1. 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（除数字外），同时将大写字符转换为小写字符，将小写字符转换为大写（别忘了cctype函数系列）
1 #include<iostream>
2 using namespace std;
3
4 int main()
5 {
6 char ch;
7 while (cin.get(ch) && ch != '@')
8 {
9 if (isupper(ch))
10 ch = tolower(ch);
11 else if (islower(ch))
12 ch = toupper(ch);
13 else if (isdigit(ch))
14 continue;
15 cout << ch;
16 }
17 system("pause");
18 return 0;
19 }
2. 编写一个程序，最多将10个donation值读到一个double数组中。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。
1 #include<iostream>
2 using namespace std;
3
4 int main()
5 {
6 double donation[10];
7 double average, sum = 0;
8 int cnt = 0, i;
9 for ( i = 0; i < 10; i++)
10 {
11 if (cin >> donation[i]) //返回bool类型的值，用来判断输入是否有效
12 {
13 sum += donation[i];
14 }
15 else
16 break;
17 }
18 average = sum / i;
19 for (int j = 0; j < i; j++)
20 {
21 if (donation[j] > average)
22 cnt++;
23 }
24 cout << "The average is " << average << endl;
25 cout << "Greater than the average is " << cnt << endl;
26
27 system("pause");
28 return 0;
29 }
3、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的操作。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
Please enter a c, p, t, or g: t
A maple is a tree.
1 #include<iostream>
2 using namespace std;
3
4 void showmenu();
5
6 int main()
7 {
8 showmenu();
9 char choice;
10 int flag = 1;
11 while (flag)
12 {
13 cin >> choice;
14 switch (choice)
15 {
16 case 'c': cout << "A maple is a carnivore." << endl;
17 flag = 0;
18 break;
19 case'p':cout << "A maple is a pianist." << endl;
20 flag = 0;
21 break;
22 case't':cout << "A maple is a tree." << endl;
23 flag = 0;
24 break;
25 case'g':cout << "A maple is a game." << endl;
26 flag = 0;
27 break;
28 default:cout << "Please enter a c, p, t, or g: ";
29 flag = 1;
30 }
31 }
32 system("pause");
33 return 0;
34 }
35
36 void showmenu()
37 {
38 cout << "Please enter one of the following choices:\n"
39 "c) carnivore p) pianist\n"
40 "t) tree g) game\n";
41 }
4、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 Programmer name structure
struct bop{
char fullname[strsize]; // real name
char title[strsize]; // job title
char bopname[strsize]; // secret BOP name
int preference; // 0 = fulname, 1 = title, 2 = bopname
}
该程序创建了一个由上述结构组成的小型数组，并将其初始化为适合的值。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.diaplay by preference
q.quit
注意，“diaplay 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将显示程序员的头衔。
1 #include<iostream>
2 #include<cstring>
3 using namespace std;
4
5 const int strsize = 20;
6 void showmenu();
7
8 struct bop
9 {
10 char fullname[strsize]; // real name
11 char title[strsize]; // job title
12 char bopname[strsize]; // secret BOP name
13 int preference; // 0 = fulname, 1 = title, 2 = bopname
14 };
15
16 int main()
17 {
18 bop list[5] = { { "Wimp Macho", "W", "Wimp", 0 }, { "Raki Rhodes", "R", "Raki", 1 }, { "Celia Laiter", "C", "Celia", 2 }, { "Hoppy Hipman", "H", "Hoppy", 0 }, { "Pat Hand", "P", "Pat", 1 } };
19 showmenu();
20 char choice;
21 int flag = 1; //注意q的处理
22 while (flag)
23 {
24 cin >> choice;
25 switch (choice)
26 {
27 case'a':for (int i = 0; i < 5; i++)
28 {
29 cout << list[i].fullname << endl;
30 }
31 break;
32 case 'b': for (int i = 0; i < 5; i++)
33 {
34 cout << list[i].title << endl;
35 }
36 break;
37 case 'c': for (int i = 0; i < 5; i++)
38 {
39 cout << list[i].bopname << endl;
40 }
41 break;
42 case'd':for (int i = 0; i < 5; i++)
43 {
44 if (list[i].preference == 0)
45 cout << list[i].fullname << endl;
46 else if (list[i].preference == 1)
47 cout << list[i].title << endl;
48 else
49 cout << list[i].bopname << endl;
50 }
51 break;
52 case'q':flag = 0;
53 cout << "Bye!" << endl;
54 break;
55 default:cout << "This is invalid input!" << endl;
56 }
57 if (flag == 0)
58 break;
59 cout << "Next choice: ";
60 }
61 system("pause");
62 return 0;
63 }
64
65 void showmenu()
66 {
67 cout << "Benevolent Order of Programmers Report" << endl
68 << "a. display by name \t" << "b.display by title" << endl
69 << "c.display by bopname\t" << "d.diaplay by preference" << endl
70 << "q.quit" << endl;
71 cout << "Enter your choice: ";
72 }
5、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38000tvarps时，所得税为5000 x 0.00 + 10000 x 0.10 + 20000 x 0.15 + 3000 x 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。
1 #include<iostream>
2 #include<cstring>
3 using namespace std;
4
5 int main()
6 {
7 double income, tax;
8 cout << "Please enter income: ";
9 while (cin >> income)
10 {
11 if (income <= 5000)
12 tax = 0;
13 else if (income > 5000 && income <= 15000)
14 tax = (income - 5000)*0.10;
15 else if (income > 15000 && income <= 35000)
16 tax = 10000 * 0.10 + (income - 15000)*0.15;
17 else if (income > 35000)
18 tax = 10000 * 0.10 + 20000 * 0.15 + (income - 35000)*0.20;
19 cout << "Tax= " << tax << endl;
20 cout << "Please enter income again: ";
21 }
22 cout << "Invalid input!" << endl;
23
24 system("pause");
25 return 0;
26 }
6.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数组中。每个结构有两个成员：用于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的捐款者姓名及其捐款数额。该列表前包含一个标题，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。
1 #include<iostream>
2 #include<string>
3 using namespace std;
4
5 struct stream
6 {
7 string name;
8 double donation;
9 };
10
11 int main()
12 {
13 int n, i, flag = 0;
14 cout << "Please enter the number of contributors: ";
15 cin >> n;
16 stream *str = new stream[n]; //创建的动态结构数组
17 for (i = 0; i < n; i++)
18 {
19 cout << "Enter name: ";
20 cin >> str[i].name;
21 cout << "Enter donation: ";
22 cin >> str[i].donation;
23 }
24 cout << endl;
25 cout << fixed;
26 cout.precision(2);
27 cout.setf(ios_base::showpoint);
28 cout << "Grand Patrons:";
29 for (i = 0; i < n; i++)
30 {
31 if (str[i].donation>10000)
32 {
33 cout << str[i].name << "\t" << str[i].donation << endl;
34 flag = 1;
35 }
36 }
37 cout << endl;
38 if (flag == 0)
39 {
40 cout << "None" << endl;
41 }
42 cout << endl;
43 cout << "Other Patrons:" << endl;
44 for (i = 0; i < n; i++)
45 {
46 if (str[i].donation <= 10000)
47 {
48 cout << str[i].name << "\t" << str[i].donation << endl;
49 flag = 1;
50 }
51 if (flag == 0)
52 {
53 cout << "None" << endl;
54 }
55 }
56 delete[]str; //记得delete
57 system("pause");
58 return 0;
59 }
7、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，有多少单词以辅音打头，还有多少单词不属于这两类。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 od lawn. q
5 words beginning with vowels
4 words beginning with consonants
2 others
1 #include<iostream>
2 #include<string>
3 using namespace std;
4
5 int main()
6 {
7 char sen[20];
8 int i, yuan = 0, fu = 0, other = 0;
9 cout << "Enter words (q to quit): " << endl;
10 cin >> sen;
11 while (strcmp(sen, "q"))
12 {
13 if (isalpha(sen[0]))
14 {
15 switch (sen[0])
16 {
17 case'a':
18 case'A':
19 case'e':
20 case'E':
21 case'i':
22 case'I':
23 case'o':
24 case'O':
25 case'u':
26 case'U':yuan++; break;
27 default:fu++;
28 }
29 }
30 else other++;
31 cin >> sen;
32 }
33 cout << yuan << " words beginning with vowels" << endl;
34 cout << fu << " words beginning with consonants" << endl;
35 cout << other << " others" << endl;
36 system("pause");
37 return 0;
38 }
8.编写一个程序，它打开一个文件夹，逐个字符的读取该文件，直到到达文件末尾，然后指出该文件中包含多少个字符。
1 #include<iostream>
2 #include<fstream>
3 using namespace std;
4
5 int main()
6 {
7 ifstream inFile;
8 char filename[60];
9 cout << "Enter name of data file: ";
10 cin.getline(filename, 60);
11 inFile.open(filename);
12 if (!inFile.is_open())
13 {
14 cout << "Could not open the file " << filename << endl;
15 cout << "Program terminatinf.\n";
16 exit(EXIT_FAILURE);
17 }
18 char value;
19 int count = 0;
20 inFile >> value;
21 while (inFile.good())
22 {
23 count++;
24 cout << value;
25 inFile >> value;
26 }
27 cout << endl;
28 if (inFile.eof())
29 cout << "End of file reached.\n";
30 else if (inFile.fail())
31 cout << "Input terminated by data mis match.\n";
32 else
33 cout << "Input terminated for unknown reason.\n";
34 if (count == 0)
35 cout << "No data processed.\n";
36 else
37 {
38 cout << "This file has " << count << " characters." << endl;
39 }
40 system("pause");
41 return 0;
42 }
9.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
1 #include<iostream>
2 #include<fstream>
3 #include<string>
4 using namespace std;
5
6 struct stream
7 {
8 string name;
9 double donation;
10 };
11
12 int main()
13 {
14 ifstream inFile; //声明ifstream对象
15 inFile.open("haha.txt"); //关联文件
16 if (!inFile.is_open()) //文件打开失败
17 {
18 cout << "Could not open the file " << endl;
19 cout << "Program terminatinf.\n";
20 exit(EXIT_FAILURE);
21 }
22 int n, i, flag = 0; //flag是决定是否打印None的标志
23 inFile >> n;
24 inFile.get(); //抵消换行
25 stream *str = new stream[n];
26 for (i = 0; i < n;i++)
27 {
28 getline(inFile, str[i].name);
29 inFile >> str[i].donation;
30 inFile.get(); //抵消换行
31 }
32 cout << "Grand Patrons:" << endl;
33 for (i = 0; i < n; i++)
34 {
35 if (str[i].donation>10000)
36 {
37 cout << str[i].name << "\t" << str[i].donation << endl;
38 flag = 1;
39 }
40 }
41 cout << endl;
42 if (flag == 0)
43 cout << "None" << endl;
44 cout << "Other Patrons:" << endl;
45 for (i = 0; i < n; i++)
46 {
47 if (str[i].donation<=10000)
48 {
49 cout << str[i].name << "\t" << str[i].donation << endl;
50 flag = 1;
51 }
52 }
53 cout << endl;
54 if (flag == 0)
55 cout << "None" << endl;
56
57 inFile.close(); //使用完该文件记得close
58 delete[]str; //用完new记得delete
59
60 system("pause");
61 return 0;
62 }
注：一开始写的时候忘记两个inFile.get()，导致两个结果都是None，找终止循环的原因，可在循环后添加：
if (inFile.eof())
cout << "End of file reached." << endl;
else if (inFile.fail())
cout << "Input terminated by data misamatch." << endl;
else
cout << "Input terminated for unknown reason." << endl;
