给出下面的一个基类框架:
class Point_1D
{ protected:
float x;//1D 点的x坐标
public:
Point_1D(float p = 0.0);
float distance( );//计算当前点到原点的距离
}
以Point_1D为基类建立一个派生类Point_2D,增加一个保护数据成员:
float y;//2D平面上点的y坐标
以Point_2D为直接基类再建立一个派生类Point_3D,增加一个保护数据成员:
float z;//3D立体空间中点的z坐标
生成上述类并编写主函数,根据输入的点的基本信息,建立点对象,并能计算该点到原点的距离。
输入格式: 测试输入包含若干测试用例,每个测试用例占一行(点的类型(1表示1D点,2表示2D点,3表示3D点) 第一个点坐标信息(与点的类型相关) 第二个点坐标信息(与点的类型相关))。当读入0时输入结束,相应的结果不要输出。
输入样例:
1 -1
2 3 4
3 1 2 2
0
输出样例:
Distance from Point -1 to original point is 1
Distance from Point(3,4) to original point is 5
Distance from Point(1,2,2) to original point is 3
以下是我的代码
#include
#include
using namespace std;
class Point_1D
{ protected:
float x;
public:
Point_1D(float p = 0.0):x(p){}
~Point_1D(){}
float distance( ){ return x>0?x:-x; }
void Print(){
cout<
}
};
class Point_2D:public Point_1D{
protected:
float y;
public:
Point_2D(float x,float q):Point_1D(x),y(q){}
~Point_2D(){}
float distance( ){
return sqrt(x*x+y*y);}
void Print(){
cout
};
class Point_3D:public Point_2D{
protected:
float z;
public:
Point_3D(float x,float y,float r):Point_2D(x,y),z(r){}
~Point_3D(){}
float distance( ){
return sqrt(x*x+y*y+z*z);}
void Print(){
cout
};
int main()
{
int choice;
float x,y,z;
cin>>choice;
while(choice!=0){
if(choice==1){
cin>>x;
Point_1D D1(x);
cout<<"Distance from Point ";
D1.Print();
cout<<" to original point is "<
}
else if(choice==2){
cin>>x>>y;
Point_2D D2(x,y);
cout<<"Distance from Point";
D2.Print();
cout<<"to original point is "<<D2.distance()<<endl;
}
else {
cin>>x>>y>>z;
Point_3D D3(x,y,z);
cout<<"Distance from Point";
D3.Print();
cout<<"to original point is "<<D3.distance()<<endl;
}
cin>>choice;
}
}