C Programming Language || Hands On Coding

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Hands-on C programming language challenges for beginners. Learn building logic by solving programs. Owner: @Pradeep_saii

India31 202 Tecnologías y Aplicaciones9 572...

📈 Análisis del canal de Telegram C Programming Language || Hands On Coding

El canal C Programming Language || Hands On Coding (@c_programming_language_coding) en el segmento lingüístico de Inglés es un actor destacado. Actualmente la comunidad reúne a 12 822 suscriptores, ocupando la posición 9 572 en la categoría Tecnologías y Aplicaciones y el puesto 31 202 en la región India.

📊 Métricas de audiencia y dinámica

Desde su creación el невідомо, el proyecto ha mostrado un crecimiento acelerado, reuniendo a 12 822 suscriptores.

Según los últimos datos del 27 agosto, 2026, el canal mantiene una actividad estable. En los últimos 30 días la variación de miembros fue de -217, y en las últimas 24 horas de -11, conservando un alto alcance.

Estado de verificación: No verificado
Tasa de interacción (ER): El promedio de interacción de la audiencia es 6.62%. Durante las primeras 24 horas tras publicar, el contenido suele obtener 2.42% de reacciones respecto al total de suscriptores.
Alcance de las publicaciones: Cada publicación recibe en promedio 849 visualizaciones. En el primer día suele acumular 310 visualizaciones.
Reacciones e interacción: La audiencia responde de forma activa: el promedio de reacciones por publicación es 2.
Intereses temáticos: El contenido se centra en temas clave como input, string, scanf("%d, array, element.

📝 Descripción y política de contenido

El autor describe el recurso como un espacio para expresar opiniones subjetivas:
“Hands-on C programming language challenges for beginners. Learn building logic by solving programs. Owner: @Pradeep_saii”

Gracias a la alta frecuencia de actualizaciones (últimos datos recibidos el 28 agosto, 2026), el canal mantiene la vigencia y un amplio alcance. La analítica demuestra que la audiencia interactúa activamente con el contenido, lo que lo convierte en un punto de referencia dentro de la categoría Tecnologías y Aplicaciones.

12 822

Suscriptores

-1124 horas

-367 días

-21730 días

849

Visitas de la publicación

~ 31024 horas

~ 40048 horas

6.62%

Tasa de compromiso

Sin datos

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Archivo de publicaciones

12 822

💻 Sum of Digits of a Number

#include <stdio.h>

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

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

  if (num < 0) {
    printf("Please enter a positive integer.n");
    return 1;
  }

  for (; num != 0; num /= 10) {
    digit = num % 10;
    sum += digit;
  }

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

  return 0;
}

📤 Output:

Input: 12345
Output: Enter a positive integer: Sum of digits = 15

Input: 9876
Output: Enter a positive integer: Sum of digits = 30

Input: 0
Output: Enter a positive integer: Sum of digits = 0

Input: -123
Output: Enter a positive integer: Please enter a positive integer.

12 822

💻 Print Multiplication Table

#include <stdio.h>

int main() {
    int num, i;

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

    for (i = 1; i <= 10; i++) {
        printf("%d * %d = %dn", num, i, num * i);
    }

    return 0;
}

📤 Output:

Input: 5
Output: Enter an integer: 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

12 822

🔧 Loops - For

12 822

💻 Find GCD/HCF of Two Numbers

#include <stdio.h>

int main() {
    int num1, num2, gcd;

    printf("Enter two integers: ");
    scanf("%d %d", &num1, &num2);

    while (num1 != num2) {
        if (num1 > num2) {
            num1 -= num2;
        } else {
            num2 -= num1;
        }
    }

    gcd = num1;

    printf("GCD = %d", gcd);

    return 0;
}

📤 Output:

Input: 12 18
Output: Enter two integers: GCD = 6
Input: 25 15
Output: Enter two integers: GCD = 5
Input: 10 10
Output: Enter two integers: GCD = 10
Input: 48 18
Output: Enter two integers: GCD = 6

12 822

💻 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;
}

📤 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

12 822

💻 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;
}

📤 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

12 822

💻 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;
}

📤 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.

12 822

💻 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;
}

📤 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.

12 822

💻 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;
}

📤 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

12 822

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12 822

💻 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;
}

📤 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

12 822

💻 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;
}

📤 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

12 822

💻 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;
}

📤 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

12 822

💻 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;
}

📤 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

12 822

💻 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;
}

📤 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

12 822

💻 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;
}

📤 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

12 822

💻 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;
}

📤 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

12 822

💻 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;
}

📤 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:

12 822

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12 822

💻 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;
}

📤 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: