Problem E: 点在圆内吗？

Description

定义一个Point类和Circle类，用于判断给定的一系列的点是否在给定的圆内。

其中，Point类：

1.有2个成员x和y，分别为其横坐标和纵坐标；1个静态成员numOfPoints，用于计算生成的点的个数。

2.具有构造函数、析构函数和拷贝构造函数，具体格式输出根据样例自行判断。

3. 具有静态方法int getNumOfPoints()，用于返回numOfPoints的值。

4. 具有int getX()和int getY()方法，用于获取横坐标和纵坐标。

Circle类：

1. 拥有Point类的对象center，表示圆心坐标。拥有radius对象，表示圆的半径；1个静态成员numOfCircles，用于指示生成了多少个圆对象。

2. 具有构造函数、析构函数和拷贝构造函数，具体格式根据样例自行判断。

3.具有静态方法int getNumOfCircles()，返回numOfCircles的值。

4. 具有getCenter()方法，返回圆心坐标。注意：根据输出结果判断返回值类型。

5. 具有bool pointInCircle(Point &)方法，用于判断给定的点是否在当前圆内。是则返回true，否则返回false。

Input

输入分多行。

第一行M>0，表示有M个测试用例。

每个测试用例又包括多行。第1行包含3个整数，分别为一个圆的横坐标、纵坐标和半径。第2行N>0，表示之后又N个点，每个点占一行，分别为其横坐标和纵坐标。

所有输入均为整数，且在int类型范围内。

Output

输出见样例。注意：在圆的边上的点，不在圆内。

Sample Input

0 0 10

2 2

11 11

10 0

1 1 20

2 2

1 1

100 100

Sample Output

The Point (0, 0) is created!Now, we have 1 points.

The Point (1, 1) is created! Now, we have 2 points.

A circle at (1, 1) and radius 1 is created! Now, we have 1 circles.

We have 2 points and 1 circles now.

The Point (0, 0) is created! Now, we have 3 points.

A Point (0, 0) is copied! Now, we have 4 points.

A Point (0, 0) is copied! Now, we have 5 points.

A circle at (0, 0) and radius 10 is created! Now, we have 2 circles.

A Point (0, 0) is erased! Now, we have 4 points.

The Point (2, 2) is created! Now, we have 5 points.

(2, 2) is in the circle at (0, 0).

The Point (11, 11) is created! Now, we have 6 points.

(11, 11) is not in the circle at (0, 0).

The Point (10, 0) is created! Now, we have 7 points.

(10, 0) is not in the circle at (0, 0).

A Point (0, 0) is erased! Now, we have 6 points.

A circle at (0, 0) and radius 10 is erased! Now, we have 1 circles.

A Point (0, 0) is erased! Now, we have 5 points.

The Point (1, 1) is created! Now, we have 6 points.

A Point (1, 1) is copied! Now, we have 7 points.

A Point (1, 1) is copied! Now, we have 8 points.

A circle at (1, 1) and radius 20 is created! Now, we have 2 circles.

A Point (1, 1) is erased! Now, we have 7 points.

The Point (2, 2) is created! Now, we have 8 points.

(2, 2) is in the circle at (1, 1).

The Point (1, 1) is created! Now, we have 9 points.

(1, 1) is in the circle at (1, 1).

The Point (100, 100) is created! Now, we have 10 points.

(100, 100) is not in the circle at (1, 1).

A Point (1, 1) is erased! Now, we have 9 points.

A circle at (1, 1) and radius 20 is erased! Now, we have 1 circles.

A Point (1, 1) is erased! Now, we have 8 points.

We have 8 points, and 1 circles.

A circle at (1, 1) and radius 1 is erased!Now, we have 0 circles.

A Point (1, 1) is erased! Now, we have 7 points.

A Point (0, 0) is erased! Now, we have 6 points.

HINT

Append Code

#include<iostream>

using namespace std;

class Point

{

private:

int x,y;

static int numOfPoints;

public:

Point(int a,int b):x(a),y(b){numOfPoints++;cout<<"The Point ("<<x<<", "<<y<<") is created! Now, we have "<<numOfPoints<<

" points.
"

;}

~Point(){numOfPoints--;cout<<"A Point ("<<x<<", "<<y<<") is erased! Now, we have "<<numOfPoints<<

" points.
"

;}

Point(const Point &q):x(q.x),y(q.y){numOfPoints++;cout<<"A Point ("<<x<<", "<<y<<") is copied! Now, we have "<<numOfPoints<<

" points.
"

;}

static int getNumOfPoints(){return numOfPoints;}

int getX(){return x;}

int getY(){return y;}

};

int Point::numOfPoints=0;

class Circle

{

private:

Point center;

int radius;

static int numOfCircles;

public:

Circle(Point b,int a):center(b),radius(a){numOfCircles++;cout<<"A circle at ("<<center.getX()<<", "<<center.getY()<<") and radius "<<radius<<" is created! Now, we have "<<numOfCircles<<

" circles.
"

;}

Circle(int a,int b,int c):center(a,b),radius(c){numOfCircles++;cout<<"A circle at ("<<center.getX()<<", "<<center.getY()<<") and radius "<<radius<<" is created! Now, we have "<<numOfCircles<<

" circles.
"

;}

~Circle(){numOfCircles--;cout<<"A circle at ("<<center.getX()<<", "<<center.getY()<<") and radius "<<radius<<" is erased! Now, we have "<<numOfCircles<<

" circles.
"

;}

static int getNumOfCircles(){return numOfCircles;}

Point &getCenter(){return center;}

bool pointInCircle(Point &q)

{

int b=(q.getX()-center.getX())*(q.getX()-center.getX())+(q.getY()-center.getY())*(q.getY()-center.getY());

if(b<radius*radius)

return 1;

return 0;

}

};

int Circle::numOfCircles=0;

int main()

{

int cases,num;

int x, y, r, px, py;

Point aPoint(0,0), *bPoint;

Circle aCircle(1,1,1);

cin>>cases;

cout<<"We have "<<Point::getNumOfPoints()<<" points and "<<Circle::getNumOfCircles()<<" circles now."<<endl;

for (int i = 0; i < cases; i++)

{

cin>>x>>y>>r;

bPoint = new Point(x,y);

Circle circle(*bPoint, r);

cin>>num;

for (int j = 0; j < num; j++)

{

cin>>px>>py;

if (circle.pointInCircle(*(new Point(px, py))))

{

cout<<"("<<px<<", "<<py<<") is in the circle at (";

cout<<circle.getCenter().getX()<<", "<<circle.getCenter().getY()<<")."<<endl;

}

else

{

cout<<"("<<px<<", "<<py<<") is not in the circle at (";

cout<<circle.getCenter().getX()<<", "<<circle.getCenter().getY()<<")."<<endl;

}

delete bPoint;

}

cout<<"We have "<<Point::getNumOfPoints()<<" points, and "<<Circle::getNumOfCircles()<<" circles."<<endl;

return 0;

}