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

allcoding1_official

allcoding1_official

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تُعد قناة allcoding1_official (@allcoding1_official) في القطاع اللغوي الإنكليزية لاعباً نشطاً. يضم المجتمع حالياً 79 303 مشتركاً، محتلاً المرتبة 1 593 في فئة التكنولوجيات والتطبيقات والمرتبة 3 842 في منطقة الهند.

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بحسب آخر البيانات بتاريخ 20 أكتوبر, 2026، تحافظ القناة على نشاط مستقر. خلال آخر 30 يوماً تغيّر عدد الأعضاء بمقدار -1 669، وفي آخر 24 ساعة بمقدار -55، مع بقاء الوصول العام مرتفعاً.

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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); } Count press

Tcs exam
Tcs exam

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; }

#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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