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allcoding1

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📈 نظرة تحليلية على قناة تيليجرام allcoding1

تُعد قناة allcoding1 (@allcoding1) في القطاع اللغوي الإنكليزية لاعباً نشطاً. يضم المجتمع حالياً 22 601 مشتركاً، محتلاً المرتبة 8 831 في فئة التعليم والمرتبة 19 534 في منطقة الهند.

📊 مؤشرات الجمهور والحراك

منذ تأسيسه في невідомо، حقق المشروع نمواً سريعاً وجمع 22 601 مشتركاً.

بحسب آخر البيانات بتاريخ 10 يونيو, 2026، تحافظ القناة على نشاط مستقر. خلال آخر 30 يوماً تغيّر عدد الأعضاء بمقدار -430، وفي آخر 24 ساعة بمقدار -7، مع بقاء الوصول العام مرتفعاً.

  • حالة التحقق: غير موثّقة
  • معدل التفاعل (ER): يبلغ متوسط تفاعل الجمهور 5.78‎%. وخلال أول 24 ساعة من النشر يحصد المحتوى عادةً 1.40‎% من ردود الفعل نسبةً إلى إجمالي المشتركين.
  • وصول المنشورات: يحصل كل منشور على متوسط 1 307 مشاهدة. وخلال اليوم الأول يجمع عادةً 317 مشاهدة.
  • التفاعلات والاستجابة: يتفاعل الجمهور بانتظام؛ متوسط التفاعلات لكل منشور يبلغ 2.
  • 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 11 يونيو, 2026) تحافظ القناة على حداثتها ومستوى وصول مرتفع. وتُظهر التحليلات تفاعلاً نشطاً من الجمهور، ما يجعلها نقطة تأثير مهمة ضمن فئة التعليم.

22 601
المشتركون
-724 ساعات
-907 أيام
-43030 أيام
أرشيف المشاركات
using System; using System.Collections.Generic; class Program { static void Main() { int N = int.Parse(Console.ReadLine()); var graph = new Dictionary>(); var indegrees = new Dictionary(); for (int i = 0; i < N; i++) { var edge = Console.ReadLine().Split(); string from = edge[0]; string to = edge[1]; if (!graph.ContainsKey(from)) graph[from] = new List(); graph[from].Add(to); if (!indegrees.ContainsKey(from)) indegrees[from] = 0; if (!indegrees.ContainsKey(to)) indegrees[to] = 0; indegrees[to]++; } var words = Console.ReadLine().Split(); var levels = new Dictionary(); var queue = new Queue(); foreach (var node in indegrees.Keys) { if (indegrees[node] == 0) { levels[node] = 1; // Root level is 1 queue.Enqueue(node); } } while (queue.Count > 0) { var current = queue.Dequeue(); int currentLevel = levels[current]; if (graph.ContainsKey(current)) { foreach (var neighbor in graph[current]) { if (!levels.ContainsKey(neighbor)) { levels[neighbor] = currentLevel + 1; queue.Enqueue(neighbor); } indegrees[neighbor]--; if (indegrees[neighbor] == 0 && !levels.ContainsKey(neighbor)) { queue.Enqueue(neighbor); } } } } int totalValue = 0; foreach (var word in words) { if (levels.ContainsKey(word)) { totalValue += levels[word]; } else { totalValue += -1; } } Console.Write(totalValue); } } String Puzzle C+

int main() { string s; cin >> s; int n = s.length(), res = 0; vector v(n); for (int i = 0; i < n; ++i) cin >> v[i]; int lw = s[0] - '0', lwv = v[0]; for (int i = 1; i < n; ++i) { if (s[i] - '0' == lw) { res += min(lwv, v[i]); lwv = max(lwv, v[i]); } else { lw = s[i] - '0'; lwv = v[i]; } } cout << res; return 0; } Form alternating string Change accordingly before submitting to avoid plagiarism

#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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