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：
#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;
}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：
#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;
}
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：
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：
#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;
}
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：
// 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：
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：
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：
#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;
}
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：
#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;
}
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献者的姓名和款项。这些信息被储存在一个动态分配的结构体数组中。每个结构体有两个成员：用来储存姓名的字符数组（或 string对象）和用来存储款项的 double成员。读取所有的数据后，程序将显示所有捐款超过 10000 的捐款者的姓名及其捐款数额。
该列表前应包含一个标题，指出下面的捐款者是重要捐款人 Grand Patrons。然后，程序将列出其他的捐款者，该列表要以 Patrons 开头。如果某种类别没有捐款者，则程序将打印单词 none。该程序只显示这两种类别，而不进行排序。
解：
#include <iostream>
#include <string>
int main() {
using namespace std;
const int Grand_Amount = 10000;
struct Patron {
string name;
double amount;
};
int contribute_num = 0;
cout << "Enter the number of contributor: ";
cin >> contribute_num;
cin.get(); // 读取输入流中的回车符
Patron *p_contribution = new Patron [contribute_num];
for (int i = 0; i < contribute_num; ++i) {
cout << "Enter the name of " << i + 1 << " contributor: ";
getline(cin, p_contribution[i].name);
cout << "Enter the amount of donation: ";
cin >> p_contribution[i].amount;
cin.get(); // 读取输入流中的回车符
}
unsigned int grand_amount_n = 0;
cout << "\nGrand patron: " << endl;
for (int i = 0; i < contribute_num; ++i) {
if (p_contribution[i].amount > Grand_Amount) {
cout << "Contributor name: " << p_contribution[i].name << "\n"
<< "Contributor amount: " << p_contribution[i].amount << endl;
++grand_amount_n;
}
}
if (grand_amount_n == 0) {
cout << "None" << endl;
}
bool is_empty = true;
cout << "\nPatrons: " << endl;
for (int i=0; i < contribute_num; ++i) {
cout << "Contributor name: " << p_contribution[i].name << "\n"
<< "Contributor amount: " << p_contribution[i].amount << endl;
is_empty = false;
}
if (is_empty) {
cout << "** None **" << endl;
}
return 0;
}
题： 编写一个程序，它每次读取一个单词，直到用户输入 q。然后，该程序指出有多少个单词以元音打头，有多少个单词以辅音打头，还有多少个单词不属于这两类。为此，方法之一是，使用 isalpha() 来区分以字母和其他字符打头的单词，然后对于通过了 isalpha() 测试的单词，使用 if 或 switch 语句来确定哪些以元音打头。
该程序的运行情况如下：
Enter words (q to quit):
The 12 awesome oxen ambled
quietly across 15 meters of lawn. q
5 words beginning with vowels
4 words beginning with consonants
2 others
解：
#include <iostream>
#include <cctype>
#include <string>
int main() {
using namespace std;
unsigned int vowels = 0;
unsigned int consonants = 0;
unsigned int other = 0;
string input;
cout << "Enter words (q to quit): " << endl;
while (cin >> input) {
if (input == "q")
break;
if (isalpha(input[0])) {
switch(toupper(input[0])) {
case 'A':;
case 'E':;
case 'I':;
case 'O':;
case 'U':
++vowels;
break;
default:
++consonants;
break;
}
} else {
++other;
}
}
cout << vowels << " words beginning with vowels.\n"
<< consonants << " words beginning with consonants.\n"
<< other << " words beginning with other letter." << endl;
return 0;
}
题： 编写一个程序，它打开一个文件文件，逐个字符地读取该文件，直到到达文件末尾，然后指出该文件中包含多少个字符。
解：
#include <iostream>
#include <fstream>
#include <string>
int main() {
using namespace std;
string fn;
ifstream in_file_handle;
unsigned int n = 0;
char ch;
cout << "Enter a file name: ";
getline(cin, fn);
in_file_handle.open(fn.c_str());
while ((ch = in_file_handle.get()) != EOF) {
++n;
}
in_file_handle.close();
cout << "There are " << n << " characters in "
<< fn << " file." << endl;
return 0;
}
题： 完成编程练习6，但从文件中读取所需的信息。该文件的第一项 应为捐款人数，余下的内容应为成对的行。在每一对中，第一行为捐款人姓名，第二行为捐款数额。即该文件类似于下面：
4 Sam Stone
2000
Freida Flass
100500
Tammy Tubbs
5000
Rich Raptor
55000解：
#include <iostream>
#include <fstream>
#include <string>
int main() {
using namespace std;
const int Grand_Amount = 10000;
string file_name;
ifstream in_file_handle;
struct Patron {
string name;
double amount;
};
int contribute_num = 0;
cout << "Enter a file name: ";
getline(cin, file_name);
in_file_handle.open(file_name.c_str());
in_file_handle >> contribute_num;
in_file_handle.get(); // 读取空白
Patron *p_contribution = new Patron [contribute_num];
for (int i = 0; i < contribute_num; ++i) {
/*
* 4 Sam Stone
* 2000
* Freida Flass
* 100500
* Tammy Tubbs
* 5000
* Rich Raptor
* 55000
*
*/
getline(in_file_handle, p_contribution[i].name);
in_file_handle >> p_contribution[i].amount;
in_file_handle.get(); // 读掉空白(包括滞留在行末的回车符)
}
in_file_handle.close();
unsigned int grand_amount_n = 0;
cout << "\nGrand patron: " << endl;
for (int i = 0; i < contribute_num; ++i) {
if (p_contribution[i].amount > Grand_Amount) {
cout << "Contributor name: " << p_contribution[i].name << "\n"
<< "Contributor amount: " << p_contribution[i].amount << endl;
++grand_amount_n;
}
}
if (grand_amount_n == 0) {
cout << "None" << endl;
}
bool is_empty = true;
cout << "\nPatrons: " << endl;
for (int i=0; i < contribute_num; ++i) {
cout << "Contributor name: " << p_contribution[i].name << "\n"
<< "Contributor amount: " << p_contribution[i].amount << endl;
is_empty = false;
}
if (is_empty) {
cout << "** None **" << endl;
}
delete [] p_contribution;
return 0;
}