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;
}