POJ C++程序设计 编程题＃1 编程作业—继承与派生

编程题＃1

来源: POJ (Coursera声明：在POJ上完成的习题将不会计入Coursera的最后成绩。)

注意： 总时间限制: 1000ms 内存限制: 65536kB

描述

写一个MyString 类，使得下面程序的输出结果是：

1. abcd-efgh-abcd-

2. abcd-

3.

4. abcd-efgh-

5. efgh-

6. c

7. abcd-

8. ijAl-

9. ijAl-mnop

10. qrst-abcd-

11. abcd-qrst-abcd- uvw xyz

about

big

me

take

abcd

qrst-abcd-

要求：MyString类必须是从C++的标准类string类派生而来。提示1：如果将程序中所有 "MyString" 用"string" 替换，那么题目的程序中除了最后两条语句编译无法通过外，其他语句都没有问题，而且输出和前面给的结果吻合。也就是说，MyString类对 string类的功能扩充只体现在最后两条语句上面。提示2: string类有一个成员函数 string substr(int start,int length); 能够求从 start位置开始，长度为length的子串

程序：

#include <iostream>
#include <cstring>
#include <cstdlib>
using namespace std;
// 在此处补充你的代码
int CompareString( const void * e1, const void * e2) {
    MyString * s1 = (MyString * ) e1;
    MyString * s2 = (MyString * ) e2;
    if( *s1 < *s2 ) return -1;
    else if( *s1 == *s2 ) return 0;
    else if( *s1 > *s2 ) return 1;
}
int main() {
    MyString s1("abcd-"),s2,s3("efgh-"),s4(s1);
    MyString SArray[4] = {"big","me","about","take"};
    cout << "1. " << s1 << s2 << s3<< s4<< endl;
    s4 = s3; s3 = s1 + s3;
    cout << "2. " << s1 << endl;
    cout << "3. " << s2 << endl;
    cout << "4. " << s3 << endl;
    cout << "5. " << s4 << endl;
    cout << "6. " << s1[2] << endl;
    s2 = s1; s1 = "ijkl-";
    s1[2] = 'A' ;
    cout << "7. " << s2 << endl;
    cout << "8. " << s1 << endl;
    s1 += "mnop";
    cout << "9. " << s1 << endl;
    s4 = "qrst-" + s2;
    cout << "10. " << s4 << endl;
    s1 = s2 + s4 + " uvw " + "xyz";
    cout << "11. " << s1 << endl;
    qsort(SArray,4,sizeof(MyString), CompareString);
    for( int i = 0;i < 4;++i )
        cout << SArray[i] << endl;
    //输出s1从下标0开始长度为4的子串
    cout << s1(0,4) << endl;
    //输出s1从下标为5开始长度为10的子串
    cout << s1(5,10) << endl;
    return 0;
}

输入

无

输出

1. abcd-efgh-abcd-

2. abcd-

3.

4. abcd-efgh-

5. efgh-

6. c

7. abcd-

8. ijAl-

9. ijAl-mnop

10. qrst-abcd-

11. abcd-qrst-abcd- uvw xyz

about

big

me

take

abcd

qrst-abcd-

样例输入

无

样例输出

1. abcd-efgh-abcd-
2. abcd-
3.
4. abcd-efgh-
5. efgh-
6. c
7. abcd-
8. ijAl-
9. ijAl-mnop
10. qrst-abcd-
11. abcd-qrst-abcd- uvw xyz
about
big
me
take
abcd
qrst-abcd-

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdlib>
 4 using namespace std;
 5 // 在此处补充你的代码
 6 class MyString:public string {
 7 public:
 8     MyString():string(){};
 9     MyString(const char *s):string(s) {};
10     MyString(const string &s):string(s) {};
11     MyString operator()(int i, int j) {
12         return this->substr(i,j);
13     }
14 };
15 int CompareString( const void * e1, const void * e2) {
16     MyString * s1 = (MyString * ) e1;
17     MyString * s2 = (MyString * ) e2;
18     if( *s1 < *s2 ) return -1;
19     else if( *s1 == *s2 ) return 0;
20     else if( *s1 > *s2 ) return 1;
21 }
22 int main() {
23     MyString s1("abcd-"),s2,s3("efgh-"),s4(s1);
24     MyString SArray[4] = {"big","me","about","take"};
25     cout << "1. " << s1 << s2 << s3<< s4<< endl;
26     s4 = s3; s3 = s1 + s3;
27     cout << "2. " << s1 << endl;
28     cout << "3. " << s2 << endl;
29     cout << "4. " << s3 << endl;
30     cout << "5. " << s4 << endl;
31     cout << "6. " << s1[2] << endl;
32     s2 = s1; s1 = "ijkl-";
33     s1[2] = 'A' ;
34     cout << "7. " << s2 << endl;
35     cout << "8. " << s1 << endl;
36     s1 += "mnop";
37     cout << "9. " << s1 << endl;
38     s4 = "qrst-" + s2;
39     cout << "10. " << s4 << endl;
40     s1 = s2 + s4 + " uvw " + "xyz";
41     cout << "11. " << s1 << endl;
42     qsort(SArray,4,sizeof(MyString), CompareString);
43     for( int i = 0;i < 4;++i )
44         cout << SArray[i] << endl;
45     //输出s1从下标0开始长度为4的子串
46     cout << s1(0,4) << endl;
47     //输出s1从下标为5开始长度为10的子串
48     cout << s1(5,10) << endl;
49     return 0;
50 }