1. 编写一个要求用户输入两个整数的程序。该程序将计算并输出这两个整数之间（包括这两个整数）所有整数的和。这里假设先输入较小的整数。例如，如果用户输入的是2和9，则程序将指出2-9之间所有整数的和为44.
#include<iostream> using namespace std; int main() { int min, max; int total = 0; cout << "Please enter two integers: "; cin >> min >> max; for (int i = min; i <= max; i++) { total += i; } cout << "The total between " << min << " and " << max << " is " << total << endl; system("pause"); return 0; }2. 使用array对象（而不是数组）和long double（而不是long long）重新编写程序清单5.4，并计算100！的值。
#include<iostream> #include<array> using namespace std; const int Size = 101; int main() { array<long double, Size> factorials; factorials[0] = factorials[1] = 1; for (int i = 2; i < Size; i++) { factorials[i] = i * factorials[i - 1]; } for (int j = 0; j < Size; j++) { cout << j << "! = " << factorials[j] << endl; } system("pause"); return 0; }3. 编写一个要求用户输入数字的程序。每次输入后，程序都将报告到目前为止，所有输入的累计和。当用户输入0时，程序结束。
#include<iostream> using namespace std; int main() { int number; int total = 0; cout << "please enter a number: "; while (cin >> number && number != 0) { total += number; cout << "The total of these number is " << total << endl; cout << "Please enter a number again(0 is quit):"; } system("pause"); return 0; } Daphne以10%的单利投资了100美元。也就是说，每一年的利润都是投资额的10%,即每年10美元： 利息 = 0.10 * 原始存款 而Cleo以5%的复利投资了100美元。也就是说，利息是当前存款（包括获得的利息）的5%： 利息 = 0.05 * 当前存款 Cleo在第一年投资100美元的盈利是5%——得到了105美元。下一年的盈利是105美元的5%——即5.25美元，依次类推。请编写一个程序，计算多少年后，Cleo的投资价值才能超过Daphne的投资价值，并显示此时两个人的投资价值。 #include<iostream> using namespace std; int main() { int year = 0; double Daphne = 100; double Cleo = 100; while (Daphne >= Cleo) { Daphne += 100 * 0.1; Cleo = Cleo * 1.05; year++; } cout << year << " years later, Cleo's money exceeds Daphne's money\n"; cout << year << " years later, Cleo's money is " << Cleo << ", and Daphne's money is " << Daphne << endl; system("pause"); return 0; }5. 假设要销售《C++ For Fools》一书。请编写一个程序，输入全年中每个月的销售量（图书数量，而不是销售额）。程序通过循环，使用初始化为月份字符串的char*数组（或string对象数组）逐月进行提示，并将输入的数据储存在一个int数组中。然后，程序计算数组中各元素的总数，并报告这一年的销售情况。
#include<iostream> #include<string> using namespace std; const int Number = 12; int main() { /*string month[Number] = { {"January"}, {"February"},{"March"},{"April"}, {"May"},{"June"},{"July"},{"August"},{"September"}, {"October"},{"November"},{"December"} };*/ char * month[Number] = { { "January" },{ "February" },{ "March" },{ "April" }, { "May" },{ "June" },{ "July" },{ "August" },{ "September" }, { "October" },{ "November" },{ "December" } }; int sales; int total = 0; for (int i = 0; i < Number; i++) { cout << month[i] << " sales are "; cin >> sales; total += sales; } cout << "The total of sales is " << total << endl; system("pause"); return 0; }6. 完成编程练习5，但这一次使用一个二维数组来存储输入——3年中每个月的销售量。程序将报告每年的销售量以及三年的总销售量。
#include<iostream> #include<string> using namespace std; const int Number = 12; const int Year = 3; int main() { /*string month[Number] = { {"January"}, {"February"},{"March"},{"April"}, {"May"},{"June"},{"July"},{"August"},{"September"}, {"October"},{"November"},{"December"} };*/ char * month[Number] = { { "January" },{ "February" },{ "March" },{ "April" }, { "May" },{ "June" },{ "July" },{ "August" },{ "September" }, { "October" },{ "November" },{ "December" } }; int sales; int total = 0; int Sales[3][12]; for (int j = 0; j < Year; j++) { int year_total = 0; for (int i = 0; i < Number; i++) { cout << j + 1 << "-th year's " << month[i] << " sales are "; cin >> sales; Sales[j][i] = sales; year_total += sales; } cout << "The total of " << j + 1 << "-th sales is " << year_total << endl; total += year_total; } cout << "The total of" << Year << " sales is " << total << endl; for (int j = 0; j < Year; j++) { for (int i = 0; i < Number; i++) { cout << Sales[j][i] << "\t"; } cout << endl; } system("pause"); return 0; }7. 设计一个名为car的结构，用它存储下述有关汽车的信息： 生产商（存储在字符数组或string对象中的字符串）、生产年份（整数）。编写一个程序，向用户询问有多少辆汽车。随后，程序使用new来创建一个由相应数量的car结构组成的动态数组。接下来，程序提示用户输入每辆车的生产商（可能由多个单词组成）和年份信息。请注意，这需要特别小心，因为它将交替读取数值和字符串（参见第4章）。最后，程序将显示每个结构的内容。该程序的运行情况如下： How many cars do you wish to catalog? 2 Car #1: Please enter the make: Hudson Hornet Please enter the year made: 1952 Car #2: Please enter the make: Kaiser Please enter the year made: 1951 Here is your collection: 1952 Hudson Hornet 1951 Kaiser
#include<iostream> #include<string> using namespace std; struct car { string name; int year; }; int main() { int number; cout << "How many cars do you wish to catalog?"; cin >> number; cin.get(); car * information = new car[number]; for (int i = 0; i < number; i++) { cout << "Car #" << i + 1 << ":" << endl; cout << "Please enter the make: "; getline(cin, information[i].name); cout << "Please enter the year made: "; cin >> information[i].year; cin.get(); } cout << "Here is your collect: " << endl; for (int i = 0; i < number; i++) { cout << information[i].year << "\t" << information[i].name << endl; } system("pause"); return 0; }8. 编写一个程序，它使用一个char数组和循环来每次读取一个单词，直到用户输入done为止。随后，该程序指出用户输入了多少个单词（不包括done在内）。下面是该程序的运行情况： Enter words (to stop, type the word done): anteater birthday category dumpster envy finagle geometry done for sure You entered a toal of 7 words. 您应该在程序中包含头文件cstring,并使用函数strcmp()来进行比较测试。
#include<iostream> #include<cstring> using namespace std; int main() { cout << "Enter words (to stop, type the word done):" << endl; char ch[20]; int count = 0; while ((cin >> ch) && strcmp(ch, "done")) count++; cout << count << endl; system("pause"); return 0; }9. 编写一个满足前一个练习中描述的程序，但使用string对象而不是字符数组。请在程序中包含头文件string，并使用关系运算符来进行比较测试。
#include <iostream> #include <string> using namespace std; int main() { cout << "Enter words (to stop, type the word done):" << endl; string str; int count = 0; while (cin >> str && str != "done") count++; cout << count << endl; system("pause"); return 0; }10. 编写一个使用嵌套循环的程序，要求用户输入一个值，指出要显示多少行。然后，程序将显示相应的行数的星号，其中第一行包括一个星号，第二行包括两个星号，依次类推。每一行包含的字符数等于用户指定的行数，在星号不够的情况下，在星号前面加上句点。该程序的运行情况如下： Enter number of rows: 5 ….* …** ..*** .**** *****
#include<iostream> using namespace std; int main() { int rows; cout << "Enter number of rows: "; cin >> rows; for (int i = 1; i <= rows; i++) { for (int j = 0; j < rows-i; j++) { cout << "."; } for (int j = 0; j < i; j++) { cout << "*"; } cout << endl; } system("pause"); return 0; }