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لا تقع ضحية للمخادعين! تيليمتريو يكتشف ويُميّز هذه القنوات 👉 إذا كنت تريد رؤية العلامة، اشترك

allcoding1

allcoding1

الذهاب إلى القناة على Telegram

إظهار المزيد

📈 نظرة تحليلية على قناة تيليجرام allcoding1

تُعد قناة allcoding1 (@allcoding1) في القطاع اللغوي الإنكليزية لاعباً نشطاً. يضم المجتمع حالياً 20 917 مشتركاً، محتلاً المرتبة 9 616 في فئة التعليم والمرتبة 19 575 في منطقة الهند.

📊 مؤشرات الجمهور والحراك

منذ تأسيسه في невідомо، حقق المشروع نمواً سريعاً وجمع 20 917 مشتركاً.

بحسب آخر البيانات بتاريخ 17 أكتوبر, 2026، تحافظ القناة على نشاط مستقر. خلال آخر 30 يوماً تغيّر عدد الأعضاء بمقدار -398، وفي آخر 24 ساعة بمقدار -5، مع بقاء الوصول العام مرتفعاً.

  • حالة التحقق: غير موثّقة
  • معدل التفاعل (ER): يبلغ متوسط تفاعل الجمهور 1.12‎%. وخلال أول 24 ساعة من النشر يحصد المحتوى عادةً 1.05‎% من ردود الفعل نسبةً إلى إجمالي المشتركين.
  • وصول المنشورات: يحصل كل منشور على متوسط 235 مشاهدة. وخلال اليوم الأول يجمع عادةً 220 مشاهدة.
  • التفاعلات والاستجابة: يتفاعل الجمهور بانتظام؛ متوسط التفاعلات لكل منشور يبلغ 1.
  • الاهتمامات الموضوعية: يركز المحتوى على مواضيع رئيسية مثل dsa, stack, namaste, javascript, learning.

📝 الوصف وسياسة المحتوى

وصف القناة غير متوفر.

بفضل وتيرة التحديث المرتفعة (أحدث البيانات بتاريخ 18 أكتوبر, 2026) تحافظ القناة على حداثتها ومستوى وصول مرتفع. وتُظهر التحليلات تفاعلاً نشطاً من الجمهور، ما يجعلها نقطة تأثير مهمة ضمن فئة التعليم.

20 917
المشتركون
-524 ساعات
-957 أيام
-39830 أيام
أرشيف المشاركات
#include<stdio.h> int main() {     int n;     scanf("%d",&n);     int arr[n];     for(int i=0;i<n;i++)     {         scanf("%d",&arr[i]);     }     int count=0;     int num=arr[0];     for(int i=1;i<n;i++)     {        if(num!=arr[i])             count++;     }     printf("%d",count); }

Tcs exam
Tcs exam

#include <iostream> #include <vector> #include <map> #include <set> #include <cmath> #include <algorithm> using namespace std; struct Line { int x1, y1, x2, y2; }; int countCells(Line line, pair<int, int> star, bool split) { if (line.x1 == line.x2) { if (split) { return min(abs(star.second - line.y1), abs(star.second - line.y2)) + 1; } else { return abs(line.y1 - line.y2) + 1; } } else { if (split) { return min(abs(star.first - line.x1), abs(star.first - line.x2)) + 1; } else { return abs(line.x1 - line.x2) + 1; } } } bool intersects(Line a, Line b, pair<int, int>& intersection) { if (a.x1 == a.x2 && b.y1 == b.y2) { if (b.x1 <= a.x1 && a.x1 <= b.x2 && a.y1 <= b.y1 && b.y1 <= a.y2) { intersection = {a.x1, b.y1}; return true; } } if (a.y1 == a.y2 && b.x1 == b.x2) { if (a.x1 <= b.x1 && b.x1 <= a.x2 && b.y1 <= a.y1 && a.y1 <= b.y2) { intersection = {b.x1, a.y1}; return true; } } return false; } int main() { int N, K; cin >> N; vector<Line> lines(N); for (int i = 0; i < N; ++i) { cin >> lines[i].x1 >> lines[i].y1 >> lines[i].x2 >> lines[i].y2; if (lines[i].x1 > lines[i].x2 || (lines[i].x1 == lines[i].x2 && lines[i].y1 > lines[i].y2)) { swap(lines[i].x1, lines[i].x2); swap(lines[i].y1, lines[i].y2); } } cin >> K; map<pair<int, int>, vector<Line>> stars; for (int i = 0; i < N; ++i) { for (int j = i + 1; j < N; ++j) { pair<int, int> intersection; if (intersects(lines[i], lines[j], intersection)) { stars[intersection].push_back(lines[i]); stars[intersection].push_back(lines[j]); } } } int asiylam = 0; for (auto& star : stars) { if (star.second.size() / 2 == K) { vector<int> intensities; for (auto& line : star.second) { intensities.push_back(countCells(line, star.first, true)); } asiylam += *min_element(intensities.begin(), intensities.end()); } } cout << asiylam << endl; return 0; } Magic Star Intensity Code C++ TCS Exam

int main() { int n, m, k, d = 1, rv = 0; cin >> n >> m; vector> f(n); for (int i = 0, u, v; i < m; ++i) { cin >> u >> v; f[u].insert(v); f[v].insert(u); } cin >> k; vector w(n, true); rv = n; while (rv < k) { vector nw(n, false); for (int i = 0; i < n; ++i) { int cnt = 0; for (int fr : f[i]) cnt += w[fr]; if (w[i] && cnt == 3) nw[i] = true; else if (!w[i] && cnt < 3) nw[i] = true; } w = nw; rv += count(w.begin(), w.end(), true); ++d; } cout << d; return 0; } Office Rostering C+ TCS exam

Industrial code
Industrial code

#include<stdio.h> int main() {     int n;     scanf("%d",&n);     int arr[n];     for(int i=0;i<n;i++)     {         scanf("%d",&arr[i]);     }     int count=0;     int num=arr[0];     for(int i=1;i<n;i++)     {        if(num!=arr[i])             count++;     }     printf("%d",count); } C Language TCS 1st Qsn --------------------------------------------------------- N=int(input()) K=int(input()) price=list(map(int,input().split())) vol=list(map(int,input().split())) maxvol=0 volu=0 maxvol=max(vol) for i in range(0,N):     if (maxvol==vol[i] and price[i]<=K):         K=K-price[i]         volu=maxvol for i in range(0,N):     for j in range(i+1,N+1):         if (price[i]<=K and price[i]==price[j]):             if (vol[i]>vol[j]):                 volu=volu+vol[i]                 K=K-price[i]             else:                 volu=volu+vol[j]                 K=K-price[j]         elif (price[i]<=K and price[i]!=price[j]):             K=K-price[i]             ------- include<stdio.h> int main() {     int n;     scanf("%d",&n);     int arr[n];     for(int i=0;i<n;i++)     {         scanf("%d",&arr[i]);     }     int count=0;     int num=arr[0];     for(int i=1;i<n;i++)     {        if(num!=arr[i])             count++;     }     printf("%d",count); } Array Code in C language

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