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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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📈 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
🚀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