Problem C: 平面上的点——Point类 (III)
Time Limit: 1 Sec
Memory Limit: 4 MB
Submit: 4828
Solved: 3278
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。实现showPoint()函数。
接口描述：
showPoint()函数按输出格式输出Point对象，调用Point::show()方法实现。
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。showPoint()函数用来输出（通过参数传入的）Point对象的值：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
1,23,32,1
Sample Output
Point : (0, 0) is created.Point : (1, 2) is created.Point : (1, 2) is copied.Point : (1, 2)Point : (1, 2) is erased.Point : (1, 2) is erased.Point : (3, 3) is created.Point : (3, 3) is copied.Point : (3, 3)Point : (3, 3) is erased.Point : (3, 3) is erased.Point : (2, 1) is created.Point : (2, 1) is copied.Point : (2, 1)Point : (2, 1) is erased.Point : (2, 1) is erased.Point : (0, 0) is copied.Point : (1, 1) is created.Point : (0, 0) is copied.Point : (1, 1) is copied.Point : (0, 0) is copied.Point : (0, 0)Point : (1, 1)Point : (0, 0)Point : (0, 0) is erased.Point : (1, 1) is erased.Point : (0, 0) is erased.Point : (1, 1) is erased.Point : (0, 0) is erased.Point : (0, 0) is erased.
HINT
思考构造函数、拷贝构造函数、析构函数的调用时机。
Append Code
All Copyright Reserved 2010-2011
SDUSTOJ TEAM
GPL2.0 2003-2011
HUSTOJ Project TEAM
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#include<iostream>
#include<iomanip>
using namespace std;
class Point
{
private:
double x,y;
public:
Point(double a,double b)
{
x=a,y=b;
cout<<"Point : ("<<setprecision(16)<<x<<", "<<setprecision(16)<<y<<")"<<" is created."<<endl;
}
Point(double q=0):x(q),y(q)
{
cout<<"Point : ("<<setprecision(16)<<x<<", "<<setprecision(16)<<y<<")"<<" is created."<<endl;
}
void show()
{
cout<<"Point : ("<<setprecision(16)<<x<<", "<<setprecision(16)<<y<<")"<<endl;
}
Point(Point &p)
{
x=p.x;
y=p.y;
cout<<"Point : ("<<setprecision(16)<<x<<", "<<setprecision(16)<<y<<")"<<" is copied."<<endl;
}
~Point()
{
cout<<"Point : ("<<setprecision(16)<<x<<", "<<setprecision(16)<<y<<")"<<" is erased."<<endl;
}
};
void showPoint(Point p)
{
p.show();
}
void showPoint(Point q1,Point q2,Point q)
{
q1.show();
q2.show();
q.show();
}
int main()
{
char c;
double a, b;
Point q;
while(std::cin>>a>>c>>b)
{
Point p(a, b);
showPoint(p);
}
Point q1(q), q2(1);
showPoint(q1, q2, q);
}
