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
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;
}