C++习题 抽象基类
时间: 1ms 内存:128M
描述:
编写一个程序,声明抽象基类Shape,由它派生出3个派生类: Circle(圆形)、Rectangle(矩形)、Triangle(三角形),用一个函数printArea分别输出以上三者的面积(结果保留两位小数),3个图形的数据在定义对象时给定。
输入:
圆的半径
矩形的边长
三角形的底与高
输出:
圆的面积
矩形的面积
三角形的面积
示例输入:
12.6
4.5 8.4
4.5 8.4
示例输出:
area of circle = 498.76
area of rectangle = 37.80
area of triangle = 18.90
提示:
参考答案(内存最优[1092]):
#include<stdio.h>
int main()
{
float a,b,c,d,e;
scanf("%f",&a);
scanf("%f %f",&b,&c);
scanf("%f %f",&d,&e);
printf("area of circle = %.2f\n",a*3.1415926535898*a);
printf("area of rectangle = %.2f\n",b*c);
printf("area of triangle = %.2f\n",d*e/2);
}
参考答案(时间最优[0]):
#include<iostream>
#include<iomanip>
using namespace std;
class Shape
{
public:
virtual double Area()=0;
};
class Circle:public Shape
{
public:
Circle(double R):r(R){}
virtual double Area()
{
return 3.1415926*r*r;
}
private:
double r;
};
class Rectangle:public Shape
{
public:
Rectangle(double a,double b):chang(a),kuan(b){};
virtual double Area()
{
return chang*kuan;
}
private:
double chang,kuan;
};
class Triangle:public Shape
{
public:
Triangle(double a,double b):di(a),gao(b){};
virtual double Area()
{
return di*gao/2;
}
private:
double di,gao;
};
void printArea(Shape& shape)
{
cout<<shape.Area()<<endl;
}
int main()
{
float r,a,b,w,h;
cout<<fixed<<setprecision(2);
cin>>r;
Circle circle(r);
cout<<"area of circle = ";
printArea(circle);
cin>>a>>b;
Rectangle rectangle(a,b);
cout<<"area of rectangle = ";
printArea(rectangle);
cin>>w>>h;
Triangle triangle(w,h);
cout<<"area of triangle = ";
printArea(triangle);
return 0;
}
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