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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.

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印度31 257技术与应用10 235...

📈 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。

12 318

订阅者

-924 小时

-837 天

-23630 天

1 337

帖子浏览量

无数据24 小时

无数据48 小时

10.85%

参与率

无数据

每日帖子数

Ads index

帖子存档

12 318

#CProgramming #DecimalConversion #BinaryOctalHex

12 318

Convert Decimal to Binary, Octal, Hex

#include <stdio.h>

void decimalToBinary(int n) {
    int binary[32];
    int i = 0;
    while (n > 0) {
        binary[i] = n % 2;
        n = n / 2;
        i++;
    }
    for (int j = i - 1; j >= 0; j--)
        printf("%d", binary[j]);
    printf("\n");
}

void decimalToOctal(int n) {
    int octal[32];
    int i = 0;
    while (n > 0) {
        octal[i] = n % 8;
        n = n / 8;
        i++;
    }
    for (int j = i - 1; j >= 0; j--)
        printf("%d", octal[j]);
    printf("\n");
}

void decimalToHexadecimal(int n) {
    char hex[32];
    int i = 0;
    while (n != 0) {
        int temp = 0;
        temp = n % 16;
        if (temp < 10) {
            hex[i] = temp + 48;
        } else {
            hex[i] = temp + 87;
        }
        n = n / 16;
        i++;
    }
    for (int j = i - 1; j >= 0; j--)
        printf("%c", hex[j]);
    printf("\n");
}

int main() {
    int decimal;
    printf("Enter a decimal number: ");
    scanf("%d", &decimal);

    printf("Binary: ");
    decimalToBinary(decimal);

    printf("Octal: ");
    decimalToOctal(decimal);

    printf("Hexadecimal: ");
    decimalToHexadecimal(decimal);

    return 0;
}

12 318

#CProgramming #SpyNumber #NumberTheory

12 318

Check Spy Number

#include <stdio.h>
#include <stdbool.h>

bool isSpyNumber(int num) {
    int sum = 0;
    int product = 1;
    int temp = num;

    while (temp > 0) {
        int digit = temp % 10;
        sum += digit;
        product *= digit;
        temp /= 10;
    }

    return sum == product;
}

int main() {
    int number;
    scanf("%d", &number);

    if (isSpyNumber(number)) {
        printf("Spy Number");
    } else {
        printf("Not Spy Number");
    }

    return 0;
}

12 318

#CProgramming #MagicNumber #Algorithm

12 318

Check if a Number is a Magic Number

#include <stdio.h>

int sum_digits(int n) {
    int sum = 0;
    while (n > 0) {
        sum += n % 10;
        n /= 10;
    }
    return sum;
}

int is_magic(int n) {
    while (n > 9) {
        n = sum_digits(n);
    }
    return (n == 1);
}

int main() {
    int num;
    scanf("%d", &num);
    if (is_magic(num)) {
        printf("Magic Number\n");
    } else {
        printf("Not a Magic Number\n");
    }
    return 0;
}

12 318

#CProgramming #NeonNumber #Algorithm

12 318

Check Neon Number

#include <stdio.h>

int main() {
  int n, square, sum = 0, temp;

  scanf("%d", &n);

  square = n * n;
  temp = square;

  while (temp > 0) {
    sum += temp % 10;
    temp /= 10;
  }

  if (sum == n) {
    printf("Neon");
  } else {
    printf("Not Neon");
  }

  return 0;
}

12 318

#CProgramming #AutomorphicNumber #InterviewQuestion

12 318

Check Automorphic Number

#include <stdio.h>
#include <stdbool.h>

int main() {
 int num, sq;
 printf("Enter a number: ");
 scanf("%d", &num);
 sq = num * num;
 int temp = num;
 int digits = 0;
 while (temp > 0) {
 digits++;
 temp /= 10;
 }
 int lastDigits = sq % (int)pow(10, digits);
 if (lastDigits == num) {
 printf("%d is an Automorphic number.\n", num);
 } else {
 printf("%d is not an Automorphic number.\n", num);
 }
 return 0;
}

12 318

#CProgramming #Divisors #Algorithms

12 318

Print All Divisors of a Number

#include <stdio.h>

void printDivisors(int n) {
  for (int i = 1; i <= n; i++) {
    if (n % i == 0) {
      printf("%d ", i);
    }
  }
  printf("\n");
}

int main() {
  int num;
  scanf("%d", &num);
  printDivisors(num);
  return 0;
}

12 318

#CProgramming #Factorial #BigNumbers

12 318

Find Factorial (Big Numbers using Arrays)

#include <stdio.h>

int multiply(int x, int res[], int res_size) {
    int carry = 0;
    for (int i = 0; i < res_size; i++) {
        int prod = res[i] * x + carry;
        res[i] = prod % 10;
        carry = prod / 10;
    }
    while (carry) {
        res[res_size] = carry % 10;
        carry = carry / 10;
        res_size++;
    }
    return res_size;
}

void factorial(int n) {
    int res[500];
    res[0] = 1;
    int res_size = 1;

    for (int x = 2; x <= n; x++)
        res_size = multiply(x, res, res_size);

    for (int i = res_size - 1; i >= 0; i--)
        printf("%d", res[i]);
    printf("\n");
}

int main() {
    int n = 100;
    factorial(n);
    return 0;
}

12 318

#CProgramming #HarshadNumber #InterviewCode

12 318

Check Harshad (Niven) Number

#include <stdio.h>
#include <stdbool.h>

bool isHarshad(int n) {
    int sum = 0;
    int temp = n;
    while (temp > 0) {
        sum += temp % 10;
        temp /= 10;
    }
    return (n % sum == 0 && sum != 0);
}

int main() {
    int num;
    scanf("%d", &num);
    if (isHarshad(num)) {
        printf("Harshad\n");
    } else {
        printf("Not Harshad\n");
    }
    return 0;
}

12 318

#CProgramming #Summation #InterviewPrep

12 318

Sum of First N Natural Numbers

#include <stdio.h>

int sum_n(int n) {
  if (n <= 0) {
    return 0;
  }
  return n * (n + 1) / 2;
}

int main() {
  int num = 10;
  int sum = sum_n(num);
  printf("Sum of first %d natural numbers: %d\n", num, sum);
  return 0;
}

12 318

#CProgramming #Algorithm #DigitalRoot

12 318

Digital Root of a Number

#include <stdio.h>

int digitalRoot(int n) {
 while (n >= 10) {
 int sum = 0;
 while (n > 0) {
 sum += n % 10;
 n /= 10;
 }
 n = sum;
 }
 return n;
}

int main() {
 int num = 12345;
 printf("Digital root of %d is %d\n", num, digitalRoot(num));
 return 0;
}