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 12 318 名订阅者,在 技术与应用 类别中位列第 10 235,并在 印度 地区排名第 31 257 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 12 318 名订阅者。
根据 13 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 -236,过去 24 小时变化为 -9,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 10.85%。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 1 337 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 14 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() {
int rows, i, j, number = 1;
printf("Enter the number of rows: ");
scanf("%d", &rows);
for (i = 0; i < rows; i++) {
for (j = 0; j <= i; j++) {
if (j == 0 || i == j) {
number = 1;
} else {
number = number * (i - j + 1) / j;
}
printf("%4d", number);
}
printf("\n");
}
return 0;
}#include <stdio.h>
int main() {
int rows, number = 1;
printf("Enter the number of rows: ");
scanf("%d", &rows);
for (int i = 1; i <= rows; i++) {
for (int j = 1; j <= i; j++) {
printf("%d ", number);
number++;
}
printf("\n");
}
return 0;
}#include <stdio.h>
void printPyramid(int rows) {
for (int i = 1; i <= rows; i++) {
for (int j = 1; j <= rows - i; j++) {
printf(" ");
}
for (int k = 1; k <= 2 * i - 1; k++) {
printf("*");
}
printf("\n");
}
}
void printInvertedPyramid(int rows) {
for (int i = rows; i >= 1; i--) {
for (int j = 1; j <= rows - i; j++) {
printf(" ");
}
for (int k = 1; k <= 2 * i - 1; k++) {
printf("*");
}
printf("\n");
}
}
void printDiamond(int rows) {
printPyramid(rows);
printInvertedPyramid(rows - 1);
}
int main() {
int rows;
printf("Enter the number of rows: ");
scanf("%d", &rows);
printf("\nPyramid:\n");
printPyramid(rows);
printf("\nInverted Pyramid:\n");
printInvertedPyramid(rows);
printf("\nDiamond:\n");
printDiamond(rows);
return 0;
}#include <stdio.h>
int countDigits(int n) {
int count = 0;
if (n == 0) return 1;
while (n != 0) {
n /= 10;
count++;
}
return count;
}
int main() {
int num;
printf("Enter an integer: ");
scanf("%d", &num);
printf("Number of digits: %d\n", countDigits(num));
return 0;
}#include <stdio.h>
int main() {
int num, sum = 0, m;
scanf("%d", &num);
while (num > 0) {
m = num % 10;
sum = sum + m;
num = num / 10;
}
printf("%d", sum);
return 0;
}#include <stdio.h>
int main() {
int n;
printf("Enter a positive integer: ");
scanf("%d", &n);
printf("Using for loop:\n");
for (int i = 1; i <= n; i++) {
printf("%d ", i);
}
printf("\n");
printf("Using while loop:\n");
int j = 1;
while (j <= n) {
printf("%d ", j);
j++;
}
printf("\n
");
printf("Using do-while loop:\n");
int k = 1;
do {
printf("%d ", k);
k++;
} while (k <= n);
printf("\n");
return 0;
}#include <stdio.h>
int main() {
int num, i;
printf("Enter a positive integer: ");
scanf("%d", &num);
printf("Divisors of %d are: ", num);
for (i = 1; i <= num; ++i) {
if (num % i == 0) {
printf("%d ", i);
}
}
printf("\n");
return 0;
}#include <stdio.h>
int main() {
int n, i, j;
printf("Enter the number of rows: ");
scanf("%d", &n);
for (i = 1; i <= n; i++) {
for (j = 1; j <= n; j++) {
printf("%4d", i * j);
}
printf("\n");
}
return 0;
}#include <stdio.h>
int main() {
int n, i;
long long t1 = 0, t2 = 1;
long long nextTerm = t1 + t2;
printf("Enter the number of terms: ");
scanf("%d", &n);
printf("Fibonacci Series: %lld, %lld, ", t1, t2);
for (i = 3; i <= n; ++i) {
printf("%lld, ", nextTerm);
t1 = t2;
t2 = nextTerm;
nextTerm = t1 + t2;
}
printf("\n");
return 0;
}#include <stdio.h>
#include <math.h>
int main() {
int num, originalNum, remainder, n = 0;
float result = 0.0;
printf("Enter an integer: ");
scanf("%d", &num);
originalNum = num;
while (originalNum != 0) {
originalNum /= 10;
++n;
}
originalNum = num;
while (originalNum != 0) {
remainder = originalNum % 10;
result += pow(remainder, n);
originalNum /= 10;
}
if ((int)result == num)
printf("%d is an Armstrong number.\n", num);
else
printf("%d is not an Armstrong number.\n", num);
return 0;
}