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.
显示更多📈 Telegram 频道 AI Coding & CSE Fundamentals 的分析概览
频道 AI Coding & CSE Fundamentals (@aicoding_csefundamentals) 英语 语言赛道中的 是活跃参与者。目前社区聚集了 12 318 名订阅者,在 技术与应用 类别中位列第 10 235,并在 印度 地区排名第 31 257 位。
📊 受众指标与增长动态
自 невідомо 创建以来,项目保持高速增长,吸引了 12 318 名订阅者。
根据 13 十一月, 2026 的最新数据,频道保持稳定运转。过去 30 天订阅人数变化为 -236,过去 24 小时变化为 -9,整体触达仍然可观。
- 认证状态: 未认证
- 互动率 (ER): 平均受众互动率为 10.85%。内容发布后 24 小时内通常能获得 N/A% 的反应,占订阅者总量。
- 帖子覆盖: 每篇帖子平均可获得 1 337 次浏览,首日通常累积 0 次浏览。
- 互动与反馈: 受众积极参与,单帖平均反应数为 11。
- 主题关注点: 内容集中在 input, string, scanf("%d, array, element 等核心主题上。
📝 描述与内容策略
作者将该频道定位为表达主观观点的平台:
“🚀 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.”
凭借高频更新(最新数据采集于 14 十一月, 2026),频道始终保持新鲜度与高覆盖。分析显示受众积极互动,使其成为 技术与应用 类别中的关键影响点。
#include <stdio.h>
#include <limits.h>
int main() {
int arr[] = {12, 34, 5, 78, 9, 101};
int n = sizeof(arr) / sizeof(arr[0]);
int max = INT_MIN;
int min = INT_MAX;
for (int i = 0; i < n; i++) {
if (arr[i] > max) {
max = arr[i];
}
if (arr[i] < min) {
min = arr[i];
}
}
printf("Max element: %d\n", max);
printf("Min element: %d\n", min);
return 0;
}#include <stdio.h>
int main() {
int n, i;
printf("Enter the number of elements: ");
scanf("%d", &n);
int arr[n];
printf("Enter %d elements:\n", n);
for (i = 0; i < n; i++) {
scanf("%d", &arr[i]);
}
printf("Array elements are: ");
for (i = 0; i < n; i++) {
printf("%d ", arr[i]);
}
printf("\n");
return 0;
}#include <stdio.h>
int reverse_number(int num, int reversed_num) {
if (num == 0) {
return reversed_num;
}
int remainder = num % 10;
reversed_num = reversed_num * 10 + remainder;
return reverse_number(num / 10, reversed_num);
}
int main() {
int number = 12345;
int reversed = reverse_number(number, 0);
printf("Original number: %d\n", number);
printf("Reversed number: %d\n", reversed);
return 0;
}#include <stdio.h>
int sumOfDigits(int n) {
if (n == 0)
return 0;
return (n % 10 + sumOfDigits(n / 10));
}
int main() {
int num;
scanf("%d", &num);
printf("%d", sumOfDigits(num));
return 0;
}#include <stdio.h>
int power(int base, int exp) {
if (exp == 0)
return 1;
else if (exp % 2 == 0) {
int temp = power(base, exp / 2);
return temp * temp;
} else {
return base * power(base, exp / 2) * power(base, exp / 2);
}
}
int main() {
int base, exp;
printf("Enter base: ");
scanf("%d", &base);
printf("Enter exponent: ");
scanf("%d", &exp);
printf("%d^%d = %d", base, exp, power(base, exp));
return 0;
}#include <stdio.h>
#include <stdbool.h>
bool isPrime(int n) {
if (n <= 1) return false;
for (int i = 2; i * i <= n; i++) {
if (n % i == 0) return false;
}
return true;
}
int main() {
int num;
printf("Enter a number: ");
scanf("%d", &num);
if (isPrime(num)) {
printf("%d is a prime number.\n", num);
} else {
printf("%d is not a prime number.\n", num);
}
return 0;
}#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;
printf("Enter two positive integers: ");
scanf("%d %d", &num1, &num2);
printf("GCD of %d and %d is %d\n", num1, num2, gcd(num1, num2));
printf("LCM of %d and %d is %d\n", num1, num2, lcm(num1, num2));
return 0;
}#include <stdio.h>
int fibonacci(int n) {
if (n <= 1)
return n;
return fibonacci(n - 1) + fibonacci(n - 2);
}
int main() {
int n = 10;
printf("Fibonacci sequence up to %d: ", n);
for (int i = 0; i < n; i++) {
printf("%d ", fibonacci(i));
}
printf("\n");
return 0;
}#include <stdio.h>
int factorial(int n) {
if (n == 0)
return 1;
else
return n * factorial(n - 1);
}
int main() {
int num = 5;
printf("Factorial of %d is %d\n", num, factorial(num));
return 0;
}