学堂在线TsinghuaX: 00740043_2X C++语言程序设计进阶第八章Lab

浏览数：64 / 时间：2015年06月08日

第一题：复数加减乘除

题目描述

求两个复数的加减乘除。

要求使用c++ class编写程序。可以创建如下class

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

class Complex{
public:
    Complex(double r = 0.0, double i = 0.0): real(r), imag(i) {};
    Complex operator+ (const Complex &c2) const;
    Complex operator- (const Complex &c2) const;
    
    /*实现下面三个函数*/
    Complex operator* (const Complex &c2) const;
    Complex operator/ (const Complex &c2) const;
    friend ostream & operator<< (ostream &out, const Complex &c);

private:
    double real;
    double imag;
};

Complex Complex::operator+ (const Complex &c2) const {
    return Complex(real + c2.real, imag + c2.imag);
}

Complex Complex::operator- (const Complex &c2) const {
    return Complex(real - c2.real, imag - c2.imag);
}


int main() {
    double real, imag;
    cin >> real >> imag;
    Complex c1(real, imag);
    cin >> real >> imag;
    Complex c2(real, imag);
    cout << c1 + c2;
    cout << c1 - c2;
    cout << c1 * c2;
    cout << c1 / c2;
}

输入描述

第一行两个double类型数，表示第一个复数的实部虚部

第二行两个double类型数，表示第二个复数的实部虚部

输出描述

输出依次计算两个复数的加减乘除，一行一个结果

输出复数先输出实部，空格，然后是虚部，

样例输入

1 1
3 -1

样例输出

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

class Complex{
public:
    Complex(double r = 0.0, double i = 0.0): real(r), imag(i) {};
    Complex operator+ (const Complex &c2) const;
    Complex operator- (const Complex &c2) const;
    Complex operator* (const Complex &c2) const;
    Complex operator/ (const Complex &c2) const;
    friend ostream & operator<< (ostream &out, const Complex &c);
    
private:
    double real;
    double imag;
};

Complex Complex::operator+ (const Complex &c2) const {
    return Complex(real + c2.real, imag + c2.imag);
}

Complex Complex::operator- (const Complex &c2) const {
    return Complex(real - c2.real, imag - c2.imag);
}

Complex Complex::operator* (const Complex &c2) const{
    return Complex(real * c2.real - imag * c2.imag, imag * c2.real + real * c2.imag);
}

Complex Complex::operator/ (const Complex &c2) const{
    return Complex((real * c2.real + imag * c2.imag) / (c2.real * c2.real + c2.imag * c2.imag), (imag * c2.real - real * c2.imag) / (c2.real * c2.real + c2.imag * c2.imag));
}

ostream & operator<< (ostream &out, const Complex &c){
    out << c.real << " "<< c.imag << endl;
    return out;
}

int main() {
    double real, imag;
    cin >> real >> imag;
    Complex c1(real, imag);
    cin >> real >> imag;
    Complex c2(real, imag);
    cout << c1 + c2;
    cout << c1 - c2;
    cout << c1 * c2;
    cout << c1 / c2;
}

第二题：圆的周长和面积

题目描述

求圆的周长和面积，已知圆类从shape抽象类继承。

要求使用c++ class编写程序。可以创建如下class

#include <iostream>
using namespace std;

const double pi = 3.14;

class Shape{
public:
    Shape(){}
    ~Shape(){}
    virtual double getArea() = 0;
    virtual double getPerim() = 0;
};

class Circle: public Shape{
public:
    Circle(double rad):radius(rad){}
    ~Circle(){}
    
    /*补充这两个函数*/
    double getArea();
    double getPerim();
private:
    double radius;
};

int main() {
    double radius;
    cin >> radius;
    Circle c(radius);
    cout << c.getArea() << " " << c.getPerim() << endl;
}

输入描述

输入圆的半径

输出描述

输出圆的周长和面积

样例输入

样例输出

314 62.8

#include <iostream>
using namespace std;

const double pi = 3.14;

class Shape{
public:
    Shape(){}
    ~Shape(){}
    virtual double getArea() = 0;
    virtual double getPerim() = 0;
};

class Circle: public Shape{
public:
    Circle(double rad):radius(rad){}
    ~Circle(){}
    double getArea()
    {
        return pi * radius * radius;
    }
    double getPerim()
    {
        return 2 * pi * radius;
    }
private:
    double radius;
};

int main() {
    double radius;
    cin >> radius;
    Circle c(radius);
    cout << c.getArea() << " " << c.getPerim() << endl;
}

第三题：三角形还是长方形？

题目描述

在多态概念中，基类的指针既可以指向基类的对象，又可以指向派生类的对象。我们可以使用dynamic_cast类型转换操作符来判断当前指针（必须是多态类型）是否能够转换成为某个目的类型的指针。

同学们先查找dynamic_cast的使用说明（如http://en.wikipedia.org/wiki/Run-time_type_information#dynamic_cast），然后使用该类型转换操作符完成下面程序（该题无输入）。

函数int getVertexCount(Shape * b)计算b的顶点数目，若b指向Shape类型，返回值为0；若b指向Triangle类型，返回值为3；若b指向Rectangle类型，返回值为4。

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

class Shape{
public:
    Shape() {}
    virtual ~Shape() {}
};

class Triangle: public Shape{
public:
    Triangle() {}
    ~Triangle() {}
};

class Rectangle: public Shape {
public:
    Rectangle() {}
    ~Rectangle() {}
};

/*用dynamic_cast类型转换操作符完成该函数*/
int getVertexCount(Shape * b){
}

int main() {
    Shape s;
    cout << getVertexCount(&s) << endl;
    Triangle t;
    cout << getVertexCount(&t) << endl;
    Rectangle r;
    cout << getVertexCount(&r) << endl;
}

答案：

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

class Shape{
public:
    Shape() {}
    virtual ~Shape() {}
};

class Triangle: public Shape{
public:
    Triangle() {}
    ~Triangle() {}
};

class Rectangle: public Shape {
public:
    Rectangle() {}
    ~Rectangle() {}
};

/*用dynamic_cast类型转换操作符完成该函数*/
int getVertexCount(Shape * b){
    if(dynamic_cast<Triangle *>(b))
        return 3;
    else if(dynamic_cast<Rectangle *>(b))
        return 4;
    else
        return 0;
}

int main() {
    Shape s;
    cout << getVertexCount(&s) << endl;
    Triangle t;
    cout << getVertexCount(&t) << endl;
    Rectangle r;
    cout << getVertexCount(&r) << endl;
}