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AI Coding & CSE Fundamentals

AI Coding & CSE Fundamentals

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πŸš€ Explore AI, Coding & Computer Science Fundamentals. πŸ’» Programming β€’ DSA β€’ DBMS β€’ OS β€’ Networks πŸ€– AI & Emerging Technologies πŸ› οΈ Projects β€’ Tutorials β€’ Learning Resources πŸ“š Your freedom to gain knowledge for free.

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πŸ“ˆ Telegram kanali AI Coding & CSE Fundamentals analitikasi

AI Coding & CSE Fundamentals (@aicoding_csefundamentals) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 12 324 obunachidan iborat bo'lib, Texnologiyalar & Aralashmalar toifasida 10 232-o'rinni va Hindiston mintaqasida 31 271-o'rinni egallagan.

πŸ“Š Auditoriya koβ€˜rsatkichlari va dinamika

Π½Π΅Π²Ρ–Π΄ΠΎΠΌΠΎ sanasidan buyon loyiha tez oβ€˜sib, 12 324 obunachiga ega boβ€˜ldi.

12 Noyabr, 2026 dagi oxirgi ma’lumotlarga koβ€˜ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -238 ga, soβ€˜nggi 24 soatda esa -8 ga oβ€˜zgardi va umumiy qamrov yuqori darajada qolmoqda.

  • Tasdiqlash holati: Tasdiqlanmagan
  • Jalb etish (ER): Auditoriya oβ€˜rtacha 10.45% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining N/A% ini tashkil etuvchi reaksiyalarni toβ€˜playdi.
  • Post qamrovi: Har bir post oβ€˜rtacha 1 288 marta koβ€˜riladi; birinchi sutkada odatda 0 ta koβ€˜rish yigβ€˜iladi.
  • Reaksiyalar va oβ€˜zaro ta’sir: Auditoriya faol: har bir postga oβ€˜rtacha 11 ta reaksiya keladi.
  • Tematik yoβ€˜nalishlar: Kontent input, string, scanf("%d, array, element kabi asosiy mavzularga jamlangan.

πŸ“ Tavsif va kontent siyosati

Muallif resursni shaxsiy fikrni ifoda etish maydoni sifatida ta’riflaydi:
β€œπŸš€ Explore AI, Coding & Computer Science Fundamentals. πŸ’» Programming β€’ DSA β€’ DBMS β€’ OS β€’ Networks πŸ€– AI & Emerging Technologies πŸ› οΈ Projects β€’ Tutorials β€’ Learning Resources πŸ“š Your freedom to gain knowledge for free.”

Yuqori yangilanish chastotasi (oxirgi ma’lumot 13 Noyabr, 2026 da olingan) sababli kanal doimo dolzarb va katta qamrovli boβ€˜lib qoladi. Analitika auditoriya kontent bilan faol hamkorlik qilishini, uni Texnologiyalar & Aralashmalar toifasidagi muhim ta’sir nuqtasiga aylantirishini koβ€˜rsatadi.

12 324
Obunachilar
-824 soatlar
-827 kun
-23830 kun
Postlar arxiv
πŸš€100 Days of LeetCode 2025 Challenge started Join below channel to be part of this challenge πŸ”—https://t.me/+utZsK17rSXRmZWQ
πŸš€100 Days of LeetCode 2025 Challenge started Join below channel to be part of this challenge πŸ”—https://t.me/+utZsK17rSXRmZWQ1

πŸ’» 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;
}
πŸ“€ 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

πŸ’» 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;
}
πŸ“€ 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

πŸ’» 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;
}
πŸ“€ Output:
Input: 1000
Input: 5
Input: 2
Output: Simple Interest = 100.000000

πŸ’» 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;
}
πŸ“€ 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

πŸš€100 Days of LeetCode 2025 Challenge started Join below channel to be part of this challenge πŸ”—https://t.me/+utZsK17rSXRmZWQ
πŸš€100 Days of LeetCode 2025 Challenge started Join below channel to be part of this challenge πŸ”—https://t.me/+utZsK17rSXRmZWQ1

πŸš€100 Days of LeetCode 2025 Challenge Join below channel to be part of this challenge πŸ”—https://t.me/+utZsK17rSXRmZWQ1
πŸš€100 Days of LeetCode 2025 Challenge Join below channel to be part of this challenge πŸ”—https://t.me/+utZsK17rSXRmZWQ1

πŸ’» 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;
}
πŸ“€ 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

πŸ’» 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;
}
πŸ“€ 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.

πŸ’» Hello World Program
#include <stdio.h>

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

πŸ”§ C Basics & Syntax

πŸš€100 Days of LeetCode 2025 Challenge Join below channel to be part of this challenge πŸ”—https://t.me/+utZsK17rSXRmZWQ1
πŸš€100 Days of LeetCode 2025 Challenge Join below channel to be part of this challenge πŸ”—https://t.me/+utZsK17rSXRmZWQ1

πŸš€100 Days of LeetCode 2025 Challenge Join below channel to be part of this challenge πŸ”—https://t.me/+utZsK17rSXRmZWQ1
πŸš€100 Days of LeetCode 2025 Challenge Join below channel to be part of this challenge πŸ”—https://t.me/+utZsK17rSXRmZWQ1

πŸš€100 Days of LeetCode 2025 Challenge Join below channel to be part of this challenge πŸ”—https://t.me/+utZsK17rSXRmZWQ1
πŸš€100 Days of LeetCode 2025 Challenge Join below channel to be part of this challenge πŸ”—https://t.me/+utZsK17rSXRmZWQ1

πŸš€100 Days of LeetCode 2025 Challenge Join below channel to be part of this challenge πŸ”—https://t.me/+utZsK17rSXRmZWQ1
πŸš€100 Days of LeetCode 2025 Challenge Join below channel to be part of this challenge πŸ”—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;
}

πŸ’‘ 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 | πŸ‘Ž Not Understood

πŸ“ 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; }

πŸ’‘ 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 | πŸ‘Ž Not Understood