《C++ Primer Plus(第六版)》(36)(第十六章 string类和标准模板库 编程题答案)

2017-01-03 07:40:20来源:CSDN作者:u012175089人点击

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16.10 编程练习

1.回文指的是顺着读和逆着读都一样的字符串。假如,“tot"和“otto”都是简短的回文。编写一个程序,让用户输入字符串,并将字符串引用传递给一个bool函数。如果字符串是回文,该函数将返回true,否则返回false。此时,不要担心诸如大小写、空格和标点符号这些复杂的问题。即这个简单的版本将拒绝“Otto”和“Madam,I'm Adam”。请查看附录F中字符串方法列表,以简化这项任务。

////  main.cpp//  HelloWorld////  Created by feiyin001 on 16/12/30.//  Copyright (c) 2016年 FableGame. All rights reserved.//#include <iostream>#include <string>#include <cstring>#include "Test.h"using namespace std;using namespace FableGame;bool check(const string& str){    string temp = str;    reverse(temp.begin(), temp.end());    return str == temp;}int main(){    cout << "Enter a Palindrome: ";    string str;    while (cin >> str) {        if (check(str)) {            cout << str << " is Palindrome" << endl;        }        else        {            cout << str << " is not Palindrome" << endl;        }        cout << "Enter a Palindrome: ";    }    return 0;}

2.与编程练习1中给出的问题相同,但要考虑诸如大小写、空格和标点符号这样的复杂问题。即“Madam, I‘m Adam”将作为回文来测试。例如,测试函数可能会将字符串缩略为“madamimadam”,然后测试倒过来是否一样。不要忘了有用的cctype库,您可能从中找到几个有用的STL函数,尽管不一定非要使用它们。

////  main.cpp//  HelloWorld////  Created by feiyin001 on 16/12/30.//  Copyright (c) 2016年 FableGame. All rights reserved.//#include <iostream>#include <string>#include <cstring>#include <cctype>#include "Test.h"using namespace std;using namespace FableGame;bool check(const string& str){    string temp0;    for (int i = 0; i < str.length(); i++) {        if (islower( str[i] )) {            temp0.push_back(str[i]);        }        else if(isupper(str[i]))        {            temp0.push_back(tolower(str[i]));        }    }    string temp = temp0;    reverse(temp.begin(), temp.end());    return temp0 == temp;}int main(){    cout << "Enter a Palindrome: ";    string str;        while ( cin && getline(cin, str)) {                if (check(str)) {            cout << str << " is Palindrome" << endl;        }        else        {            cout << str << " is not Palindrome" << endl;        }        cout << "Enter a Palindrome: ";    }    return 0;}

3.修改程序清单16.3,使之从文件中读取单词。一种方案是,使用vector<string>对象而不是string数组。这样便可以使用push_back()将数据文件中的单词复制到vector<string>对象中,并使用size()来确定单词列表的长度。优于程序应该每次从文件中读取一个单词,因此应该使用>>而不是getline()。文件中包含的单词应该用空格、制表符或换行符分隔。

原来的程序:程序清单16.3

////  main.cpp//  HelloWorld////  Created by feiyin001 on 16/12/30.//  Copyright (c) 2016年 FableGame. All rights reserved.//#include <iostream>#include <string>#include <cstring>#include <cctype>#include <cstdlib>#include <ctime>#include "Test.h"using namespace std;using namespace FableGame;const int NUM = 26;const string wordlist[NUM] = {    "apiary", "beetle", "cereal", "danger",    "ensign", "florid", "garage", "health",    "insult", "jackal", "keeper", "loaner",    "manage", "nonce", "onset", "plaid",    "quilt", "remote", "stolid", "train",    "useful", "valid", "whence", "xenon",     "yearn", "zippy"};int main(){	    srand((int)time(0));    char play;    cout << "Will you play a word game?<y/n> ";    cin >> play;    play = tolower(play);    while ( play == 'y') {        string target = wordlist[rand() % NUM];        size_t length = target.length();        string attemp(length, '-');        string badchars;        int guesses = 6;        cout << "Guess my secret word. Is has " << length            << " letters, and you guess/none lettle at a time. You get " << guesses << " wrong guesses./n";                cout << "Your word: " << attemp << endl;        while (guesses > 0 && attemp != target) {            char letter;            cout << "Guess a letter: ";            cin >> letter;            if (badchars.find(letter) != string::npos || attemp.find(letter) != string::npos) {                cout << "You already guessed that. Try again./n";                continue;            }            size_t loc = target.find(letter);            if (loc == string::npos) {                cout << "Oh, bad guess!/n";                --guesses;                badchars += letter;            }            else            {                cout << "Good guess!/n";                attemp[loc] = letter;                loc = target.find(letter, loc + 1);                while (loc != string::npos) {                    attemp[loc] = letter;                    loc = target.find(letter, loc + 1);                }            }            cout << "Your word: " << attemp << endl;            if (attemp != target) {                if (badchars.length() > 0) {                    cout << "Bad choices: " << badchars << endl;                }                cout << guesses << " bad guess left/n";            }        }        if (guesses > 0) {            cout << "That's right!/n";        }        else        {            cout << "Sorry, the word is " << target << "./n";        }        cout << "Will you play another?<y/n> ";        cin >> play;        play = tolower(play);    }        cout << "Bye/n";    return 0;}
修改后的程序:

main.cpp

////  main.cpp//  HelloWorld////  Created by feiyin001 on 16/12/30.//  Copyright (c) 2016年 FableGame. All rights reserved.//#include <iostream>#include <string>#include <cstring>#include <cctype>#include <cstdlib>#include <ctime>#include <fstream>#include "Test.h"#include <vector>using namespace std;using namespace FableGame;int main(){    vector<string> wordlist;    ifstream ifs;    ifs.open("/Users/feiyin001/Documents/tttttt.txt");    string str;    while (ifs && ifs >> str) {        wordlist.push_back(str);    }    ifs.close();        srand((int)time(0));    char play;    cout << "Will you play a word game?<y/n> ";    cin >> play;    play = tolower(play);    while ( play == 'y') {        string target = wordlist[rand() % wordlist.size()];        size_t length = target.length();        string attemp(length, '-');        string badchars;        int guesses = 6;        cout << "Guess my secret word. Is has " << length            << " letters, and you guess/none lettle at a time. You get " << guesses << " wrong guesses./n";                cout << "Your word: " << attemp << endl;        while (guesses > 0 && attemp != target) {            char letter;            cout << "Guess a letter: ";            cin >> letter;            if (badchars.find(letter) != string::npos || attemp.find(letter) != string::npos) {                cout << "You already guessed that. Try again./n";                continue;            }            size_t loc = target.find(letter);            if (loc == string::npos) {                cout << "Oh, bad guess!/n";                --guesses;                badchars += letter;            }            else            {                cout << "Good guess!/n";                attemp[loc] = letter;                loc = target.find(letter, loc + 1);                while (loc != string::npos) {                    attemp[loc] = letter;                    loc = target.find(letter, loc + 1);                }            }            cout << "Your word: " << attemp << endl;            if (attemp != target) {                if (badchars.length() > 0) {                    cout << "Bad choices: " << badchars << endl;                }                cout << guesses << " bad guess left/n";            }        }        if (guesses > 0) {            cout << "That's right!/n";        }        else        {            cout << "Sorry, the word is " << target << "./n";        }        cout << "Will you play another?<y/n> ";        cin >> play;        play = tolower(play);    }        cout << "Bye/n";    return 0;}

4.编写一个具有老式风格接口的函数,其原型如下:

int reduce(long ar[], int n);

实参应是数组名和数组中的元素个数。该函数对数组进行排序,删除重复的值,返回缩减后数组中的元素数目。请使用STL函数编写该函数(如果决定使用通过的unique()函数,请注意它将返回结果区间的结尾)。使用一个小程序测试该函数。

////  main.cpp//  HelloWorld////  Created by feiyin001 on 16/12/30.//  Copyright (c) 2016年 FableGame. All rights reserved.//#include <iostream>#include <string>#include <cstring>#include <cctype>#include <cstdlib>#include <ctime>#include <fstream>#include "Test.h"#include <vector>using namespace std;using namespace FableGame;int reduce(long ar[], int n){    sort(ar, ar + n);    long* p = unique(ar, ar + n);    return (int)(p - ar);}int main(){    long a[ 10] = {11, 2, 31, 31, 41, 41, 5, 7, 8, 10};        for (int i = 0; i < 10; ++i) {        cout << a[i] << " ";    }    cout << endl;        int num = reduce(a, 10);        for (int i = 0; i < num; ++i) {        cout << a[i] << " ";    }    cout << endl;    return 0;}

5.问题与编程练习4相同,但要编写一个模板函数:

template<class T>

in reduce(T ar[], int n);

在一个使用long实例和string实例的小程序中测试该函数。

////  main.cpp//  HelloWorld////  Created by feiyin001 on 16/12/30.//  Copyright (c) 2016年 FableGame. All rights reserved.//#include <iostream>#include <string>#include <cstring>#include <cctype>#include <cstdlib>#include <ctime>#include <fstream>#include "Test.h"#include <vector>using namespace std;using namespace FableGame;template<class T>int reduce(T ar[], int n){    sort(ar, ar + n);    T* p = unique(ar, ar + n);    return (int)(p - ar);}int main(){    {        long a[ 10] = {11, 2, 31, 31, 41, 41, 5, 7, 8, 10};                for (int i = 0; i < 10; ++i) {            cout << a[i] << " ";        }        cout << endl;                int num = reduce(a, 10);                for (int i = 0; i < num; ++i) {            cout << a[i] << " ";        }        cout << endl;    }        {        string a[ 10] = {"yuandan2017", "fable", "kuaidi", "caonima", "niubi",            "fancy3d", "fanrenxiuzhen", "fangchangjia", "laosiji", "shuang11"};                for (int i = 0; i < 10; ++i) {            cout << a[i] << " ";        }        cout << endl;                int num = reduce(a, 10);                for (int i = 0; i < num; ++i) {            cout << a[i] << " ";        }        cout << endl;    }    return 0;}


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