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-

---

 #include <iostream>
 #include <cstring>
 #include <cstdlib>
 using namespace std;
 // 在此处补充你的代码
 class MyString:public string {
 public:
     MyString():string(){};
     MyString(const char *s):string(s) {};
     MyString(const string &s):string(s) {};
     MyString operator()(int i, int j) {
         return this->substr(i,j);
     }
 };
 int CompareString( const void * e1, const void * e2) {
     MyString * s1 = (MyString * ) e1;
     MyString * s2 = (MyString * ) e2;
     if( *s1 < *s2 ) return -;
     else if( *s1 == *s2 ) return ;
     else if( *s1 > *s2 ) return ;
 }
 int main() {
     MyString s1("abcd-"),s2,s3("efgh-"),s4(s1);
     MyString SArray[] = {"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[] << endl;
     s2 = s1; s1 = "ijkl-";
     s1[] = '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,,sizeof(MyString), CompareString);
     for( int i = ;i < ;++i )
         cout << SArray[i] << endl;
     //输出s1从下标0开始长度为4的子串
     cout << s1(,) << endl;
     //输出s1从下标为5开始长度为10的子串
     cout << s1(,) << endl;
     return ;
 }