题目描述
完成一维点类Point、二维平面点类Point_2D、三维空间点类Point_3D的类定义的程序填空。
并根据主函数要求完成每个点到原点的距离结果输出。
假设所有点坐标都是整数，输出的距离结果要求精确到小数点后2位。
类声明
#include"iostream" #include<iomanip> #include<cmath> using namespace std; class Point { protected: int x; public: Point(int); void distance(); }; class Point_2D:public Point { protected: int y; public: Point_2D(int,int); void distance(); }; class Point_3D:public Point_2D { protected: int z; public: Point_3D(int,int,int); void distance(); };
主函数
int main() { int num,tx,ty,tz; cin>>num; while(num) { switch(num) { case 1: { cin>>tx; Point p1(tx); p1.distance(); break; } case 2: { cin>>tx>>ty; Point_2D p2(tx,ty); p2.distance(); break; } case 3: { cin>>tx>>ty>>tz; Point_3D p3(tx,ty,tz); p3.distance(); break; } } cin>>num; } }
输入
每行输入一个点的信息，第一个参数表示点的维度，后面参数表示点坐标数值
以输入0结束
输出
输出每个点到原点的距离具体看样例
输入样例1
1 -3 2 2 3 3 3 4 5 0
输出样例1
Distance between [-3] and [0] = 3.00 Distance between [2, 3] and [0, 0] = 3.61 Distance between [3, 4, 5] and [0, 0, 0] = 7.07
思路分析
这道题唯一吸引人的地方就在于只有当你本地跑出理想结果，满怀信心地提交代码却意味看到答案错误的抓狂，你是无论如何都想不到，会有那么多奇奇怪怪不应该存在的空格在输出中，然而，就是少了空格。
AC代码
代码语言：javascript复制Point::Point(int x):x(x){}
void Point::distance()
{
float dist=x*x;
cout<<"Distance between ["<<x<<"] and [0] = "<<fixed<<setprecision(2)<<sqrt(dist)<<endl;
}
Point_2D::Point_2D(int x,int y):Point(x),y(y){}
void Point_2D::distance()
{
float dist=x*x y*y;
cout<<"Distance between ["<<x<<", "<<y<<"] and [0, 0] = "<<fixed<<setprecision(2)<<sqrt(dist)<<endl;
}
Point_3D::Point_3D(int x,int y,int z):Point_2D(x,y),z(z){}
void Point_3D::distance()
{
float dist=x*x y*y z*z;
cout<<"Distance between ["<<x<<", "<<y<<", "<<z<<"] and [0, 0, 0] = "<<fixed<<setprecision(2)<<sqrt(dist)<<endl;
}
