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（数字除外），
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）
#include <iostream>
int main()
{using namespace std;char ch;cout << "Enter characters:(enter @ to stop)" << endl;while ((ch = cin.get()) != '@') {if (isupper(ch))ch = tolower(ch);else if(islower(ch)){ch = toupper(ch);}if(!isdigit(ch)){cout << ch << endl;}}return 0;
}2.编写一个程序，最多将10个donation值读入一个double数组中
（如果你愿意，也可以使用模板类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。
#include <iostream>
const int num = 10;
int main() {using namespace std;double donation[num];double sum = 0;double avg;cout << "Enter a number(Stop when entering an unnumber):" << endl;for (int i = 0; i < num; i++) {cin >> donation[i];/*if(!isdigit(donation[i])){break;}*/sum += donation[i];}avg = sum / num;cout << "The sum of all numbers is:" << avg << endl;cout << "The number that is larger than avg is:" << endl;for (int i = 0; i < num; i++) {if (donation[i] > avg) {cout << donation[i] << endl;}}return 0;
}
3.编写一个菜单驱动程序的雏形。
该程序显示一个提供四个选项的菜单--每个选项用一个字母表标记。
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有效的字母，
直到用户这样选择为止。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
Please enter a c,p,t or g:t
A maple is a tree.
#include <iostream>
int main() {using namespace std;char ch;cout << "Please enter one of the following choices:" << endl;cout << "c) carnivore p) pianist" << endl;cout << "t) tree g)game" << endl;cin >> ch;while (isalpha(ch)) {switch (ch){case 'c':cout << "A maple is a carnivore." << endl;break;case 'p':cout << "A maple is a pianist." << endl;break;case 't':cout << "A maple is a tree." << endl;break;case 'g':cout << "A maple is a game." << endl;break;}cout << "Please enter a c,p,t or g:";cin >> ch;}return 0;
}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 Programmers name structure
struct bop{char fullname[strsize]; //real namechar title[strzie]; //job titlechar bopname[strsize]; //secret BOP nameint preference; //0=fullname,1=title,2=bopname
};
该程序创建一个由上述结构组成的小型数组，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.diaplay by title
c.display by bopname d.display by preference
q.quit
注意，“display by preference”并不意味显示成员的偏好，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。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 choice: 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!#include <iostream>
const int strsize = 20;int main() {using namespace std;struct bop {char fullname[strsize]; //real namechar title[strsize]; //job titlechar bopname[strsize]; //secret BOP nameint preference; //0=fullname,1=title,2=bopname};cout << "Benevolent Order of Programmers Report" << endl;cout << "a.display by name b.display by title" << endl;cout << "c.display by bopname d.display by preference" << endl;cout << "q.quit" << endl;bop programmers[5] = { {"Wimp Macho", "Junior Programmer", "WM",0},{"Raki Rhodes", "Junior Programmer", "RR", 1},{"Celia Laiter", "Junior Programmer", "MIPS", 2},{"Hoppy Hipman", "Analyst Trainee", "HH", 1},{"Pat Hand", "Junior Programmer", "LOOPY", 2} };cout << "Enter your choice:" << endl;char choice;cin >> choice;while (choice == 'a' || choice == 'b' || choice == 'c' || choice == 'd' || choice == 'q') {if (choice == 'a') {for (int i = 0; i < 5; i++) {cout << programmers[i].fullname << endl;}}else if (choice == 'b') {for (int i = 0; i < 5; i++) {cout << programmers[i].title << endl;}}else if (choice == 'c') {for (int i = 0; i < 5; i++) {cout << programmers[i].bopname << endl;}}else if (choice == 'd') {for (int i = 0; i < 5; i++) {cout << programmers[i].preference << endl;}}else if (choice == 'q') {cout << "Bye!" << endl;break;}cout << "Next choice:" << endl;cin >> choice;}return 0;
}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38000 tvarps时,所得税为5000×0.00+10000×0.10+2000×0.15+3000×0.20,即 4600tvarps。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。
#include <iostream>int main() {using namespace std;double tax, income;while (true) {cout << "Enter your income (negative number or non-numeric to exit): ";if (!(cin >> income)) {// 如果输入不是数字，则结束循环break;}if (income < 0) {// 如果收入为负数，则结束循环break;}// 计算所得税if (income <= 5000) {tax = 0;}else if (income <= 15000) {tax = (income - 5000) * 0.1;}else if (income <= 35000) {tax = 10000 * 0.1 + (income - 15000) * 0.15;}else {tax = 10000 * 0.1 + 20000 * 0.15 + (income - 35000) * 0.2;}// 输出所得税cout << "Your tax is: " << tax << " tvarps." << endl;}cout << "Thank you for using the tax calculator." << endl;
}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赠者数目，然后要求用户输入每一个捐献者的姓名和款项。这些信息被存储在一个动态分配的结构数组中.每个数据结构有两个成员：用来存储姓名的字符串数组(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的捐款着姓名及其捐款数额。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Partons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。
#include <iostream>
#include <string>struct Donor {string name;double money;
};
void getKind(Donor donor[], int num, bool isGrand) {int count = 0;for (int i = 0; i < num; i++) {if (isGrand) {if (donor[i].money > 10000) {count++;cout << donor[i].name << ":" << donor[i].money <<endl;}}else {if (donor[i].money <= 10000) {count++;cout << donor[i].name;}}}if (count == 0) {cout << "none" << endl;}
}
int main() {using namespace std;int num;cout << "Enter the number of donator:" << endl;cin >> num;Donor* donor = new Donor[num];for (int i = 0; i < num; i++) {cout << "Please enter the donor's name:";cin >> donor[i].name;cout << "Please enter the donation amount:";cin >> donor[i].money;}cout << "Grand Patrons:" << endl;getKind(donor, num, true);cout << "Patrons:" << endl;getKind(donor, num, false);return 0;
}7.编写一个程序,它每次读取一个单词，直到用户只输入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母打头的单词，然后对于通过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 oxem ambled
quiety across 15 meters of lawn. q
5 words beginning with vowels
4 words beginning with consonants
2 others
#include <iostream>
#include <string>int main() {using namespace std;string word;int vowelNum = 0;int consonantNum = 0;int othersNum = 0;cout << "Enter words (q to quit):" << endl;while (cin >> word) {if (word == "q") {break;}else {if (isalpha(word[0])){switch (word[0]){case 'a':case 'e':case 'i':case 'o':case 'u':case 'A':case 'E':case 'I':case 'O':case 'U':vowelNum++;break;default:consonantNum++;break;}}else {othersNum++;}}}cout << vowelNum << " words beginning with vowels." << endl;cout << consonantNum << " words beginning with consonant." << endl;cout << othersNum << " words beginning with others." << endl;return 0;
}8.编写一个程序，他打开一个文件，逐个字地读取该文件，直到到达文件末尾，然后指出该文件中包含多少个字符。
#include <iostream>
#include <string>
#include <fstream>int main() {using namespace std; string filename = "1.txt";char c;fstream file;int count = 0;file.open(filename);if (!file.is_open()) {cout << "Can't open!" << endl;}while (file.get(c)) {if (isalpha(c)) {count++;}}cout << "The number of characters is " << count << endl;file.close();return 0;
}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：
#include <iostream>
#include <string>
#include <fstream>
using namespace std;struct Donor {string name;double money;
};
void getKind(Donor donor[], int num, bool isGrand) {int count = 0;for (int i = 0; i < num; i++) {if (isGrand) {if (donor[i].money > 10000) {count++;cout << donor[i].name << ":" << donor[i].money <<endl;}}else {if (donor[i].money <= 10000) {count++;cout << donor[i].name;}}}if (count == 0) {cout << "none" << endl;}
}
int main() {string fileName = "1.txt";fstream file(fileName);if (!file.is_open()) {cout << "Can't open!" << endl;}while (file.get()) {}int num = 0;(file >> num).get();Donor* donor = new Donor[num];for (int i = 0; i < num; i++) {file >> donor[i].money;file >> donor[i].name;}cout << "Grand Patrons:" << endl;getKind(donor, num, true);cout << "Patrons:" << endl;getKind(donor, num, false);return 0;
}
