en
Feedback
C Programming Codes

C Programming Codes

Open in Telegram

C Programming Codes || Quizzes || DSA Learn along with the community Any queries admin - @Pradeep_saii

Show more

πŸ“ˆ Analytical overview of Telegram channel C Programming Codes

Channel C Programming Codes (@c_programming_codes) in the English language segment is an active participant. Currently, the community unites 13 430 subscribers, ranking 9 534 in the Technologies & Applications category and 32 075 in the India region.

πŸ“Š Audience metrics and dynamics

Since its creation on Π½Π΅Π²Ρ–Π΄ΠΎΠΌΠΎ, the project has demonstrated rapid growth, gathering an audience of 13 430 subscribers.

According to the latest data from 11 June, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -239 over the last 30 days and by -9 over the last 24 hours, overall reach remains high.

  • Verification status: Not verified
  • Engagement rate (ER): The average audience engagement rate is 9.78%. Within the first 24 hours after publication, content typically collects N/A% reactions from the total number of subscribers.
  • Post reach: On average, each post receives 0 views. Within the first day, a publication typically gains 0 views.
  • Reactions and interaction: The audience actively supports content: the average number of reactions per post is 0.
  • Thematic interests: Content is focused on key topics such as input, string, scanf("%d, array, element.

πŸ“ Description and content policy

The author describes the resource as a platform for expressing subjective opinions:
β€œC Programming Codes || Quizzes || DSA Learn along with the community Any queries admin - @Pradeep_saii”

Thanks to the high frequency of updates (latest data received on 12 June, 2026), the channel maintains relevance and a high level of publication reach. Analytics show that the audience actively interacts with content, making it an important point of influence in the Technologies & Applications category.

13 430
Subscribers
-924 hours
-577 days
-23930 days
Posts Archive
πŸ’» Find First N Fibonacci Numbers
#include <stdio.h>

int main() {
    int n, i;
    long long first = 0, second = 1, next;

    printf("Enter the number of Fibonacci numbers to generate: ");
    scanf("%d", &n);

    printf("First %d Fibonacci numbers are:
", n);

    i = 0;
    while (i < n) {
        printf("%lld ", first);
        next = first + second;
        first = second;
        second = next;
        i++;
    }

    printf("
");
    return 0;
}
πŸ“€ Output:
Input: 10
Output: Enter the number of Fibonacci numbers to generate: First 10 Fibonacci numbers are:
0 1 1 2 3 5 8 13 21 34

πŸ’» Print Multiplication Table
#include <stdio.h>

int main() {
    int num, i = 1;

    printf("Enter a number: ");
    scanf("%d", &num);

    while (i <= 10) {
        printf("%d * %d = %dn", num, i, num * i);
        i++;
    }

    return 0;
}
πŸ“€ Output:
Input: 5
Output: Enter a number: 5 * 1 = 5
5 * 2 = 10
5 * 3 = 15
5 * 4 = 20
5 * 5 = 25
5 * 6 = 30
5 * 7 = 35
5 * 8 = 40
5 * 9 = 45
5 * 10 = 50

πŸ’» Check if Number is Armstrong
#include <stdio.h>
#include <math.h>

int main() {
    int number, originalNumber, remainder, n = 0;
    float result = 0.0;

    printf("Enter an integer: ");
    scanf("%d", &number);

    originalNumber = number;

    // Count number of digits
    while (originalNumber != 0) {
        originalNumber /= 10;
        ++n;
    }

    originalNumber = number;

    // Calculate result
    while (originalNumber != 0) {
        remainder = originalNumber % 10;
        result += pow(remainder, n);
        originalNumber /= 10;
    }

    // Check if number is Armstrong
    if ((int)result == number)
        printf("%d is an Armstrong number.", number);
    else
        printf("%d is not an Armstrong number.", number);

    return 0;
}
πŸ“€ Output:
Input: 153
Output: 153 is an Armstrong number.

Input: 121
Output: 121 is not an Armstrong number.

Input: 370
Output: 370 is an Armstrong number.

Input: 1634
Output: 1634 is an Armstrong number.

Input: 123
Output: 123 is not an Armstrong number.

πŸ’» Check if Number is Palindrome
#include <stdio.h>

int main() {
    int n, reversed = 0, remainder, original;

    printf("Enter an integer: ");
    scanf("%d", &n);

    original = n;

    while (n != 0) {
        remainder = n % 10;
        reversed = reversed * 10 + remainder;
        n /= 10;
    }

    if (original == reversed)
        printf("%d is a palindrome.n", original);
    else
        printf("%d is not a palindrome.n", original);

    return 0;
}
πŸ“€ Output:
Input: 121
Output: 121 is a palindrome.

Input: 123
Output: 123 is not a palindrome.

Input: 12321
Output: 12321 is a palindrome.

Input: 10
Output: 10 is not a palindrome.

πŸ’» Sum of Digits of a Number
#include <stdio.h>

int main() {
    int num, sum = 0, digit;

    printf("Enter a number: ");
    scanf("%d", &num);

    while (num > 0) {
        digit = num % 10;
        sum += digit;
        num /= 10;
    }

    printf("Sum of digits: %dn", sum);

    return 0;
}
πŸ“€ Output:
Input: 12345
Output: Enter a number: Sum of digits: 15

Input: 9876
Output: Enter a number: Sum of digits: 30

Input: 0
Output: Enter a number: Sum of digits: 0

Input: 1
Output: Enter a number: Sum of digits: 1

Solve LeetCode problems consistently in structured manner Join hereπŸ‘‡ https://t.me/+L6Z9gjVIVEQ4NmRl

πŸ’» Count Digits in a Number
#include <stdio.h>

int main() {
    int number, count = 0;

    printf("Enter an integer: ");
    scanf("%d", &number);

    if (number == 0) {
        count = 1;
    } else {
        while (number != 0) {
            number /= 10;
            count++;
        }
    }

    printf("Number of digits: %dn", count);

    return 0;
}
πŸ“€ Output:
Input: 12345
Output: Number of digits: 5

Input: 0
Output: Number of digits: 1

Input: -987
Output: Number of digits: 3

Input: 10
Output: Number of digits: 2

πŸ’» Reverse a Number
#include <stdio.h>

int main() {
  int n, reversed = 0, remainder;

  printf("Enter an integer: ");
  scanf("%d", &n);

  while (n != 0) {
    remainder = n % 10;
    reversed = reversed * 10 + remainder;
    n /= 10;
  }

  printf("Reversed number = %d", reversed);

  return 0;
}
πŸ“€ Output:
Input: 123
Output: Reversed number = 321

Input: -456
Output: Reversed number = -654

Input: 0
Output: Reversed number = 0

Input: 1200
Output: Reversed number = 21

πŸ’» Factorial of a Number
#include <stdio.h>

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

  printf("Enter an integer: ");
  scanf("%d", &n);

  if (n < 0) {
    printf("Factorial is not defined for negative numbers.n");
  } else {
    int i = 1;
    while (i <= n) {
      factorial *= i;
      i++;
    }
    printf("Factorial of %d = %llun", n, factorial);
  }

  return 0;
}
πŸ“€ Output:
Input: 5
Output: Enter an integer: Factorial of 5 = 120

Input: -2
Output: Enter an integer: Factorial is not defined for negative numbers.

Input: 0
Output: Enter an integer: Factorial of 0 = 1

πŸ’» Sum of Odd Numbers from 1 to N
#include <stdio.h>

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

  printf("Enter a positive integer: ");
  scanf("%d", &n);

  i = 1;
  while (i <= n) {
    if (i % 2 != 0) {
      sum += i;
    }
    i++;
  }

  printf("Sum of odd numbers from 1 to %d is: %dn", n, sum);

  return 0;
}
πŸ“€ Output:
Input: 10
Output: Sum of odd numbers from 1 to 10 is: 25

Input: 5
Output: Sum of odd numbers from 1 to 5 is: 9

Input: 1
Output: Sum of odd numbers from 1 to 1 is: 1

Input: 2
Output: Sum of odd numbers from 1 to 2 is: 1

πŸ’» Sum of Even Numbers from 1 to N
#include <stdio.h>

int main() {
    int N, i, sum = 0;

    printf("Enter the value of N: ");
    scanf("%d", &N);

    i = 2;
    while (i <= N) {
        sum += i;
        i += 2;
    }

    printf("Sum of even numbers from 1 to %d is: %dn", N, sum);

    return 0;
}
πŸ“€ Output:
Input: 10
Output: Enter the value of N: Sum of even numbers from 1 to 10 is: 30
Input: 5
Output: Enter the value of N: Sum of even numbers from 1 to 5 is: 6
Input: 20
Output: Enter the value of N: Sum of even numbers from 1 to 20 is: 110
Input: 1
Output: Enter the value of N: Sum of even numbers from 1 to 1 is: 0

πŸ’» Sum of First N Natural Numbers
#include <stdio.h>

int main() {
    int n, i = 1, sum = 0;

    printf("Enter a positive integer: ");
    scanf("%d", &n);

    while (i <= n) {
        sum += i;
        i++;
    }

    printf("Sum of first %d natural numbers = %dn", n, sum);

    return 0;
}
πŸ“€ Output:
Input: 5
Output: Enter a positive integer: Sum of first 5 natural numbers = 15

Input: 10
Output: Enter a positive integer: Sum of first 10 natural numbers = 55

Input: 1
Output: Enter a positive integer: Sum of first 1 natural numbers = 1

Input: 0
Output: Enter a positive integer: Sum of first 0 natural numbers = 0

πŸ’» Print Odd Numbers from 1 to N
#include <stdio.h>

int main() {
    int n, i = 1;

    printf("Enter the value of N: ");
    scanf("%d", &n);

    while (i <= n) {
        if (i % 2 != 0) {
            printf("%d ", i);
        }
        i++;
    }

    printf("n");

    return 0;
}
πŸ“€ Output:
Input: 10
Output: Enter the value of N: 1 3 5 7 9

Input: 1
Output: Enter the value of N: 1

Input: 2
Output: Enter the value of N: 1

Input: 15
Output: Enter the value of N: 1 3 5 7 9 11 13 15

πŸ’» Print Even Numbers from 1 to N
#include <stdio.h>

int main() {
  int N, i = 2;

  printf("Enter the value of N: ");
  scanf("%d", &N);

  while (i <= N) {
    printf("%d ", i);
    i += 2;
  }

  printf("n");

  return 0;
}
πŸ“€ Output:
Input: 10
Output: 2 4 6 8 10

Input: 15
Output: 2 4 6 8 10 12 14

Input: 1
Output:

Input: 0
Output:

Input: -5
Output:

πŸ’» Print Numbers from N to 1
#include <stdio.h>

int main() {
  int n;

  printf("Enter a number: ");
  scanf("%d", &n);

  while (n >= 1) {
    printf("%d ", n);
    n--;
  }
  printf("n");

  return 0;
}
πŸ“€ Output:
Input: 5
Output: 5 4 3 2 1
Input: 1
Output: 1
Input: 10
Output: 10 9 8 7 6 5 4 3 2 1
Input: 0
Output:
Input: -3
Output:

πŸ’» Print Numbers from 1 to N
#include <stdio.h>

int main() {
  int N, i = 1;

  printf("Enter the value of N: ");
  scanf("%d", &N);

  while (i <= N) {
    printf("%d ", i);
    i++;
  }
  printf("n");

  return 0;
}
πŸ“€ Output:
Input: 5
Output: 1 2 3 4 5

πŸ”§ Loops - While

πŸ’» Unit Converter Using Switch
#include <stdio.h>

int main() {
  int choice;
  float value, result;

  printf("Unit Convertern");
  printf("1. Celsius to Fahrenheitn");
  printf("2. Fahrenheit to Celsiusn");
  printf("Enter your choice (1 or 2): ");
  scanf("%d", &choice);

  switch (choice) {
    case 1:
      printf("Enter temperature in Celsius: ");
      scanf("%f", &value);
      result = (value * 9 / 5) + 32;
      printf("Temperature in Fahrenheit: %.2fn", result);
      break;
    case 2:
      printf("Enter temperature in Fahrenheit: ");
      scanf("%f", &value);
      result = (value - 32) * 5 / 9;
      printf("Temperature in Celsius: %.2fn", result);
      break;
    default:
      printf("Invalid choice!n");
  }

  return 0;
}
πŸ“€ Output:
Unit Converter
1. Celsius to Fahrenheit
2. Fahrenheit to Celsius
Enter your choice (1 or 2): Input: 1
Enter temperature in Celsius: Input: 25
Temperature in Fahrenheit: 77.00
Unit Converter
1. Celsius to Fahrenheit
2. Fahrenheit to Celsius
Enter your choice (1 or 2): Input: 2
Enter temperature in Fahrenheit: Input: 98.6
Temperature in Celsius: 37.00
Unit Converter
1. Celsius to Fahrenheit
2. Fahrenheit to Celsius
Enter your choice (1 or 2): Input: 3
Invalid choice!

πŸ’» Season Identification from Month
#include <stdio.h>

int main() {
  int month;

  printf("Enter month number (1-12): ");
  scanf("%d", &month);

  switch (month) {
    case 12:
    case 1:
    case 2:
      printf("Wintern");
      break;
    case 3:
    case 4:
    case 5:
      printf("Springn");
      break;
    case 6:
    case 7:
    case 8:
      printf("Summern");
      break;
    case 9:
    case 10:
    case 11:
      printf("Autumnn");
      break;
    default:
      printf("Invalid month number.n");
  }

  return 0;
}
πŸ“€ Output:
Input: 1
Output: Winter

Input: 4
Output: Spring

Input: 7
Output: Summer

Input: 10
Output: Autumn

Input: 13
Output: Invalid month number.