uz
Feedback
C Programming Language || Hands On Coding

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 12 809 obunachidan iborat bo'lib, Texnologiyalar & Aralashmalar toifasida 9 558-o'rinni va Hindiston mintaqasida 30 933-o'rinni egallagan.

📊 Auditoriya ko‘rsatkichlari va dinamika

невідомо sanasidan buyon loyiha tez o‘sib, 12 809 obunachiga ega bo‘ldi.

29 Avgust, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -215 ga, so‘nggi 24 soatda esa -7 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.

  • Tasdiqlash holati: Tasdiqlanmagan
  • Jalb etish (ER): Auditoriya o‘rtacha 7.22% 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 925 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 2 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 30 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.

12 809
Obunachilar
-724 soatlar
-317 kunlar
-21530 kunlar
Postlar arxiv
#Cprogramming #Palindrome #InterviewQuestion

Check Palindrome Number
#include <stdio.h>
#include <stdbool.h>

bool isPalindrome(int num) {
    int reversedNum = 0, originalNum = num, remainder;

    while (num != 0) {
        remainder = num % 10;
        reversedNum = reversedNum * 10 + remainder;
        num /= 10;
    }

    return originalNum == reversedNum;
}

int main() {
    int number;
    scanf("%d", &number);
    if (isPalindrome(number)) {
        printf("Palindrome");
    } else {
        printf("Not Palindrome");
    }
    return 0;
}

#CProgramming #BitwiseOperators #PowerOfTwo

Check Power of 2 (Iterative and Bitwise)
#include <stdbool.h>

bool isPowerOfTwoIterative(unsigned int n) {
 if (n == 0) return false;
 while (n % 2 == 0) {
 n /= 2;
 }
 return (n == 1);
}

bool isPowerOfTwoBitwise(unsigned int n) {
 return (n != 0) && ((n & (n - 1)) == 0);
}

#CProgramming #Algorithms #Exponentiation

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;
}

#CProgramming #GCD #Arrays

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;
}

#CProgramming #LCM #GCD

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;
}

#CProgramming #GCD #EuclideanAlgorithm

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;
}

#CProgramming #StrongNumber #InterviewPrep

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;
}

#CProgramming #DigitsCount #InterviewPrep

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;
}

#CProgramming #PrimeNumbers

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;
}

#CProgramming #DigitsCount #InterviewPrep

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;
}