c++ primer 第六版pdf+代码实现

# 第六章 《分支语句和逻辑运算符》 编程练习题之我解 ## 6.1 **题：** 编写一个程序，读取键盘输入，直到遇到 `@`符号为止，并回显输入（数字除外），同时将大写字符转换为小写，将小写字符转换为大 写（别忘了`cctype`函数系列）。 **解：** ```Cpp #include <iostream> #include <cctype> int main() { using namespace std; char ch; cout << "Enter any charater: "; while ((ch=cin.get()) != '@') { if (isdigit(ch)) { continue; } else if (islower(ch)) { ch = toupper(ch); } else if (isupper(ch)) { ch = tolower(ch); } cout << ch; } cout << "** done **" << endl; return 0; } ``` ## 6.2 **题：** 编写一个程序，最多将10个 `donation` 值读入到一个 `double` 数组中（如果您愿意，也可使用模板类 `array` ）。程序遇到非数字输入时将结束输入，并报告这些数字的平均值以及数组中有多少个数字大于平均值。 **解：** ```Cpp #include <iostream> #include <array> int main() { using namespace std; const unsigned int size = 10; array<double, size> donation; double sum_value = 0; unsigned int large_avg = 0, n = 0; cout << "Enter 10 double value or type non-digital value to exit: "; while ((n < size) && (cin >> donation[n])) { sum_value += donation[n]; ++n; } double avg = sum_value / n; for (int i=0; i < n; i++) { if (donation[i]>avg) ++large_avg; } cout << "The average value is: " << avg << ", there are " << large_avg << " larger than average value." << endl; return 0; } ``` ## 6.3 **题：** 编写一个菜单驱动程序的雏形。该程序显示一个提供4个选项的菜单——每个选项用一个字母标记。如果用户使用有效选项之外的字母进行响应，程序将提示用户输入一个有效的字母，直到用户这样做为止。然后，该程序使用一条 `switch` 语句，根据用户的选择执行一个简单操作。该程序的运行情况如下： ```bash Please enter one of the following choices: c) carnivore p) pianist t) tree g) game f Please enter a c, p, t, or g: q A maple is a tree. ``` **解：** ```Cpp #include <iostream> int main() { using namespace std; cout << "Please enter one of the following choice: \n"; cout << "c) carnivore\tp) pianist.\n" << "t) tree\tg) game" << endl; bool exit = false; char c; while (!exit && (cin >> c)) { switch (c) { case 'c': cout << "Tiger is a carnivore." << endl; exit = true; break; case 'p': cout << "Mozart is a great pianst." << endl; exit = true; break; case 't': cout << "A maple is a tree." << endl; exit = true; break; case 'g': cout << "Supper Mario is a great game." << endl; exit = true; break; default: cout << "Please enter c, p, t, or g: q" << endl; break; } } return 0; } ``` ## 6.4 **题：** 加入 `Benevolent Order of Programmer` 后，在BOP大会上，人们便可以通过加入者的真实姓名、头衔或秘密BOP姓名来了解他（她）。请编写一个程序，可以使用真实姓名、头衔、秘密姓名或成员偏好来列出成员。编写该程序时，请使用下面的结构： ```Cpp // Benevolent order of programmers strcture struct bop { char fullname[strsize]; // real name char title[strsize]; // job title char bopname[strsize]; // secret BOP name int preference; // 0 = fullname, 1 = title, 2 = bopname }; ``` 该程序创建一个有上述结构体组成的小型数组，并将其初始化为适当的值。另外，该程序使用一个循环，让用户在下面的选项中进行选择： ```bash a. display by name b. display by title c. display by bopname d. display by preference q. quit ``` 注意，`display by preference` 并不意味着显示成员的偏好，而是意味着根据成员的偏好来列出成员。例如，如果偏好号为 1，则选择 d 将显示成员的头衔。该程序的运行情况如下： ```bash Benevolent order of Programmers report. a. display by name b. display by title c. display by bopname d. display by preference q. quit Enter your choices: a Wimp Macho Raki Rhodes Celia Laiter Hoppy Hipman Pat Hand Next choice： d Wimp Macho Junior Programmer MIPS Analyst Trainee LOOPY Next choice: q Bye! ``` **解：** ```Cpp #include <iostream> int main() { using namespace std; const int strsize = 80; struct Bop { char fullname[strsize]; // real name char title[strsize]; // job title char bopname[strsize]; // secret BOP name int preference; // 0 = fullname, 1 = title, 2 = bopname }; const int size = 5; const Bop bops[size] = { {"Wimp Macho", "bbb", "c", 0}, {"Raki Rhodes", "2XXXX", "3XXXXX", 1}, {"Celia Laiter", "2AAAA", "3AAAAA", 2}, {"Hoppy Hipman", "2BBBB", "3BBBBB", 0}, {"Pat Hand", "4CCCC", "3CCCCC", 1} }; cout << "Benevolent order of Programmers report.\n"; cout << "a. display by name b. display by title\n" << "c. display by bopname d. display by preference\n" << "q. quit" << endl; char ch; while (cin >> ch) { if (ch == 'q') { break; } for (int i=0; i < size; ++i) { switch (ch) { case 'a': cout << bops[i].fullname << "\n"; break; case 'b': cout << bops[i].title << "\n"; break; case 'c': cout << bops[i].bopname << "\n"; break; case 'd': cout << bops[i].preference << "\n"; break; default: break; } } cout << "Next choice: "; } cout << "** Done **" << endl; return 0; } ``` ## 6.5 **题：** 在 `Neutronia` 王国，货币单位是 `tvarp`，收入所得税的计算方式如下： - 5000 tvarps：不收税; - 5001～15000 tvarps：10%; - 15001～35000 tvarps：15%; - 35000 tvarps以上：20%; 例如，收入为 `38000 tvarps` 时，所得税为 `5000 * 0.00 + 10000 * 0.10 + 20000 * 0.15 + 3000 * 0.20`， 即 `4600 tvarps`。请编写一个程序，使用循环来 要求用户输入收入，并报告所得税。当用户输入负数或非数字时，循环将结束。 **解：** ```Cpp #include <iostream> int main() { using namespace std; const double tax_rate1 = 0.1; const double tax_rate2 = 0.15; const double tax_rate3 = 0.20; double income = 0.0, tax = 0.0; cout << "Please enter your income: "; while ((cin >> income) && (income > 0)) { if (income <= 5000) { tax = 0.0; } else if (income <= 15000 ) { tax = (income - 5000) * tax_rate1; } else if (income <= 35000) { tax = (15000 - 5000) * tax_rate1 + (income - 15000) * tax_rate2; } else { tax = (15000 - 5000) * tax_rate1 + (35000 - 15000) * tax_rate2 + (income - 35000) * tax_rate3; } cout << "Income = " << income << ", tax = " << tax << endl; cout << "Please enter your income again or enter a negative value to quit: "; } return 0; } ``` ## 6.6 **题：** 编写一个程序，记录捐助给 “维护合法权利团体” 的资金。该程序要求用户输入捐献者数目，然后要求用户输入每一个捐�