实验9：Problem D: 从点到面

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Problem D: 从点到面

Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 529 Solved: 332
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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

append.cc,

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All Copyright Reserved 2010-2011 SDUSTOJ TEAM
GPL2.0 2003-2011 HUSTOJ Project TEAM
Anything about the Problems, Please Contact Admin:admin

#include<iostream>
using namespace std;
class Point{friend class Rectangle;
public:int x,y;Point(int a,int b):x(a),y(b){cout<<"A point ("<<x<<", "<<y<<") is created!"<<endl;}~Point(){cout<<"A point ("<<x<<", "<<y<<") is erased!"<<endl;}Point(const Point &p){x=p.x;y=p.y;cout<<"A point ("<<x<<", "<<y<<") is copied!"<<endl;}int getX(){return x;}int getY(){return y;}};
class Rectangle{friend class Point;
public:Point leftTop,rightBottom;Rectangle(Point p,Point q):leftTop(p),rightBottom(q){cout<<"A rectangle ("<<leftTop.x<<", "<<leftTop.y<<") to ("<<rightBottom.x<<", "<<rightBottom.y<<") is created!"<<endl;}Rectangle(int a,int b,int c,int d):leftTop(a,b),rightBottom(c,d){cout<<"A rectangle ("<<leftTop.x<<", "<<leftTop.y<<") to ("<<rightBottom.x<<", "<<rightBottom.y<<") is created!"<<endl;}~Rectangle(){cout<<"A rectangle ("<<leftTop.x<<", "<<leftTop.y<<") to ("<<rightBottom.x<<", "<<rightBottom.y<<") is erased!"<<endl;}Point &getLeftTop(){return leftTop;}Point getRightBottome(){return rightBottom;}int getArea(){return (rightBottom.x-leftTop.x)*(leftTop.y-rightBottom.y);}
};
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;
}

转载于:https://www.cnblogs.com/auto1945837845/p/5447296.html