C Programming Language || Hands On Coding
Kanalga Telegram’da o‘tish
Hands-on C programming language challenges for beginners. Learn building logic by solving programs. Owner: @Pradeep_saii
Ko'proq ko'rsatish📈 Telegram kanali C Programming Language || Hands On Coding analitikasi
C Programming Language || Hands On Coding (@c_programming_language_coding) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 12 822 obunachidan iborat bo'lib, Texnologiyalar & Aralashmalar toifasida 9 562-o'rinni va Hindiston mintaqasida 31 207-o'rinni egallagan.
📊 Auditoriya ko‘rsatkichlari va dinamika
невідомо sanasidan buyon loyiha tez o‘sib, 12 822 obunachiga ega bo‘ldi.
26 Avgust, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -210 ga, so‘nggi 24 soatda esa -2 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.
- Tasdiqlash holati: Tasdiqlanmagan
- Jalb etish (ER): Auditoriya o‘rtacha 12.56% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 2.42% ini tashkil etuvchi reaksiyalarni to‘playdi.
- Post qamrovi: Har bir post o‘rtacha 1 612 marta ko‘riladi; birinchi sutkada odatda 310 ta ko‘rish yig‘iladi.
- Reaksiyalar va o‘zaro ta’sir: Auditoriya faol: har bir postga o‘rtacha 4 ta reaksiya keladi.
- Tematik yo‘nalishlar: Kontent input, string, scanf("%d, array, element kabi asosiy mavzularga jamlangan.
📝 Tavsif va kontent siyosati
Muallif resursni shaxsiy fikrni ifoda etish maydoni sifatida ta’riflaydi:
“Hands-on C programming language challenges for beginners. Learn building logic by solving programs.
Owner: @Pradeep_saii”
Yuqori yangilanish chastotasi (oxirgi ma’lumot 27 Avgust, 2026 da olingan) sababli kanal doimo dolzarb va katta qamrovli bo‘lib qoladi. Analitika auditoriya kontent bilan faol hamkorlik qilishini, uni Texnologiyalar & Aralashmalar toifasidagi muhim ta’sir nuqtasiga aylantirishini ko‘rsatadi.
12 822
Obunachilar
-224 soatlar
-357 kunlar
-21030 kunlar
Postlar arxiv
💻 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;
}
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.
💻 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;
}
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
💻 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;
}
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
💻 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;
}
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
💻 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;
}
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
💻 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;
}
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.
💻 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;
}
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.
💻 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;
}
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
💻 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;
}
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
💻 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;
}
Input: 123 Output: Reversed number = 321 Input: -456 Output: Reversed number = -654 Input: 0 Output: Reversed number = 0 Input: 1200 Output: Reversed number = 21
💻 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;
}
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
💻 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;
}
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
💻 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;
}
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
💻 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;
}
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
💻 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;
}
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
💻 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;
}
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:
💻 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;
}
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:
