Problem B: 平面上的点——Point类 (II)

Problem B: 平面上的点——Point类 (II)

Time Limit: 1 Sec  Memory Limit: 4 MB
Submit: 12244  Solved: 6148
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Description

在数学上，平面直角坐标系上的点用X轴和Y轴上的两个坐标值唯一确定。现在我们封装一个“Point类”来实现平面上的点的操作。

根据“append.cc”，完成Point类的构造方法和show()方法，输出各Point对象的构造和析构次序。

接口描述：
Point::show()方法：按输出格式输出Point对象。

Input

输入多行，每行为一组坐标“x,y”，表示点的x坐标和y坐标，x和y的值都在double数据范围内。

Output

输出每个Point对象的构造和析构行为。对每个Point对象，调用show()方法输出其值：X坐标在前，Y坐标在后，Y坐标前面多输出一个空格。每个坐标的输出精度为最长16位。输出格式见sample。

C语言的输入输出被禁用。

Sample Input

1,2 3,3 2,1

Sample Output

Point : (0, 0) is created. Point : (1, 2) is created. Point : (1, 2) Point : (1, 2) is erased. Point : (3, 3) is created. Point : (3, 3) Point : (3, 3) is erased. Point : (2, 1) is created. Point : (2, 1) Point : (2, 1) is erased. Point : (0, 0) is copied. Point : (1, 1) is created. Point : (0, 0) Point : (1, 1) Point : (0, 0) Point : (1, 1) is erased. Point : (0, 0) is erased. Point : (0, 0) is erased.

#include <iostream>
#include <iomanip>
using namespace std;
class Point
{
    double x;
    double y;
public:
    Point():x(0),y(0)
    {
        cout<<setprecision(16)<<"Point : ("<<x<<", "<<y<<") is created."<<endl;
    }
    Point(double a,double b):x(a),y(b)
    {
        cout<<setprecision(16)<<"Point : ("<<x<<", "<<y<<") is created."<<endl;
    }
    Point(double a):x(a),y(1)
    {
        cout <<setprecision(16)<<"Point : ("<<x<<", "<<y<<")"<<" is created."<<endl;
    }
    Point(const Point& p)
    {
        x=p.x; y=p.y;
        cout <<setprecision(16)<<"Point : ("<<p.x<<", "<<p.y<<")"<<" is copied."<<endl;
    }
    ~Point()
    {
        cout<<setprecision(16)<<"Point : ("<<x<<", "<<y<<") is erased."<<endl;
    }
    void show()
    {
        cout<<setprecision(16)<<"Point : ("<<x<<", "<<y<<")"<<endl;
    }
};
 
int main()
{
    char c;
    double a, b;
    Point q;
    while(std::cin>>a>>c>>b)
    {
        Point p(a, b);
        p.show();
    }
    Point q1(q), q2(1);
    q1.show();
    q2.show();
    q.show();
}

 

标签：erased,输出,show,created,Point,II,坐标,Problem
来源： https://www.cnblogs.com/Begin-Again/p/12825250.html