AI Coding & CSE Fundamentals
🚀 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.
显示更多📈 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。
#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
#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
#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
#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
#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
#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.
#include <stdio.h>
int main() {
printf("Hello, World!n");
return 0;
}
📤 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".
─────────────────────────────
Have you Understood? Drop a reaction:
❤️ Understood | 👎 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.