Description

一個矩形可以由左上角和右下角的頂點而唯一確定。現在請定義兩個類：Point和Rectangle。

其中Point類有x和y兩個屬性（均為int類型），表示二維空間內一個點的橫縱坐標，並具有相應的構造函數、析構函數和拷貝構造函數。此外，還有getX()和getY()方法用以得到一個點的坐標值。

Rectangle類有leftTop和rightBottom兩個屬性（均為Point類的對象），表示一個矩形的左上角和右下角的兩個點，並具有相應的構造函數、析構函數。此外，還有getLeftTop()、getRightBottom()方法用於獲取相應的左上角點、右下角點，getArea()方法用以獲取面積。

Input

輸入有多行。

第一行是一個正整數M，表示後面有M個測試用例。

每個測試用例占一行，包括4個正整數，分別為左上角的橫坐標、縱坐標，右下角的橫坐標、縱坐標。

註意：

1.請根據輸出樣例判斷兩個類中相應方法的書寫方法。

2. 假定屏幕的左下角為坐標原點。

Output

輸出見樣例。

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

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;
}

#include <iostream>
using namespace std;
class Point
{
public :
    double x_, y_;
    Point(double x=0):x_(x), y_(x)
    {
        cout<<"A point ("<<x_<<", "<<y_<<") is created!"<<endl;
    }
    Point(double x, double y):x_(x), y_(y)
    {
        cout<<"A point ("<<x_<<", "<<y_<<") is created!"<<endl;
    }
    Point(const Point &p)
    {
        x_=p.x_; y_=p.y_;
        cout<<"A point ("<<x_<<", "<<y_<<") is copied!"<<endl;
    }
    double getX(){return x_;}
    double getY(){return y_;}
    ~Point()
    {
        cout<<"A point ("<<x_<<", "<<y_<<") is erased!"<<endl;
    }
};
class Rectangle
{
public :
    Point leftTop, rightBottom;
    double x1, y1, x2, y2;
    Rectangle(double a, double b, double c, double d):leftTop(a,b), rightBottom(c,d)
    {
        x1=a; y1=b; x2=c; y2=d;
        cout<<"A rectangle ("<<a<<", "<<b<<") to ("<<c<<", "<<d<<") is created!"<<endl;
    }
    Point &getLeftTop()
    {
        return leftTop;
    }
    Point getRightBottome()//本題純屬胡鬧，為了一個出現copy一個不出現copy一個加&一個不加&。
    {
        return rightBottom;
    }
    double getArea()
    {
        return(x1-x2)*(y2-y1);
    }
    ~Rectangle()
    {
         cout<<"A rectangle ("<<x1<<", "<<y1<<") to ("<<x2<<", "<<y2<<") is erased!"<<endl;
    }

};
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 C: 從點到面