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AI Coding & CSE Fundamentals

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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📈 Аналитический обзор Telegram-канала AI Coding & CSE Fundamentals

Канал AI Coding & CSE Fundamentals (@aicoding_csefundamentals) языкового сегмента Английский является активным участником. Сейчас сообщество объединяет 12 324 подписчиков, занимая 10 232 место в категории Технологии и приложения и 31 271 место в регионе Индия.

📊 Показатели аудитории и динамика

С момента создания невідомо проект демонстрирует стремительный рост, собрав аудиторию из 12 324 подписчиков.

Согласно последним данным от 12 ноября, 2026, канал показывает стабильную активность. За последние 30 дней изменение числа участников составило -238, а за последние 24 часа — -8, при этом общий охват остаётся высоким.

  • Статус верификации: Не верифицирован
  • Уровень вовлечённости (ER): Средний показатель вовлечённости аудитории составляет 10.45%. В первые 24 часа после публикации контент обычно набирает N/A% реакций от общего числа подписчиков.
  • Охват публикаций: В среднем каждый пост получает 1 288 просмотров. В течение первых суток публикация набирает 0 просмотров.
  • Реакции и взаимодействия: Аудитория активно поддерживает контент: среднее количество реакций на один пост — 11.
  • Тематические интересы: Контент сосредоточен на ключевых темах, таких как input, string, scanf("%d, array, element.

📝 Описание и контентная политика

Автор описывает ресурс как площадку для выражения субъективного мнения:
“🚀 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.”

Благодаря высокой частоте обновлений (последние данные получены 13 ноября, 2026) канал поддерживает актуальность и высокий уровень охвата публикаций. Аналитика показывает, что аудитория активно взаимодействует с контентом, что делает его важной точкой влияния в категории Технологии и приложения.

12 324
Подписчики
-824 часа
-827 дней
-23830 дней
Архив постов
🚀100 Days of LeetCode 2025 Challenge started Join below channel to be part of this challenge 🔗https://t.me/+utZsK17rSXRmZWQ
🚀100 Days of LeetCode 2025 Challenge started Join below channel to be part of this challenge 🔗https://t.me/+utZsK17rSXRmZWQ1

💻 Area of Circle Calculator
#include <stdio.h>

int main() {
    float radius, area;
    const float PI = 3.14159;

    printf("Enter the radius of the circle: ");
    scanf("%f", &radius);

    area = PI * radius * radius;

    printf("The area of the circle is: %fn", area);

    return 0;
}
📤 Output:
Input: 5
Output: Enter the radius of the circle: The area of the circle is: 78.539757

Input: 2.5
Output: Enter the radius of the circle: The area of the circle is: 19.634938

💻 Compound Interest Calculator
#include <stdio.h>
#include <math.h>

int main() {
  float principal, rate, time, compound_interest, amount;
  int compounding_frequency;

  printf("Enter principal amount: ");
  scanf("%f", &principal);

  printf("Enter annual interest rate (as a decimal, e.g., 0.05 for 5%%): ");
  scanf("%f", &rate);

  printf("Enter time in years: ");
  scanf("%f", &time);

  printf("Enter compounding frequency (e.g., 1 for annually, 12 for monthly): ");
  scanf("%d", &compounding_frequency);

  amount = principal * pow(1 + (rate / compounding_frequency), compounding_frequency * time);
  compound_interest = amount - principal;

  printf("Compound Interest: %.2fn", compound_interest);

  return 0;
}
📤 Output:
Enter principal amount: 1000
Input: 1000
Enter annual interest rate (as a decimal, e.g., 0.05 for 5%): 0.05
Input: 0.05
Enter time in years: 5
Input: 5
Enter compounding frequency (e.g., 1 for annually, 12 for monthly): 1
Input: 1
Compound Interest: 276.28

💻 Simple Interest Calculator
#include <stdio.h>

int main() {
  float principal, rate, time, interest;

  printf("Enter principal amount: ");
  scanf("%f", &principal);

  printf("Enter rate of interest: ");
  scanf("%f", &rate);

  printf("Enter time in years: ");
  scanf("%f", &time);

  interest = (principal * rate * time) / 100;

  printf("Simple Interest = %fn", interest);

  return 0;
}
📤 Output:
Input: 1000
Input: 5
Input: 2
Output: Simple Interest = 100.000000

💻 Variable Swapping (Three Variables)
#include <stdio.h>

int main() {
  int a, b, c, temp;

  printf("Enter the values of a, b, and c: ");
  scanf("%d %d %d", &a, &b, &c);

  printf("Before swapping: a = %d, b = %d, c = %dn", a, b, c);

  temp = a;
  a = b;
  b = c;
  c = temp;

  printf("After swapping: a = %d, b = %d, c = %dn", a, b, c);

  return 0;
}
📤 Output:
Input: 10 20 30
Output: Enter the values of a, b, and c: Before swapping: a = 10, b = 20, c = 30
After swapping: a = 20, b = 30, c = 10

🚀100 Days of LeetCode 2025 Challenge started Join below channel to be part of this challenge 🔗https://t.me/+utZsK17rSXRmZWQ
🚀100 Days of LeetCode 2025 Challenge started Join below channel to be part of this challenge 🔗https://t.me/+utZsK17rSXRmZWQ1

🚀100 Days of LeetCode 2025 Challenge Join below channel to be part of this challenge 🔗https://t.me/+utZsK17rSXRmZWQ1
🚀100 Days of LeetCode 2025 Challenge Join below channel to be part of this challenge 🔗https://t.me/+utZsK17rSXRmZWQ1

💻 Variable Swapping (Two Variables)
#include <stdio.h>

int main() {
  int a, b, temp;

  printf("Enter the value of a: ");
  scanf("%d", &a);

  printf("Enter the value of b: ");
  scanf("%d", &b);

  printf("Before swapping: a = %d, b = %dn", a, b);

  temp = a;
  a = b;
  b = temp;

  printf("After swapping: a = %d, b = %dn", a, b);

  return 0;
}
📤 Output:
Input: 5
Input: 10
Output: Enter the value of a: Enter the value of b: Before swapping: a = 5, b = 10
After swapping: a = 10, b = 5

💻 Basic Calculator (Add, Subtract, Multiply, Divide)
#include <stdio.h>

int main() {
    float num1, num2, result;
    char operator;

    printf("Enter first number: ");
    scanf("%f", &num1);

    printf("Enter an operator (+, -, *, /): ");
    scanf(" %c", &operator);

    printf("Enter second number: ");
    scanf("%f", &num2);

    switch (operator) {
        case '+':
            result = num1 + num2;
            printf("%.2f + %.2f = %.2fn", num1, num2, result);
            break;
        case '-':
            result = num1 - num2;
            printf("%.2f - %.2f = %.2fn", num1, num2, result);
            break;
        case '*':
            result = num1 * num2;
            printf("%.2f * %.2f = %.2fn", num1, num2, result);
            break;
        case '/':
            if (num2 == 0) {
                printf("Error! Division by zero.n");
            } else {
                result = num1 / num2;
                printf("%.2f / %.2f = %.2fn", num1, num2, result);
            }
            break;
        default:
            printf("Error! Invalid operator.n");
    }

    return 0;
}
📤 Output:
Input: 10
Input: +
Input: 5
Output: 10.00 + 5.00 = 15.00

Input: 20
Input: -
Input: 7
Output: 20.00 - 7.00 = 13.00

Input: 4
Input: *
Input: 6
Output: 4.00 * 6.00 = 24.00

Input: 15
Input: /
Input: 3
Output: 15.00 / 3.00 = 5.00

Input: 8
Input: /
Input: 0
Output: Error! Division by zero.

Input: 9
Input: %
Input: 2
Output: Error! Invalid operator.

💻 Hello World Program
#include <stdio.h>

int main() {
    printf("Hello, World!n");
    return 0;
}
📤 Output:
Hello, World!

🔧 C Basics & Syntax

🚀100 Days of LeetCode 2025 Challenge Join below channel to be part of this challenge 🔗https://t.me/+utZsK17rSXRmZWQ1
🚀100 Days of LeetCode 2025 Challenge Join below channel to be part of this challenge 🔗https://t.me/+utZsK17rSXRmZWQ1

🚀100 Days of LeetCode 2025 Challenge Join below channel to be part of this challenge 🔗https://t.me/+utZsK17rSXRmZWQ1
🚀100 Days of LeetCode 2025 Challenge Join below channel to be part of this challenge 🔗https://t.me/+utZsK17rSXRmZWQ1

🚀100 Days of LeetCode 2025 Challenge Join below channel to be part of this challenge 🔗https://t.me/+utZsK17rSXRmZWQ1
🚀100 Days of LeetCode 2025 Challenge Join below channel to be part of this challenge 🔗https://t.me/+utZsK17rSXRmZWQ1

🚀100 Days of LeetCode 2025 Challenge Join below channel to be part of this challenge 🔗https://t.me/+utZsK17rSXRmZWQ1
🚀100 Days of LeetCode 2025 Challenge Join below channel to be part of this challenge 🔗https://t.me/+utZsK17rSXRmZWQ1

Print days of the week using switch statement
#include <stdio.h>

int main() {
    int dayNumber;

    printf("Enter a number (1-7): ");
    scanf("%d", &dayNumber);

    switch (dayNumber) {
        case 1:
            printf("Sundayn");
            break;
        case 2:
            printf("Mondayn");
            break;
        case 3:
            printf("Tuesdayn");
            break;
        case 4:
            printf("Wednesdayn");
            break;
        case 5:
            printf("Thursdayn");
            break;
        case 6:
            printf("Fridayn");
            break;
        case 7:
            printf("Saturdayn");
            break;
        default:
            printf("Invalid inputn");
    }

    return 0;
}

💡 Approach Step 1: Get integer input from the user representing the day of the week (1 for Sunday, 2 for Monday, ..., 7 for Saturday). Step 2: Create a switch statement. The switch expression will be the integer input from the user. Step 3: Inside the switch statement, create case labels for each day of the week (1 to 7). Step 4: Within each case label, use printf to print the corresponding day of the week (e.g., case 1: printf("Sunday"); break;). Include a break statement after each printf to exit the switch statement after a match. Step 5: Add a default case to handle invalid input (numbers outside the range 1-7). Print an error message like "Invalid input". ───────────────────────────── Have you Understood? Drop a reaction: ❤️ Understood | 👎 Not Understood

📝 Print days of the week using switch statement Write a C program that takes an integer input representing a day of the week (1 for Sunday, 2 for Monday, etc.). Using a switch statement, print the corresponding day of the week's name based on the input number.

Implement a simple calculator using `switch` statement for operations (+, -, , /) #include <stdio.h> int main() { float num1, num2, result; char operator; printf("Enter first number: "); scanf("%f", &num1); printf("Enter operator (+, -, , /): "); scanf(" %c", &operator); printf("Enter second number: "); scanf("%f", &num2); switch (operator) { case '+': result = num1 + num2; break; case '-': result = num1 - num2; break; case '': result = num1 num2; break; case '/': if (num2 == 0) { printf("Error: Division by zero is not allowed.\n"); return 1; } result = num1 / num2; break; default: printf("Error: Invalid operator.\n"); return 1; } printf("Result: %.2f\n", result); return 0; }

💡 Approach Step 1: Include Header and Declare Variables: Include the standard input/output library (`stdio.h`). Declare variables to store the two numbers (e.g., `num1`, `num2` as `float` for decimal support), the operator (e.g., `operator` as `char`), and the result (e.g., `result` as `float`). Step 2: Get User Input: Prompt the user to enter the two numbers and the operator (+, -, , /). Use `scanf` to read these values from the console. Step 3: Implement the `switch` Statement: Use a `switch` statement to perform the calculation based on the operator entered by the user. Each `case` will correspond to a specific operator. Step 4: Handle Each Case: Inside each `case`, perform the corresponding arithmetic operation. For example, in the `case '+'`, calculate `result = num1 + num2`. Step 5: Handle Division by Zero: In the `case '/'`, add a check to prevent division by zero. If `num2` is zero, print an error message and possibly exit the `switch` statement or program. Step 6: Provide a `default` Case: Include a `default` case in the `switch` statement to handle invalid operator inputs. Print an error message if the operator is not recognized. Step 7: Print the Result: After the `switch` statement (assuming no errors occurred), print the calculated `result` to the console with appropriate formatting. Step 8: Return 0:* At the end of the `main` function, return 0 to indicate successful execution of the program. ───────────────────────────── Have you Understood\? Drop a reaction: ❤️ Understood | 👎 Not Understood