AI Coding & CSE Fundamentals
π Explore AI, Coding & Computer Science Fundamentals. π» Programming β’ DSA β’ DBMS β’ OS β’ Networks π€ AI & Emerging Technologies π οΈ Projects β’ Tutorials β’ Learning Resources π Your freedom to gain knowledge for free.
Show moreπ Analytical overview of Telegram channel AI Coding & CSE Fundamentals
Channel AI Coding & CSE Fundamentals (@aicoding_csefundamentals) in the English language segment is an active participant. Currently, the community unites 12 324 subscribers, ranking 10 232 in the Technologies & Applications category and 31 271 in the India region.
π Audience metrics and dynamics
Since its creation on Π½Π΅Π²ΡΠ΄ΠΎΠΌΠΎ, the project has demonstrated rapid growth, gathering an audience of 12 324 subscribers.
According to the latest data from 12 November, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -238 over the last 30 days and by -8 over the last 24 hours, overall reach remains high.
- Verification status: Not verified
- Engagement rate (ER): The average audience engagement rate is 10.45%. Within the first 24 hours after publication, content typically collects N/A% reactions from the total number of subscribers.
- Post reach: On average, each post receives 1 288 views. Within the first day, a publication typically gains 0 views.
- Reactions and interaction: The audience actively supports content: the average number of reactions per post is 11.
- Thematic interests: Content is focused on key topics such as input, string, scanf("%d, array, element.
π Description and content policy
The author describes the resource as a platform for expressing subjective opinions:
βπ Explore AI, Coding & Computer Science Fundamentals.
π» Programming β’ DSA β’ DBMS β’ OS β’ Networks
π€ AI & Emerging Technologies
π οΈ Projects β’ Tutorials β’ Learning Resources
π Your freedom to gain knowledge for free.β
Thanks to the high frequency of updates (latest data received on 13 November, 2026), the channel maintains relevance and a high level of publication reach. Analytics show that the audience actively interacts with content, making it an important point of influence in the Technologies & Applications category.
#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;
}
π€ 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
#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;
}
π€ 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
#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;
}
π€ Output:
Input: 1000 Input: 5 Input: 2 Output: Simple Interest = 100.000000
#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;
}
π€ 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
#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;
}
π€ 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
#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;
}
π€ 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.
#include <stdio.h>
int main() {
printf("Hello, World!n");
return 0;
}
π€ Output:
Hello, World!
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;
}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".
βββββββββββββββββββββββββββββ
Have you Understood? Drop a reaction:
β€οΈ Understood | π Not Understoodswitch 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.