AI Coding & CSE Fundamentals
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🚀 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 12 324 名订阅者，在 技术与应用 类别中位列第 10 232，并在 印度 地区排名第 31 271 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 12 324 名订阅者。
根据 12 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 -238，过去 24 小时变化为 -8，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 10.45%。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 1 288 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 13 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。
12 324
订阅者
-824 小时
-827 天
-23830 天
帖子存档
💻 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: 120 Output: 120 is not an Armstrong number. Input: 1634 Output: 1634 is an Armstrong number. Input: 1234 Output: 1234 is not an Armstrong number.
💻 Check if Number is Palindrome
#include <stdio.h>
int main() {
int num, reversed_num = 0, remainder, original_num;
printf("Enter an integer: ");
scanf("%d", &num);
original_num = num;
for (; num != 0; num /= 10) {
remainder = num % 10;
reversed_num = reversed_num * 10 + remainder;
}
if (original_num == reversed_num)
printf("%d is a palindrome.n", original_num);
else
printf("%d is not a palindrome.n", original_num);
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: 12345 Output: 12345 is not a palindrome. Input: 1 Output: 1 is a palindrome. Input: 0 Output: 0 is a palindrome.
💻 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:
