标签：false let for private radius 判断 led sam dex
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
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;
}
Problem E: 点在圆内吗？
标签：false let for private radius 判断 led sam dex
原文地址：http://www.cnblogs.com/TogetherLaugh/p/6544727.html
