AI Coding & CSE Fundamentals
🚀 Explore AI, Coding & Computer Science Fundamentals. 💻 Programming • DSA • DBMS • OS • Networks 🤖 AI & Emerging Technologies 🛠️ Projects • Tutorials • Learning Resources 📚 Your freedom to gain knowledge for free.
Ko'proq ko'rsatish📈 Telegram kanali AI Coding & CSE Fundamentals analitikasi
AI Coding & CSE Fundamentals (@aicoding_csefundamentals) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 12 318 obunachidan iborat bo'lib, Texnologiyalar & Aralashmalar toifasida 10 235-o'rinni va Hindiston mintaqasida 31 257-o'rinni egallagan.
📊 Auditoriya ko‘rsatkichlari va dinamika
невідомо sanasidan buyon loyiha tez o‘sib, 12 318 obunachiga ega bo‘ldi.
13 Noyabr, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -236 ga, so‘nggi 24 soatda esa -9 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.
- Tasdiqlash holati: Tasdiqlanmagan
- Jalb etish (ER): Auditoriya o‘rtacha 10.85% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining N/A% ini tashkil etuvchi reaksiyalarni to‘playdi.
- Post qamrovi: Har bir post o‘rtacha 1 337 marta ko‘riladi; birinchi sutkada odatda 0 ta ko‘rish yig‘iladi.
- Reaksiyalar va o‘zaro ta’sir: Auditoriya faol: har bir postga o‘rtacha 11 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:
“🚀 Explore AI, Coding & Computer Science Fundamentals.
💻 Programming • DSA • DBMS • OS • Networks
🤖 AI & Emerging Technologies
🛠️ Projects • Tutorials • Learning Resources
📚 Your freedom to gain knowledge for free.”
Yuqori yangilanish chastotasi (oxirgi ma’lumot 14 Noyabr, 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.
#include <stdio.h>
int main() {
int rows, i, j, number = 1;
printf("Enter the number of rows: ");
scanf("%d", &rows);
for (i = 0; i < rows; i++) {
for (j = 0; j <= i; j++) {
if (j == 0 || i == j) {
number = 1;
} else {
number = number * (i - j + 1) / j;
}
printf("%4d", number);
}
printf("\n");
}
return 0;
}#include <stdio.h>
int main() {
int rows, number = 1;
printf("Enter the number of rows: ");
scanf("%d", &rows);
for (int i = 1; i <= rows; i++) {
for (int j = 1; j <= i; j++) {
printf("%d ", number);
number++;
}
printf("\n");
}
return 0;
}#include <stdio.h>
void printPyramid(int rows) {
for (int i = 1; i <= rows; i++) {
for (int j = 1; j <= rows - i; j++) {
printf(" ");
}
for (int k = 1; k <= 2 * i - 1; k++) {
printf("*");
}
printf("\n");
}
}
void printInvertedPyramid(int rows) {
for (int i = rows; i >= 1; i--) {
for (int j = 1; j <= rows - i; j++) {
printf(" ");
}
for (int k = 1; k <= 2 * i - 1; k++) {
printf("*");
}
printf("\n");
}
}
void printDiamond(int rows) {
printPyramid(rows);
printInvertedPyramid(rows - 1);
}
int main() {
int rows;
printf("Enter the number of rows: ");
scanf("%d", &rows);
printf("\nPyramid:\n");
printPyramid(rows);
printf("\nInverted Pyramid:\n");
printInvertedPyramid(rows);
printf("\nDiamond:\n");
printDiamond(rows);
return 0;
}#include <stdio.h>
int countDigits(int n) {
int count = 0;
if (n == 0) return 1;
while (n != 0) {
n /= 10;
count++;
}
return count;
}
int main() {
int num;
printf("Enter an integer: ");
scanf("%d", &num);
printf("Number of digits: %d\n", countDigits(num));
return 0;
}#include <stdio.h>
int main() {
int num, sum = 0, m;
scanf("%d", &num);
while (num > 0) {
m = num % 10;
sum = sum + m;
num = num / 10;
}
printf("%d", sum);
return 0;
}#include <stdio.h>
int main() {
int n;
printf("Enter a positive integer: ");
scanf("%d", &n);
printf("Using for loop:\n");
for (int i = 1; i <= n; i++) {
printf("%d ", i);
}
printf("\n");
printf("Using while loop:\n");
int j = 1;
while (j <= n) {
printf("%d ", j);
j++;
}
printf("\n
");
printf("Using do-while loop:\n");
int k = 1;
do {
printf("%d ", k);
k++;
} while (k <= n);
printf("\n");
return 0;
}#include <stdio.h>
int main() {
int num, i;
printf("Enter a positive integer: ");
scanf("%d", &num);
printf("Divisors of %d are: ", num);
for (i = 1; i <= num; ++i) {
if (num % i == 0) {
printf("%d ", i);
}
}
printf("\n");
return 0;
}#include <stdio.h>
int main() {
int n, i, j;
printf("Enter the number of rows: ");
scanf("%d", &n);
for (i = 1; i <= n; i++) {
for (j = 1; j <= n; j++) {
printf("%4d", i * j);
}
printf("\n");
}
return 0;
}#include <stdio.h>
int main() {
int n, i;
long long t1 = 0, t2 = 1;
long long nextTerm = t1 + t2;
printf("Enter the number of terms: ");
scanf("%d", &n);
printf("Fibonacci Series: %lld, %lld, ", t1, t2);
for (i = 3; i <= n; ++i) {
printf("%lld, ", nextTerm);
t1 = t2;
t2 = nextTerm;
nextTerm = t1 + t2;
}
printf("\n");
return 0;
}#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;
}