C Programming Codes

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📈 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 13 412 subscribers, ranking 9 552 in the Technologies & Applications category and 32 040 in the India region.

📊 Audience metrics and dynamics

Since its creation on невідомо, the project has demonstrated rapid growth, gathering an audience of 13 412 subscribers.

According to the latest data from 13 June, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -228 over the last 30 days and by -2 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.

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The author describes the resource as a platform for expressing subjective opinions:
“C 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 14 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.

13 412

Subscribers

-224 hours

-497 days

-22830 days

1 313

Post views

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No data48 hours

9.78%

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Posts per day

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Posts Archive

13 408

#CProgramming #Algorithms #Exponentiation

13 408

Fast Exponentiation (Binary Exponentiation)

#include <stdio.h>

long long power(long long base, long long exp, long long mod) {
    long long res = 1;
    base %= mod;
    while (exp > 0) {
        if (exp % 2 == 1)
            res = (res * base) % mod;
        base = (base * base) % mod;
        exp /= 2;
    }
    return res;
}

int main() {
    long long base = 2;
    long long exp = 10;
    long long mod = 1000000007;
    long long result = power(base, exp, mod);
    printf("%lld\n", result);
    return 0;
}

13 408

#CProgramming #GCD #Arrays

13 408

GCD of Array Elements

#include <stdio.h>

int gcd(int a, int b) {
    if (b == 0) {
        return a;
    }
    return gcd(b, a % b);
}

int gcd_array(int arr[], int n) {
    int result = arr[0];
    for (int i = 1; i < n; i++) {
        result = gcd(result, arr[i]);
    }
    return result;
}

int main() {
    int arr[] = {12, 18, 24, 30};
    int n = sizeof(arr) / sizeof(arr[0]);
    int result = gcd_array(arr, n);
    printf("GCD of the array is: %d\n", result);
    return 0;
}

13 408

#CProgramming #LCM #GCD

13 408

LCM of Two Numbers

#include <stdio.h>

int gcd(int a, int b) {
    if (b == 0)
        return a;
    return gcd(b, a % b);
}

int lcm(int a, int b) {
    return (a * b) / gcd(a, b);
}

int main() {
    int num1, num2;
    scanf("%d %d", &num1, &num2);
    printf("%d\n", lcm(num1, num2));
    return 0;
}

13 408

#CProgramming #GCD #EuclideanAlgorithm

13 408

GCD of Two Numbers (Euclidean Algorithm)

#include <stdio.h>

int gcd(int a, int b) {
  while (b != 0) {
    int temp = b;
    b = a % b;
    a = temp;
  }
  return a;
}

int main() {
  int num1, num2;
  printf("Enter two integers: ");
  scanf("%d %d", &num1, &num2);
  printf("GCD of %d and %d is %d\n", num1, num2, gcd(num1, num2));
  return 0;
}

13 408

#CProgramming #StrongNumber #InterviewPrep

13 408

Check Strong Number

#include <stdio.h>

int factorial(int n) {
    if (n == 0)
        return 1;
    else
        return n * factorial(n - 1);
}

int isStrong(int num) {
    int sum = 0, temp = num, digit;
    while (temp > 0) {
        digit = temp % 10;
        sum += factorial(digit);
        temp /= 10;
    }
    return (sum == num);
}

int main() {
    int number;
    scanf("%d", &number);
    if (isStrong(number))
        printf("Strong Number\n");
    else
        printf("Not Strong Number\n");
    return 0;
}

13 408

#CProgramming #DigitsCount #InterviewPrep

13 408

Count Digits Without % or /

#include <stdio.h>

int countDigits(int n) {
  int count = 0;
  if (n == 0) return 1;
  while (n > 0) {
    n = n - (n / 10) * 10;
    n = n / 10;
    count++;
  }
  return count;
}

int main() {
  int num = 12345;
  int digitCount = countDigits(num);
  printf("Number of digits in %d is %d\n", num, digitCount);
  return 0;
}

13 408

#CProgramming #PrimeNumbers

13 408

Check Prime Number in Range

#include <stdio.h>
#include <stdbool.h>
#include <math.h>

bool isPrime(int num) {
 if (num <= 1) return false;
 for (int i = 2; i <= sqrt(num); i++) {
 if (num % i == 0) return false;
 }
 return true;
}

void printPrimesInRange(int start, int end) {
 for (int i = start; i <= end; i++) {
 if (isPrime(i)) {
 printf("%d ", i);
 }
 }
 printf("\n");
}

int main() {
 int start, end;
 printf("Enter the start of the range: ");
 scanf("%d", &start);
 printf("Enter the end of the range: ");
 scanf("%d", &end);
 printf("Prime numbers in the range %d to %d are: ", start, end);
 printPrimesInRange(start, end);
 return 0;
}

13 408

#CProgramming #DigitsCount #InterviewPrep

13 408

Count Digits Without % or /

#include <stdio.h>

int countDigits(int n) {
  int count = 0;
  if (n == 0) return 1;
  while (n > 0) {
    n = n - (n / 10) * 10;
    n = n / 10;
    count++;
  }
  return count;
}

int main() {
  int num = 12345;
  int digitCount = countDigits(num);
  printf("Number of digits in %d is %d\n", num, digitCount);
  return 0;
}

13 408

#CProgramming #PerfectNumber #Algorithm

13 408

Check Perfect Number

#include <stdio.h>
#include <stdbool.h>

bool isPerfect(int num) {
 if (num <= 1) return false;
 int sum = 1;
 for (int i = 2; i * i <= num; i++) {
 if (num % i == 0) {
 sum += i;
 if (i * i != num) {
 sum += num / i;
 }
 }
 }
 return sum == num;
}

int main() {
 int number;
 scanf("%d", &number);
 if (isPerfect(number)) {
 printf("%d is a perfect number.\n", number);
 } else {
 printf("%d is not a perfect number.\n", number);
 }
 return 0;
}

13 408

#CProgramming #ArmstrongNumber #InterviewPrep

13 408

Check Armstrong Number (n digits)

#include <stdio.h>
#include <math.h>

int main() {
    int num, originalNum, remainder, n = 0;
    float result = 0.0;

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

    originalNum = num;

    while (originalNum != 0) {
        originalNum /= 10;
        ++n;
    }

    originalNum = num;

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

    if ((int)result == num)
        printf("%d is an Armstrong number.\n", num);
    else
        printf("%d is not an Armstrong number.\n", num);

    return 0;
}