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;
}