Growing Reverse Number Pattern in C++

What You’ll Learn

Program 61 prints a growing reverse-number pattern: starting from an integer like 86523, build the reverse one digit at a time and print each partial result — 3, 32, 325, and so on. This tutorial covers modulo, reverse building, a live preview, worked C++ examples, edge cases, and O(d) complexity.

📝 Problem & Approach

Given starting integer num = 86523, build its reverse one digit at a time and print each partial reverse until all digits are processed.

// num = 86523
//3
//32
//325
//3256
//32568

Inputs & Outputs

Minimal workflow

reverse = 0
while num is not 0:
    digit = num % 10
    reverse = reverse * 10 + digit
    print reverse
    num = num / 10

Approach comparison

🔮 Live Preview

Enter a positive integer between 10 and 99999999 and draw the growing reverse pattern in the browser.

Press "Draw pattern".

Examples Gallery

Three complete C programs — fixed starting value, user input with long, and a compact num = 123 trace. Click View Output to reveal sample console results.

#include <iostream>
using namespace std;

int main() {
    int num = 86523;
    int reverse = 0;

    while (num != 0) {
        reverse = reverse * 10;
        reverse = reverse + (num % 10);
        cout << reverse << "\n";
        num = num / 10;
    }

    return 0;
}

3
32
325
3256
32568

#include <iostream>
#include <cstdlib>
using namespace std;

int main() {
    long long num;

    cout << "Enter a number: ";
    cin >> num;
    if (cin.fail()) {
        cout << "Please enter a valid integer.\n";
        return 1;
    }

    num = llabs(num);
    long long reverse = 0;

    while (num != 0) {
        reverse = reverse * 10 + (num % 10);
        cout << reverse << "\n";
        num /= 10;
    }

    return 0;
}

Enter a number: 86523
3
32
325
3256
32568

#include <iostream>
using namespace std;

int main() {
    int num = 123;
    int reverse = 0;

    while (num != 0) {
        reverse = reverse * 10 + (num % 10);
        cout << reverse << "\n";
        num /= 10;
    }

    return 0;
}

3
32
321

🔎 Worked Walkthrough — num = 86523

Trace each loop iteration: extract the digit, update reverse, print, then shrink num.

Total lines printed: 5 = digit count of the starting number. Final reverse equals the full reversed number.

🎯 Practice Problems

Try these variations to lock in the pattern.

Notes

⏱️ Time and Space Complexity

💡 Best Practices