Problem C: 动物要吃饭
Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 1356 Solved: 1226
[Submit][Status][Web Board]
Description
定义Animal类，只有一个纯虚函数eat。
定义Dog、Cat、Sheep、Chicken四个类，它们都是Animal的子类。
每个类重载eat方法，每个方法输出的内容见样例。
Input
一系列0~3之内的整数。
Output
每个输入对应一行输出，0、1、2、3分别代表Dog、Cat、Sheep、Chicken。
Sample Input
01232310
Sample Output
Dog eats bone.Cat eats fish.Sheep eats grass.Chicken eats worm.Sheep eats grass.Chicken eats worm.Cat eats fish.Dog eats bone.
HINT
Append Code
append.cc,
[Submit][Status][Web Board]
?
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
|
#include <iostream>
#include <iostream>
#include <iomanip>
#include <stdio.h>
#include <cstring>
#include <cmath>
#include <list>
#include <vector>
using namespace std;
class Animal
{
public:
virtual void eat()const = 0;
};
class Dog:public Animal
{
public:
void eat()const{cout<<"Dog eats bone."<<endl;}
};
class Cat:public Animal
{
public:
void eat()const{cout<<"Cat eats fish."<<endl;}
};
class Sheep:public Animal
{
public:
void eat()const{cout<<"Sheep eats grass."<<endl;}
};
class Chicken:public Animal
{
public:
void eat()const{cout<<"Chicken eats worm."<<endl;}
};
int main()
{
int c;
vector<Animal*> animals;
vector<Animal*>::iterator itr;
while(cin>>c)
{
switch(c)
{
case 0 :
animals.push_back(new Dog());
break;
case 1:
animals.push_back(new Cat());
break;
case 2:
animals.push_back(new Sheep());
break;
case 3:
animals.push_back(new Chicken());
}
}
for (itr = animals.begin(); itr != animals.end(); itr++)
(*itr)->eat();
return 0;
}
|