Problem A: 让动物们叫起来吧！

Posted 2020-09-16 好奇怪的妹妹

Description

Tom家里养了很多动物，比如有鸭子、火鸡和公鸡。它们的叫声都不相同。现在，请编写类Animal、Cock、Turkey和Duck，根据给出的main()函数及样例分析每个类的属性、行为及相互关系，以模仿Tom家的情况。

提示：动物们都有自己的名字。

Input

输入有多行。第一行正整数M表示之后有M个测试用例，每个测试用例包括2部分：前一部分是动物的名字，后一部分是动物的类型（用A、B、C分别表示鸭子、火鸡和公鸡）。

Output

输出有M行，每个测试用例对应一样。见样例。

Sample Input

3 Baby C Rubby B Tobby A

Sample Output

Baby is a cock, and it can crow. Rubby is a turkey, and it can gobble. Tobby is a duck, and it can quack.

HINT

Append Code

append.cc,

int main()

{

    int cases;

    string name;

    char type;

    Animal *animal;

    cin>>cases;

    for (int i = 0; i < cases; i++)

    {

       cin>>name>>type;

       switch(type)

       {

        case ‘A‘:

            animal = new Duck(name);

            break;

        case ‘B‘:

            animal = new Turkey(name);

            break;

        case ‘C‘:

            animal = new Cock(name);

            break;

       }

       animal->sound();

    }

    return 0;

}

 

代码

#include <iostream>

using namespace std;

 

class Animal

{

 

    public:

    Animal(){}

    virtual void sound(){}

};

class Cock:public Animal

{

    string name1;

public:

    Cock(string n):name1(n){}

    void sound()

    {

        cout<<name1<<" is a cock, and it can crow."<<endl;

    }

};

class Turkey:public Animal

{

    string name2;

public:

    Turkey(string  n):name2(n){}

    void sound()

    {

        cout<<name2<<" is a turkey, and it can gobble."<<endl;

    }

};

class Duck:public Animal

{

    string name3;

public:

    Duck(string n):name3(n){}

    void sound()

    {

        cout<<name3<<" is a duck, and it can quack."<<endl;

    }

 

};

int main()

{

    int cases;

    string name;

    char type;

    Animal *animal;

    cin>>cases;

    for (int i = 0; i < cases; i++)

    {

       cin>>name>>type;

       switch(type)

       {

        case ‘A‘:

            animal = new Duck(name);

            break;

        case ‘B‘:

            animal = new Turkey(name);

            break;

        case ‘C‘:

            animal = new Cock(name);

            break;

       }

       animal->sound();

    }

    return 0;

}