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

2

0 0 10

3

2 2

11 11

10 0

1 1 20

3

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.\n"

;}

    

~Point(){numOfPoints--;cout<<

"A Point ("

<<x<<

", "

<<y<<

") is erased! Now, we have "

<<numOfPoints<<

" points.\n"

;}

    

Point(

const

Point &q):x(q.x),y(q.y){numOfPoints++;cout<<

"A Point ("

<<x<<

", "

<<y<<

") is copied! Now, we have "

<<numOfPoints<<

" points.\n"

;}

    

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.\n"

;}

    

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.\n"

;}

    

~Circle(){numOfCircles--;cout<<

"A circle at ("

<<center.getX()<<

", "

<<center.getY()<<

") and radius "

<<radius<<

" is erased! Now, we have "

<<numOfCircles<<

" circles.\n"

;}

   

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;

}