Learning C++ Programming from Scratch, Day 403: 1083 – Palindrome Numbers; Problem Set Answers; Fourth Method

1083 – Palindrome Numbers

This program addresses an interesting mathematical problem: Calculating how many times a number needs to undergo the “reverse and add” operation to become a palindrome (a number that reads the same forwards and backwards, such as 121).

How does the program work?

1. First, the user inputs a number (for example, 57).

2. The program performs the following actions:

• Reverses the number (57→75).

• Checks if it is a palindrome (57≠75, it is not).

• If it is not, it adds the original number and the reversed number (57+75=132).

• Records this as the first operation.

Repeats this process for the new number:

• 132→231.

• 132≠231, continue adding (132+231=363).

• Records the second operation.

Now 363 is a palindrome (363 reversed is still 363), stop!

Finally, informs the user: 2 operations were used.

Special Note::

• The program uses the <span>ji</span> variable to record the number of operations.

• It will stop when the number becomes 1 (this is a special setting).

• The final output of <span>ji-1</span> is because the first check does not count as a real operation.

Reference Code

#include<iostream>  // Include input-output library, allows using cin/coutusing namespace std; // Use standard namespace to avoid writing std::int main() // Main function of the program, execution starts here{    // Define variables:    int n;    // Store the user input number    int ni;   // Store the result of the reversed number    int ce;   // Temporary variable to hold the number during reversal    int ji=0; // Counter to record the number of operations
    cin>>n; // Read the user input number from the keyboard
    // Main loop: continue while n is not equal to 1 (this is a special setting)    while(n!=1){        ni=0;  // Clear the reverse result before each iteration        ce=n;   // Save the current number to temporary variable ce        ji++;   // Increment operation count
        // Number reversal process (core algorithm)        while(ce!=0){          // Continue while ce still has digits            ni=ni*10+ce%10;    // Add the last digit of ce to the end of ni            ce=ce/10;          // Remove the last digit from ce        }
        // Check if it is a palindrome        if(n!=ni){       // If it is not a palindrome            n=n+ni;      // Add the number and its reversed number        }    }
    cout<<ji-1; // Output the number of operations (subtract 1 because the first check does not count as an operation)    return 0;   // Program ends normally}

1. What does the program do?

This program calculates how many times you need to repeatedly “add its reverse” to any given number to obtain a palindrome. For example:

• Input 56 → 56+65=121 → 1 operation.

• Input 57 → 57+75=132 → 132+231=363 → 2 operations.

2. Program Structure Analysis

① Preparation Stage

• <span>#include<iostream></span>: Equivalent to preparing paper and pen, allowing the program to input and output.

• <span>using namespace std</span>: Equivalent to saying “the cin/cout I write later are from the standard library”.

② Variable Explanation

• <span>n</span>: The number provided by the user (which will change continuously).

• <span>ni</span>: The appearance of the number after reversal.

• <span>ce</span>: Temporary variable, helps with the reversal calculation.

• <span>ji</span>: A “notebook” for recording the number of operations (starting from 0).

③ Core Algorithm

1. Number reversal (like looking in a mirror):

• <span>ce%10</span>: Extract the last digit (123%10=3).

• <span>ce/10</span>: Remove the last digit (123/10=12).

• <span>ni=ni*10+digit</span>: Append the digit to the right of ni.

Palindrome check:

• Compare <span>n</span> and <span>ni</span> for equality.

• If different, add:<span>n = n + ni</span>

3. Example of Execution Flow

Assuming input is 57:

1. First loop:

• Reverse 57→75.

• 57≠75 → Add to get 132.

• ji=1.

Second loop:

• Reverse 132→231.

• 132≠231 → Add to get 363.

• ji=2.

Third loop:

• Reverse 363→363.

• 363=363 → Stop.

Output ji-1=1 (this may be controversial, see notes).

4. Important Details

• Why use ji-1: Because the first check has not performed any addition operation yet.

• while(n!=1): This is a special setting; normally it should be <span>while(n!=ni)</span>.

• Number Reversal Principle:

  Example: Reverse 123
  Step 1: ni=0 * 10+3=3, ce=12
  Step 2: ni=3 * 10+2=32, ce=1 
  Step 3: ni=32 * 10+1=321, ce=0
  
  
  
  

5. Cautions

1. Potential Issues:

• In the original code, <span>while(n!=1)</span> may be a bug; it should be <span>while(n!=ni)</span>.

• Some numbers may require many steps to become a palindrome.

Improvement Suggestions:

• A maximum step limit can be added to avoid infinite loops.

• Add intermediate step printing for easier debugging.

• Handle cases of negative number inputs.

6. Learning Suggestions

Beginners can practice as follows:

1. Simulate the program’s operation with paper and pen (choose a small number).

2. Try modifying the code to print each operation process.

3. Consider: What happens if the input is 1?

4. Challenge: How to make the program handle larger numbers?

This program effectively demonstrates:

• The use of loop structures.

• Techniques for manipulating digit positions.

• The basic ideas of algorithm design.

I hope this explanation helps you fully understand! If you have any questions, I recommend running the program and observing its behavior.