C++ Lecture 6: Loop Applications and Problem-Solving

C++ Lecture 6: Loop Applications and Problem-Solving

This lecture applies for, while, and do-while loops to practical programming problems, including factorials, number sums, digit counting, and menu-driven programs.

1. Learning Objectives

By the end of this lecture, students will be able to:

• Calculate factorials using loops.
• Find the sum of a sequence of numbers.
• Count the digits of a number.
• Calculate the sum of digits.
• Reverse a number using a loop.
• Develop a simple menu-driven program.

2. Example 1: Calculate the Factorial of a Number

The factorial of a positive integer is the product of all positive integers from 1 up to that number.

For example:

\[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \]

Remember that \(0! = 1\).

C++ Code

#include <iostream>
using namespace std;

int main()
{
    int n;
    long long factorial = 1;

    cout << "Enter a number (0 to 20): ";
    cin >> n;

    if (n < 0 || n > 20)
    {
        cout << "Please enter a number from 0 to 20.";
    }
    else
    {
        for (int i = 1; i <= n; i++)
        {
            factorial = factorial * i;
        }

        cout << "Factorial = " << factorial;
    }

    return 0;
}

Understand the Code

• factorial = 1 initializes the product.
• The loop starts from 1 and continues up to n.
• factorial = factorial * i multiplies the running result by each number.
• The range check prevents negative input and values above 20, which exceed the range of a typical 64-bit signed long long factorial calculation.

For input 5, the calculation is:

Output:

Enter a number (0 to 20): 5
Factorial = 120

Modify the Code

1. Use a while loop instead of a for loop.
2. Display the multiplication steps as well as the answer.
3. Test the program with input 0.

3. Example 2: Calculate the Sum of Digits

Suppose the user enters 456. The sum of its digits is:

\[ 4 + 5 + 6 = 15 \]

C++ Code

#include <iostream>
using namespace std;

int main()
{
    int n;
    int sum = 0;

    cout << "Enter a positive integer: ";
    cin >> n;

    if (n < 0)
    {
        cout << "Please enter a non-negative integer.";
    }
    else
    {
        while (n > 0)
        {
            int digit = n % 10;
            sum = sum + digit;
            n = n / 10;
        }

        cout << "Sum of digits = " << sum;
    }

    return 0;
}

Understand the Code

Two operators are important:

• n % 10 gives the last digit.
• n / 10 removes the last digit when both operands are integers.

For example, if n = 456:

When n becomes 0, the loop stops.

Output:

Enter a positive integer: 456
Sum of digits = 15

Modify the Code

1. Find the sum of digits of 1234.
2. Count how many digits are even.
3. Count how many digits are odd.

4. Example 3: Count the Digits of a Number

For example, the number 58321 contains five digits.

C++ Code

#include <iostream>
using namespace std;

int main()
{
    int n;
    int count = 0;

    cout << "Enter an integer: ";
    cin >> n;

    if (n < 0)
    {
        n = -n;
    }

    if (n == 0)
    {
        count = 1;
    }
    else
    {
        while (n > 0)
        {
            count++;
            n = n / 10;
        }
    }

    cout << "Number of digits = " << count;

    return 0;
}

Understand the Code

• count = 0 initializes the digit counter.
• count++ increases the count by 1 for each digit.
• n = n / 10 removes the last digit.
• Zero needs special handling because it has one digit, even though the loop condition n > 0 would initially be false.

Sample output:

Enter an integer: 58321
Number of digits = 5

Modify the Code

1. Test the program with 700.
2. Test it with 0.
3. Test it with -12345.

5. Example 4: Reverse a Number

A number is reversed by taking its last digit and adding it to a new number.

For example:

1234 becomes 4321.

C++ Code

#include <iostream>
using namespace std;

int main()
{
    int n;
    int reverse = 0;

    cout << "Enter a positive integer: ";
    cin >> n;

    if (n < 0)
    {
        cout << "Please enter a non-negative integer.";
    }
    else
    {
        while (n > 0)
        {
            int digit = n % 10;
            reverse = reverse * 10 + digit;
            n = n / 10;
        }

        cout << "Reversed number = " << reverse;
    }

    return 0;
}

Understand the Code

The key statement is:

reverse = reverse * 10 + digit;

Multiplying reverse by 10 shifts its existing digits one place to the left, making room for the new last digit.

For input 1234:

Output:

Enter a positive integer: 1234
Reversed number = 4321

Note: Leading zeros disappear when a number is stored as an integer. For example, reversing 1200 produces 21.

Modify the Code

1. Reverse 56789.
2. Test the program with 1005.
3. Modify the program to check whether a number is a palindrome, such as 121.

6. Example 5: Menu-Driven Calculator

A menu-driven program repeatedly displays options until the user chooses to exit.

C++ Code

#include <iostream>
using namespace std;

int main()
{
    int choice;
    double a, b;

    do
    {
        cout << "\n--- Simple Calculator ---" << endl;
        cout << "1. Addition" << endl;
        cout << "2. Subtraction" << endl;
        cout << "3. Exit" << endl;
        cout << "Enter your choice: ";
        cin >> choice;

        if (choice == 1 || choice == 2)
        {
            cout << "Enter two numbers: ";
            cin >> a >> b;

            if (choice == 1)
            {
                cout << "Result = " << a + b << endl;
            }
            else
            {
                cout << "Result = " << a - b << endl;
            }
        }
        else if (choice == 3)
        {
            cout << "Calculator closed." << endl;
        }
        else
        {
            cout << "Invalid choice." << endl;
        }

    }
    while (choice != 3);

    return 0;
}

Understand the Code

• The do-while loop displays the menu at least once.
• choice stores the selected option.
• The if-else statements determine which operation to perform.
• The menu repeats until the user enters 3.

Modify the Code

1. Add multiplication as option 3 and move Exit to option 4.
2. Add division as another operation.
3. Display a warning if the user tries to divide by zero.

7. Coding Exercises

Solve the following tasks independently.

Level 1 — Basic

1. Print all numbers from 1 to 100.

2. Calculate the sum of numbers from 1 to n.

3. Print the multiplication table of a number.

4. Count the digits of an integer.

5. Calculate the sum of the digits of an integer.

6. Reverse an integer using a loop.

7. Calculate the factorial of a number.

8. Check whether a number is a palindrome.

9. Display all even numbers between two user-entered limits.

10. Create a menu-driven calculator using do-while.

0 of 10 completed

8. Quick Revision Quiz

1. What does `n % 10` return for n = 456?

45

6

456

2. What is the value of 5!?

25

60

120

3. Which loop is suitable for a menu that should appear at least once?

do-while

if only

No loop

4. What does integer division `456 / 10` produce in C++?

45.6

45

46

Check Answers

9. Lesson Summary

• Loops automate repeated calculations.
• Factorial problems use multiplication and a running product.
• Digit problems commonly use % 10 and / 10.
• Nested loops handle repeated rows and columns.
• break stops a loop, while continue skips an iteration.
• do-while is useful for menus that should appear at least once.

Suggested next lecture: Choosing the right loop (for vs while vs do-while) and solving mixed loop problems with trace tables.