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📈 Аналитический обзор Telegram-канала allcoding1

Канал allcoding1 (@allcoding1) языкового сегмента Английский является активным участником. Сейчас сообщество объединяет 22 601 подписчиков, занимая 8 831 место в категории Образование и 19 534 место в регионе Индия.

📊 Показатели аудитории и динамика

С момента создания невідомо проект демонстрирует стремительный рост, собрав аудиторию из 22 601 подписчиков.

Согласно последним данным от 10 июня, 2026, канал показывает стабильную активность. За последние 30 дней изменение числа участников составило -430, а за последние 24 часа — -7, при этом общий охват остаётся высоким.

  • Статус верификации: Не верифицирован
  • Уровень вовлечённости (ER): Средний показатель вовлечённости аудитории составляет 5.78%. В первые 24 часа после публикации контент обычно набирает 1.40% реакций от общего числа подписчиков.
  • Охват публикаций: В среднем каждый пост получает 1 307 просмотров. В течение первых суток публикация набирает 317 просмотров.
  • Реакции и взаимодействия: Аудитория активно поддерживает контент: среднее количество реакций на один пост — 2.
  • Тематические интересы: Контент сосредоточен на ключевых темах, таких как dsa, stack, namaste, javascript, learning.

📝 Описание и контентная политика

Описание канала не предоставлено.

Благодаря высокой частоте обновлений (последние данные получены 11 июня, 2026) канал поддерживает актуальность и высокий уровень охвата публикаций. Аналитика показывает, что аудитория активно взаимодействует с контентом, что делает его важной точкой влияния в категории Образование.

22 601
Подписчики
-724 часа
-907 дней
-43030 день
Архив постов
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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allcoding1 - Статистика и аналитика Telegram-канала @allcoding1