()方法用以得到一个点的坐标值。

Rectangle类有leftTop和rightBottom两个属性（均为Point类的对象），表示一个矩形的左上角和右下角的两个点，并具有相应的构造函数、析构函数。此外，还有getLeftTop()、getRightBottom()方法用于获取相应的左上角点、右下角点，getArea()方法用以获取面积。

Input

输入有多行。

第一行是一个正整数M，表示后面有M个测试用例。

每个测试用例占一行，包括4个正整数，分别为左上角的横坐标、纵坐标，右下角的横坐标、纵坐标。

注意：

1.请根据输出样例判断两个类中相应方法的书写方法。

2. 假定屏幕的左下角为坐标原点。

Output

输出见样例。

Sample Input

1
10 10 20 0

Sample Output

A point (10, 10) is created!
A point (20, 0) is created!
A rectangle (10, 10) to (20, 0) is created!
Area: 100
Left top is (10, 10)
A point (20, 0) is copied!
A point (20, 0) is copied!
Right bottom is (20, 0)
A point (20, 0) is erased!
A point (20, 0) is erased!
A rectangle (10, 10) to (20, 0) is erased!
A point (20, 0) is erased!
A point (10, 10) is erased!

HINT

Append Code

#include<iostream>
using namespace std;
class Point
{
private:
    int x,y;
public:
    Point(int a,int b){x=a,y=b;cout<<"A point ("<<x<<", "<<y<<") is created!\n";}
    ~Point(){cout<<"A point ("<<x<<", "<<y<<") is erased!\n";}
    Point(const Point &p){x=p.x,y=p.y;cout<<"A point ("<<x<<", "<<y<<") is copied!\n";}
    int getX(){return x;}
    int getY(){return y;}
};
class Rectangle
{
private:
    Point leftTop,rightBottom;//leftTop.x(a),leftTop.y(b),rightBottom.x(c),rightBottom.y(d)
public:
    Rectangle(int a,int b,int c,int d):leftTop(a,b),rightBottom(c,d){cout<<"A rectangle ("<<leftTop.getX()<<", "<<leftTop.getY()<<") to ("<<rightBottom.getX()<<", "<<rightBottom.getY()<<") is created!\n";}//
    //Rectangle(Point a,Point b):leftTop(a),rightBottom(b){cout<<"A rectangle ("<<leftTop.getX()<<", "<<leftTop.getY()<<") to ("<<rightBottom.getX()<<", "<<rightBottom.getY()<<") is created!\n";}
    ~Rectangle(){cout<<"A rectangle ("<<leftTop.getX()<<", "<<leftTop.getY()<<") to ("<<rightBottom.getX()<<", "<<rightBottom.getY()<<") is erased!\n";}
    Point &getLeftTop(){return leftTop;}
    Point getRightBottome(){return rightBottom;}
    int getArea(){return (rightBottom.getX()-leftTop.getX())*(leftTop.getY()-rightBottom.getY());}
};
int main()
{
    int cases;
    int x1, y1, x2, y2;
 
    cin>>cases;
    for (int i = 0; i < cases; i++)
    {
        cin>>x1>>y1>>x2>>y2;
        Rectangle rect(x1,y1,x2,y2);
        cout<<"Area: "<<rect.getArea()<<endl;
        cout<<"Left top is ("<<rect.getLeftTop().getX()<<", "<<rect.getLeftTop().getY()<<")"<<endl;
        cout<<"Right bottom is ("<<rect.getRightBottome().getX()<<", "<<rect.getRightBottome().getY()<<")"<<endl;
    }
    return 0;
}

Problem D: 从点到面


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