C Programming Language || Hands On Coding
Hands-on C programming language challenges for beginners. Learn building logic by solving programs. Owner: @Pradeep_saii
Show moreπ Analytical overview of Telegram channel C Programming Language || Hands On Coding
Channel C Programming Language || Hands On Coding (@c_programming_language_coding) in the English language segment is an active participant. Currently, the community unites 12 822 subscribers, ranking 9 562 in the Technologies & Applications category and 31 207 in the India region.
π Audience metrics and dynamics
Since its creation on Π½Π΅Π²ΡΠ΄ΠΎΠΌΠΎ, the project has demonstrated rapid growth, gathering an audience of 12 822 subscribers.
According to the latest data from 26 August, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -210 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 12.56%. Within the first 24 hours after publication, content typically collects 2.42% reactions from the total number of subscribers.
- Post reach: On average, each post receives 1 612 views. Within the first day, a publication typically gains 310 views.
- Reactions and interaction: The audience actively supports content: the average number of reactions per post is 4.
- Thematic interests: Content is focused on key topics such as input, string, scanf("%d, array, element.
π Description and content policy
The author describes the resource as a platform for expressing subjective opinions:
βHands-on C programming language challenges for beginners. Learn building logic by solving programs.
Owner: @Pradeep_saiiβ
Thanks to the high frequency of updates (latest data received on 27 August, 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.
#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;
}
π€ 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.
#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;
}
π€ 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
#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;
}
π€ 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
#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;
}
π€ 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
#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;
}
π€ 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
#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;
}
π€ 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.
#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;
}
π€ 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.
#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;
}
π€ 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
#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;
}
π€ 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
#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;
}
π€ 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
#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;
}
π€ 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
#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;
}
π€ 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
#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;
}
π€ 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
#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;
}
π€ 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
#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;
}
π€ 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
#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;
}
π€ 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:
#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;
}
π€ 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:
