ru
Feedback
ACCENTURE EXAM SOLUTIONS

ACCENTURE EXAM SOLUTIONS

Открыть в Telegram

🔥Guys plz Stop fearing for daily exams 📝 👨‍💻 @srksvk is here to help you all at lowest cost possible.💪 🌀 ” Our Only Aim Is To Let Get Placed To You In A Reputed Company 🔥Effort from our side = 💯 📱Main Channel: @coding_are 📱Tel I'd : @srksvk

Больше

📈 Аналитический обзор Telegram-канала ACCENTURE EXAM SOLUTIONS

Канал ACCENTURE EXAM SOLUTIONS (@coding_are) языкового сегмента Английский является активным участником. Сейчас сообщество объединяет 14 129 подписчиков, занимая 14 097 место в категории Образование и 28 067 место в регионе Индия.

📊 Показатели аудитории и динамика

С момента создания невідомо проект демонстрирует стремительный рост, собрав аудиторию из 14 129 подписчиков.

Согласно последним данным от 26 сентября, 2026, канал показывает стабильную активность. За последние 30 дней изменение числа участников составило -112, а за последние 24 часа — -6, при этом общий охват остаётся высоким.

  • Статус верификации: Не верифицирован
  • Уровень вовлечённости (ER): Средний показатель вовлечённости аудитории составляет 3.64%. В первые 24 часа после публикации контент обычно набирает 1.54% реакций от общего числа подписчиков.
  • Охват публикаций: В среднем каждый пост получает 514 просмотров. В течение первых суток публикация набирает 218 просмотров.
  • Реакции и взаимодействия: Аудитория активно поддерживает контент: среднее количество реакций на один пост — 2.
  • Тематические интересы: Контент сосредоточен на ключевых темах, таких как placement, gaurntee, suree, capgemini, infosy.

📝 Описание и контентная политика

Автор описывает ресурс как площадку для выражения субъективного мнения:
“🔥Guys plz Stop fearing for daily exams 📝 👨‍💻 @srksvk is here to help you all at lowest cost possible.💪 🌀 ” Our Only Aim Is To Let Get Placed To You In A Reputed Company 🔥Effort from our side = 💯 📱Main Channel: @coding_are 📱Tel I'd : @srks...”

Благодаря высокой частоте обновлений (последние данные получены 27 сентября, 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.

14 128
Подписчики
-624 часа
-127 дней
-11230 дней
Архив постов
#include <bits/stdc++.h> using namespace std; #define loop(i, a, n) for (lli i = (a); i < (n); ++i) #define loopD(i, a, n) for (lli i = (a); i >= (n); --i) #define all(c) (c).begin(), (c).end() #define rall(c) (c).rbegin(), (c).rend() #define sz(a) ((lli)a.size()) #define YES cout << "YES" << endl; #define NO cout << "NO" << endl; #define endl '\n' #define fastio std::ios::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL); #define pb push_back #define pp pop_back() #define fi first #define si second #define v(a) vector<int>(a) #define vv(a) vector<vector<int>>(a) #define present(c, x) ((c).find(x) != (c).end()) #define set_bits __builtin_popcountll #define MOD 1000000007 #define int long long typedef long long lli; typedef vector<int> vi; typedef vector<vi> vvi; typedef pair<lli, lli> pll; typedef pair<int, int> pii; typedef unordered_map<int, int> umpi; typedef map<int, int> mpi; typedef vector<pii> vp; typedef vector<lli> vll; typedef vector<vll> vvll; struct Line { int x1, y1, x2, y2; bool vertical() const { return x1 == x2; } bool horizontal() const { return y1 == y2; } bool diagonal() const { return abs(x2 - x1) == abs(y2 - y1); } }; int N, K; vector<Line> lines; map<pair<int, int>, vector<int>> ptsMap; void add(const Line& line, int idx) { int x1 = line.x1, y1 = line.y1; int x2 = line.x2, y2 = line.y2; if (line.vertical()) { int yStart = min(y1, y2); int yEnd = max(y1, y2); for(int y = yStart; y <= yEnd; y++) { ptsMap[{x1, y}].push_back(idx); } } else if (line.horizontal()) { int xStart = min(x1, x2); int xEnd = max(x1, x2); for(int x = xStart; x <= xEnd; x++) { ptsMap[{x, y1}].push_back(idx); } } else if (line.diagonal()) { int steps = abs(x2 - x1); int dx = (x2 - x1) / steps; int dy = (y2 - y1) / steps; for(int i = 0; i <= steps; i++) { int x = x1 + i * dx; int y = y1 + i * dy; ptsMap[{x, y}].push_back(idx); } } } int ff(int x1, int y1, int x2, int y2) { if(x1 == x2) return abs(y1 - y2); if(y1 == y2) return abs(x1 - x2); if(abs(x1 - x2) == abs(y1 - y2)) return abs(x1 - x2); return 0; } int solve(const pair<int, int>& pt, const vector<int>& lst) { vector<int> d; for(auto lIdx : lst) { const Line& ln = lines[lIdx]; bool oneSided = (pt.first == ln.x1 && pt.second == ln.y1) || (pt.first == ln.x2 && pt.second == ln.y2); if(oneSided) { int ex = (pt.first == ln.x1 && pt.second == ln.y1) ? ln.x2 : ln.x1; int ey = (pt.first == ln.x1 && pt.second == ln.y1) ? ln.y2 : ln.y1; d.push_back(ff(pt.first, pt.second, ex, ey)); } else { d.push_back(ff(pt.first, pt.second, ln.x1, ln.y1)); d.push_back(ff(pt.first, pt.second, ln.x2, ln.y2)); } } return d.empty() ? 0 : *min_element(d.begin(), d.end()); } void solve() { cin >> N; lines.resize(N); for(int i = 0; i < N; i++) { cin >> lines[i].x1 >> lines[i].y1 >> lines[i].x2 >> lines[i].y2; add(lines[i], i); } cin >> K; int total = 0; for(auto &[pt, lst] : ptsMap) { if(sz(lst) == K) { total += solve(pt, lst); } } cout << total; } int32_t main() { solve(); return 0; } // magic stars intensity ac Full accepted ☺️

After 16k i will upload one more code answer with all tets caes passed 🔥🥳🥳🥳 Show share fast guys

Fast guy's next answer are ready

Guys now you turn to share my group So, share group and add your friend Please guy's https://t.me/codeing_are

All are correct code with verified ☺️ So do correctly everyone

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+

photo content
+1

Guys now you turn to share my group So, share group and add your friend Please guy's https://t.me/codeing_are

photo content

int main() { string s; cin >> s; int n = s.length(), res = 0; vector<int> 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 @codeing_are

Gya i will upload all correct answer 🔥

Tcs code vitaa answer?👇👇👇👇👇

Cisco exam help successfully done by Remote access 🔥🔥🔥🔥🔥🔥🔥🔥🔥 3/3 code done with all tests caes passed 🔥🔥🔥🔥 Conta
+2
Cisco exam help successfully done by Remote access 🔥🔥🔥🔥🔥🔥🔥🔥🔥 3/3 code done with all tests caes passed 🔥🔥🔥🔥 Contact for placement exam @srksvk

Capgemini on campus exam help successfully completed by Remote access 🔥🔥🔥🔥 40/40 MCQ done with all tests caes passed 🔥🔥
+1
Capgemini on campus exam help successfully completed by Remote access 🔥🔥🔥🔥 40/40 MCQ done with all tests caes passed 🔥🔥🔥 All gameing round done ✅✅✅✅✅ Contact for placement exam @srksvk

Ibm on campus exam successfully completed by Remote access 🔥🔥🔥🔥🔥🔥 1/1 code done with all tests caes passed 🔥🔥🔥🔥🔥 C
Ibm on campus exam successfully completed by Remote access 🔥🔥🔥🔥🔥🔥 1/1 code done with all tests caes passed 🔥🔥🔥🔥🔥 Contact for placement exam @srksvk

Cognizant communication round help available Contact fast and book your slots Contact @srksvk 200% sure clearance grauntee 🔥
Cognizant communication round help available Contact fast and book your slots Contact @srksvk 200% sure clearance grauntee 🔥 Note - if not cleard then I will refund whole money

Those who needs help to clear Cognizant communication round contact me immediately 💯 % clearance guaranteed (If not cleared
Those who needs help to clear Cognizant communication round contact me immediately 💯 % clearance guaranteed (If not cleared money will be refunded) Contact for placement exam @srksvk

hasheldn exam cleard 🎉🎉🎉🎉🎉🎉🎉🎉🎉🎉🔥🔥🎉🎉🔥🔥🎉🎉🔥🔥🎉🎉🔥🔥 Got interview mail 🔥🔥🔥🔥🔥🔥😀 Helped proof 👇👇👇👇
+1
hasheldn exam cleard 🎉🎉🎉🎉🎉🎉🎉🎉🎉🎉🔥🔥🎉🎉🔥🔥🎉🎉🔥🔥🎉🎉🔥🔥 Got interview mail 🔥🔥🔥🔥🔥🔥😀 Helped proof 👇👇👇👇👇👇👇 Contact for placement exam @srksvk

Saks exam successfully completed by Remote access to 🔥🔥🔥🔥🔥🔥🔥🔥 2/2 codes +1 SQL done with all tests caes passed 🔥🔥🔥
+1
Saks exam successfully completed by Remote access to 🔥🔥🔥🔥🔥🔥🔥🔥 2/2 codes +1 SQL done with all tests caes passed 🔥🔥🔥🔥🔥 Contact @srksvk

Saks exam successfully completed by Remote access to 🔥🔥🔥🔥🔥🔥🔥🔥 2/2 codes +1 SQL done with all tests caes passed 🔥🔥🔥
+1
Saks exam successfully completed by Remote access to 🔥🔥🔥🔥🔥🔥🔥🔥 2/2 codes +1 SQL done with all tests caes passed 🔥🔥🔥🔥🔥 Contact @srksvk