1.编写一个要求用户输入两个证书的小程序。该程序将计算并输出两个整数之间（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，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.

该题使用一个for循环就可以解决，代码如下：

// ex1.cpp -- calculate the sum of between two integers' all integer
#include<iostream>
int main()
{
    using namespace std;
    int i,j;
    int sum = 0;
    cout << "Enter two integers: ";
    cin >> i >> j;
    for(;i<=j;i++)
        sum += i;
    cout << "The sum of between two integers' all integers is ";
    cout << sum << endl;
    return 0;
}

运行结果如下：

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。

该题要求使用array对象，数据类型为long double，
array<long double,100> factories
代码如下：

// ex2.cpp -- calculate 100!
#include<iostream>
#include<array>
const int arsize = 101;
int main()
{
    using namespace std;
    array<long double,arsize> factories;
    factories[0] = 1;
    factories[1] = 1;
    for(int i = 2; i < arsize; ++i)
        factories[i] = factories[i-1] * i;
    cout << "100! = " << factories[100] << endl;
    return 0;
}

运行结果如下：

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。

本题不知道循环次数，显然应该用while循环。
代码如下：

// ex3.cpp -- calculate sum
#include<iostream>
int main()
{
    using namespace std;
    double i;
    double sum = 0;
    cout << "Enter a figure, input 0 will end.\n";
    cin >> i;
    while(i)
    {
        sum += i;
        cout << "sum = " << sum << endl;
        cin >> i;
    }
    return 0;
}

运行结果如下：

4. 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=0.10x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=0.05x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，依此类推。请编写一个程序，计算多少年之后，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。
本题首先分别计算两人的盈利:
Daphne: profit_d = 100 + 10*year;
Cleo: profit_c = 100*1.05^year;
写成循环，代码如下：

// ex4.cpp -- calculate profit
#include<iostream>
const int money_init = 100;
int main()
{
    using namespace std;
    double profit_c = money_init;
    int profit_d = money_init;
    int year = 0;
    while(profit_c<=profit_d)
    {
        ++ year;
        profit_d += 10;
        profit_c += 0.05 * profit_c;
    }
    cout << "After " << year << " years, ";
    cout << "Cleo over Daphne.\n";
    cout << "Cleo's profit: " << profit_c << endl;
    cout << "Daphne's profit: " << profit_d << endl;
    return 0;
}

运行结果如下：

5. 假设要销售《C++ForFools》一书。请编写换一个程序，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（或者string对象数组进行逐月提示），并将输入的数据存储在一个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。

本题使用string对象数组构建十二个月的字符，然后用for循环来进行输入和数据处理，代码如下：

// ex5.cpp --calculate a year sales of book
#include<iostream>
#include<string>
const int Months = 12;
int main()
{
    using namespace std;
    string month[Months]={"January", "February","March","April","May",
    "June","July","August","September","October","November","December"};
    // const char *month[Months] = {"January", "February","March","April","May",
    // "June","July","August","September","October","November","December"};
    long sum = 0;
    int sale[Months];
    for(int i = 0; i < Months; ++i)
    {
        cout << "Enter the amount of sale in " << month[i] << ": ";
        cin >> sale[i];
        sum += sale[i];
    }
    cout << "The sale volume of this year is " << sum << endl;
    return 0;
}

注：使用char * 定义数组时初始化时，应该加const.
运行结果如下：

6. 完成编程练习5，但这一次使用一个二位数组来输入——3年中每个月的销售量。程序将报告每年销售量以及三年的总销售量。

本题涉及到二维数组的定义和循环遍历。
代码如下：

// ex6.cpp --calculate three year sales of book
#include<iostream>
#include<string>
const int Months = 12;
const int Years = 3;
int main()
{
    using namespace std;
    string month[Months]={"January", "February","March","April","May",
    "June","July","August","September","October","November","December"};
    const char* year[Years] = {"2019", "2020", "2021"};
    // const char *month[Months] = {"January", "February","March","April","May",
    // "June","July","August","September","October","November","December"};
    long sum_year = 0;
    long sum = 0;
    int sale[Months][Years];
    for(int i = 0; i < Years; ++i)
    {
        for(int j = 0; j < Months; ++j)
        {
            cout << "Enter the amount of sale in " << year[i] << ",";
            cout << month[j] << ": ";
            cin >> sale[i][j];
            sum_year += sale[i][j];
        }
        cout << "The sales in " << year[i] << ": " << sum_year << endl;
        sum += sum_year;
        sum_year = 0;
    }
    cout << "The sales of three years: " << sum << endl;
    return 0;
}

运行结果如下：

Enter the amount of sale in 2019,January: 1000
Enter the amount of sale in 2019,February: 1001
Enter the amount of sale in 2019,March: 1002
Enter the amount of sale in 2019,April: 1003
Enter the amount of sale in 2019,May: 1004
Enter the amount of sale in 2019,June: 1005
Enter the amount of sale in 2019,July: 1006
Enter the amount of sale in 2019,August: 1007
Enter the amount of sale in 2019,September: 1008
Enter the amount of sale in 2019,October: 1009
Enter the amount of sale in 2019,November: 1010
Enter the amount of sale in 2019,December: 1011
The sales in 2019: 12066
Enter the amount of sale in 2020,January: 2001
Enter the amount of sale in 2020,February: 2002
Enter the amount of sale in 2020,March: 2003
Enter the amount of sale in 2020,April: 1004
Enter the amount of sale in 2020,May: 2005
Enter the amount of sale in 2020,June: 2006
Enter the amount of sale in 2020,July: 2007
Enter the amount of sale in 2020,August: 2008
Enter the amount of sale in 2020,September: 2009
Enter the amount of sale in 2020,October: 2010
Enter the amount of sale in 2020,November: 2011
Enter the amount of sale in 2020,December: 2012 
The sales in 2020: 23078
Enter the amount of sale in 2021,January: 3001
Enter the amount of sale in 2021,February: 3002
Enter the amount of sale in 2021,March: 3003
Enter the amount of sale in 2021,April: 3004
Enter the amount of sale in 2021,May: 3005
Enter the amount of sale in 2021,June: 2006
Enter the amount of sale in 2021,July: 3007
Enter the amount of sale in 2021,August: 3008
Enter the amount of sale in 2021,September: 3009
Enter the amount of sale in 2021,October: 3010
Enter the amount of sale in 2021,November: 3011
Enter the amount of sale in 2021,December: 3012
The sales in 2021: 35078
The sales of three years: 70222

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（存储在字符数组或者string对象的字符串）、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
本题首先定义一个car结构，使用new创建一个动态数组，然后按照题目要求进行输入和输出。

• 注意string输入一行时用getline(cin,string),且回车符会被读取
• 在数字与字符串交替读取时，注意读取数字会保留回车符，因此需要用cin.get()

代码如下：

// ex7.cpp -- struct,new,for
#include<iostream>
#include<string>
#include<cstring>

struct car
{
    std::string make;
    int year;
};

int main()
{
    using namespace std;
    char temp[20];
    int temp_y;
    int n;
    cout << "How many cars do you wish to catalog? ";
    (cin >> n).get();
    car* Car = new car[n];
    for(int i = 0; i < n; ++i)
    {
        cout << "Car #" << (i+1) << ":\n"; 
        cout << "Please enter the make: ";
        getline(cin,Car[i].make);
        cout << "Please enter the year made: ";
        (cin >> Car[i].year).get();
    }
    for(int i = 0; i < n; ++i)
        cout << (Car+i)->year << " " << Car[i].make << endl;
    return 0;
}

运行结果如下:

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）。下面是程序的运行情况：

Enter words(to stop, typer the word done):
anteater birthday category dumpster
envy finagle geometry done for sure
You entered a total of 7 words.
您应在程序中包含头文件cstring，并使用函数strcmp来进行比较测试。
本题每次读取一个单词的意思，笔者刚开始误解了，以为要自己写一个while循环，按一个字符一个字符输入，遇到空格 ' '停止，重新开始读取下一个单词，存在许多问题，如临时存储数组的更新等等，导致程序一直调试错误，这里挖个坑吧，等把C++这本书学会了，再来用这次的理解做一下这道题。目前使用cin>>直接按照单词读取，实际上题目也是这么要求的，代码如下：

// ex8.cpp -- get the number of enter words
#include<iostream>
#include<cstring>
int main()
{
    using namespace std;
    char temp[20]; 
    int count = 0;
    int i = 0;
    cout << "Enter words(to stop, type the word done): \n";
    cin >> temp;
    while(strcmp(temp,"done")) // or while(strcmp(temp,"done")!=0)
    {
		/* use while or other to get cin >> temp*/
        // cin >> temp[i];
        // while(temp[i] != ' ')
        // {
        //     i++;
        //     cin >> temp[i];
        // }
        // i = 0;
        cin >> temp;
        count++;
    }
    cout << "You entered a total of " << count << " words.\n";
    return 0;
}

注：准确的分析题目比编程更重要。
运行结果如下：

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。

本题与第八题类似，需要修改的代码就是单词声明，与while的test-conditon, 代码如下：

// ex9.cpp -- get the number of enter words by using string
#include<iostream>
#include<string>
int main()
{
    using namespace std;
    string word; 
    int count = 0;
    int i = 0;
    cout << "Enter words(to stop, type the word done): \n";
    cin >> word;
    while(word != "done")
    {
        cin >> word;
        count++;
    }
    cout << "You entered a total of " << count << " words.\n";
    return 0;
}

运行结果如下：

10. 编写一个使用循环嵌套的程序，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，依此类推。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，在星号的前面加上句点。该程序的运行情况如下：

Enter number of rows: 5
....*
...**
..***
.****
*****
根据图形分析，首先需要一个for循环输出row行字符，在上述for循环里嵌套for循环输入row-当前行数的.，接着输出当前行数个的*。代码如下：

// ex10.cpp -- draw figure
#include<iostream>
int main()
{
    using namespace std;
    int rows;
    cout << "Enter number of rows: ";
    cin >> rows;
    for(int i = 1; i <= rows; ++i)
    {
        for(int j = rows -i; j > 0; --j)
        {
            cout <<".";
        }  
        for(int k = 1; k <=i; ++k)
            cout <<"*";
        cout << endl;
    }
    return 0;
}

运行结果如下：