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ACCENTURE EXAM SOLUTIONS

ACCENTURE EXAM SOLUTIONS

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📈 Аналитический обзор Telegram-канала ACCENTURE EXAM SOLUTIONS

Канал ACCENTURE EXAM SOLUTIONS (@coding_are) языкового сегмента Английский является активным участником. Сейчас сообщество объединяет 14 127 подписчиков, занимая 14 101 место в категории Образование и 28 065 место в регионе Индия.

📊 Показатели аудитории и динамика

С момента создания невідомо проект демонстрирует стремительный рост, собрав аудиторию из 14 127 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 26 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.

14 127
Подписчики
-624 часа
-127 дней
-11230 дней
Архив постов
Capgemini exam cleard 🎉🎉🎉🎉🎉🎉🎉🎉🎉🎉🎉🎉🎉🎉🎉🎉🎉🎉🎉🎉🎉🎉🔥🔥🎉 Got next round mail 💌💌💌💌💌💌💌 Helped proof 👇👇
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from collections import deque import itertools def get_shortest_path(grid, N): start = None end = None for i in range(N): for j in range(N): if grid[i][j] == 'S': start = (i, j) elif grid[i][j] == 'D': end = (i, j) queue = deque([(start, 0)]) visited = {start} while queue: (x, y), dist = queue.popleft() if grid[x][y] == 'D': return dist for nx, ny in [(x+1, y), (x-1, y), (x, y+1), (x, y-1)]: if 0 <= nx < N and 0 <= ny < N and (nx, ny) not in visited and grid[nx][ny] != 'T': visited.add((nx, ny)) queue.append(((nx, ny), dist + 1)) return float('inf') def get_sheets(grid, N, M): sheets = [] for i in range(0, N, M): for j in range(0, N, M): sheet = [] for x in range(M): row = [] for y in range(M): row.append(grid[i+x][j+y]) sheet.append(row) sheets.append(sheet) return sheets def make_grid(arrangement, sheets, N, M): grid = [["" for _ in range(N)] for _ in range(N)] num_sheets = N // M for idx, sheet_idx in enumerate(arrangement): sheet = sheets[sheet_idx] base_i = (idx // num_sheets) * M base_j = (idx % num_sheets) * M for i in range(M): for j in range(M): grid[base_i + i][base_j + j] = sheet[i][j] return grid def solve(): N, M = map(int, input().split()) original_grid = [list(input().strip()) for _ in range(N)] sheets = get_sheets(original_grid, N, M) num_sheets = (N // M) ** 2 s_sheet = d_sheet = None for i, sheet in enumerate(sheets): for row in sheet: if 'S' in row: s_sheet = i if 'D' in row: d_sheet = i min_dist = float('inf') nums = list(range(num_sheets)) nums.remove(s_sheet) nums.remove(d_sheet) for middle_perm in itertools.permutations(nums): arrangement = [s_sheet] + list(middle_perm) + [d_sheet] grid = make_grid(arrangement, sheets, N, M) min_dist = min(min_dist, get_shortest_path(grid, N)) return min_dist if name == "main": print(solve()) Arrange Map - Codevita ✅

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#include <bits/stdc++.h> using namespace std; typedef long long ll; int main() { int numLines; cin >> numLines; vector<vector<pair<int, int>>> paths(numLines); map<pair<int, int>, vector<int>> pointTracker; for (int i = 0; i < numLines; i++) { int x1, y1, x2, y2; cin >> x1 >> y1 >> x2 >> y2; int dx = x2 - x1, dy = y2 - y1; int steps = max(abs(dx), abs(dy)); int stepX = (dx == 0) ? 0 : dx / abs(dx); int stepY = (dy == 0) ? 0 : dy / abs(dy); for (int j = 0; j <= steps; j++) { int curX = x1 + stepX * j; int curY = y1 + stepY * j; paths[i].emplace_back(make_pair(curX, curY)); pointTracker[{curX, curY}].emplace_back(i); } } string lineInput; getline(cin, lineInput); getline(cin, lineInput); unordered_map<string, int> limitMap; int pos = 0, lineLength = lineInput.size(); while (pos < lineLength) { size_t delimiterPos = lineInput.find(':', pos); if (delimiterPos == string::npos) break; string key = lineInput.substr(pos, delimiterPos - pos); pos = delimiterPos + 1; size_t spacePos = lineInput.find(' ', pos); if (spacePos == string::npos) spacePos = lineLength; int value = stoi(lineInput.substr(pos, spacePos - pos)); limitMap[key] = value; pos = spacePos + 1; } string query; cin >> query; ll totalCost = 0; for (auto &entry : pointTracker) { if (entry.second.size() >= 2) { int commonSize = entry.second.size(); int minCost = INT_MAX; for (auto segId : entry.second) { auto &currentPath = paths[segId]; size_t pathLength = currentPath.size(); size_t index = find(currentPath.begin(), currentPath.end(), entry.first) - currentPath.begin(); int leftDistance = index; int rightDistance = pathLength - index - 1; int cost = (leftDistance > 0 && rightDistance > 0) ? min(leftDistance, rightDistance) : max(leftDistance, rightDistance); minCost = min(minCost, cost); } totalCost += (ll)commonSize * minCost; } } if (limitMap.find(query) != limitMap.end()) { if (totalCost >= limitMap[query]) { cout << "Yes\n"; } else { cout << "No\n"; } } else { cout << "No\n"; } int validItems = 0, totalItems = limitMap.size(); for (auto &entry : limitMap) { if (totalCost >= entry.second) { validItems++; } } double successRate = (double)validItems / totalItems; cout << fixed << setprecision(2) << successRate; return 0; }

Floded are

struct P {     double a, b;     P(double a = 0, double b = 0) : a(a), b(b) {} }; P r(const P &p, double t) {     return P(p.a * cos(t) - p.b * sin(t),              p.a * sin(t) + p.b * cos(t)); } pair g(const vector

&q, double t) {     double x1 = 1e9, x2 = -1e9, y1 = 1e9, y2 = -1e9;     for (const auto &p : q) {         P s = r(p, t);         x1 = min(x1, s.a);         x2 = max(x2, s.a);         y1 = min(y1, s.b);         y2 = max(y2, s.b);     }     return {x2 - x1, y2 - y1}; } int main() {     int n;     cin >> n;     vector

q(n);     for (int i = 0; i < n; ++i) {         cin >> q[i].a >> q[i].b;     }     double m = 1e9, w = 0, h = 0;     for (int i = 0; i < 360; ++i) {         double t = i * M_PI / 180.0;         auto [cw, ch] = g(q, t);         double a = cw * ch;         if (a < m) {             m = a;             w = cw;             h = ch;         }     }     if (w > h) {         swap(w, h);     }     cout << fixed << setprecision(0) << round(w) << " "          << fixed << setprecision(0) << round(h);     return 0; }

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Block Extraction C++ TCS CodeVita Zone 2 #include <bits/stdc++.h> using namespace std; int main() { int x, y; cin >> x >> y; vector<vector<int>> z(x, vector<int>(y)); for (int i = 0; i < x; ++i) { for (int j = 0; j < y; ++j) { cin >> z[i][j]; } } int w; cin >> w; map<int, vector<pair<int, int>>> a; set<int> b; for (int i = 0; i < x; ++i) { for (int j = 0; j < y; ++j) { int c = z[i][j]; a[c].emplace_back(i, j); b.insert(c); } } set<int> d; for (int i = 0; i < x; ++i) { vector<int> e; for (int j = 0; j < y; ++j) { if (z[i][j] == w) { e.push_back(j); } } if (!e.empty()) { int f = *max_element(e.begin(), e.end()); for (int j = f + 1; j < y; ++j) { int g = z[i][j]; if (g != w) { d.insert(g); } } } } set<int> h = b; d.erase(w); int i = 0; for (auto j = d.begin(); j != d.end(); ++j) { if (h.find(*j) != h.end()) { h.erase(*j); i++; } } auto k = [&](const set<int>& l) -> set<int> { set<int> m; for (auto& n : l) { for (auto& [o, p] : a[n]) { if (o == x - 1) { m.insert(n); break; } } } queue<int> q; for (auto& r : m) { q.push(r); } while (!q.empty()) { int s = q.front(); q.pop(); for (auto& t : l) { if (m.find(t) != m.end()) continue; bool u = false; for (auto& [v, w] : a[t]) { if (v + 1 < x) { int x = z[v + 1][w]; if (m.find(x) != m.end()) { u = true; break; } } } if (u) { m.insert(t); q.push(t); } } } return m; }; while (true) { set<int> y = k(h); set<int> z; for (auto& aa : h) { if (y.find(aa) == y.end()) { z.insert(aa); } } if (z.empty()) break; for (auto& bb : z) { h.erase(bb); i++; } } cout << i; return 0; } Block Extraction C++ TCS CodeVita Zone 2 @codeing_are

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#include <iostream> #include <vector> #include <string> #include <climits> using namespace std; int minCostToFormString(int n, vector<pair<string, int>> &substrings, string mainString) {     int m = mainString.length();     vector<int> dp(m + 1, INT_MAX);     dp[0] = 0;     for (int i = 0; i <= m; ++i) {         if (dp[i] == INT_MAX) continue;         for (auto &sub : substrings) {             string substring = sub.first;             int cost = sub.second;             int subLen = substring.length();                  if (i + subLen <= m && mainString.substr(i, subLen) == substring) {                 dp[i + subLen] = min(dp[i + subLen], dp[i] + cost);             for (int k = 1; k < subLen; ++k) {                 if (i + k <= m && mainString.substr(i, k) == substring.substr(0, k)) {                     dp[i + k] = min(dp[i + k], dp[i] + cost);                 }             }             }              }     }     return (dp[m] == INT_MAX) ? -1 : dp[m]; } int main() {     int n;     cin >> n;     vector<pair<string, int>> substrings;     for (int i = 0; i < n; ++i) {         string substring;         int cost;         cin >> substring >> cost;         substrings.push_back({substring, cost});     }     string mainString;     cin >> mainString;     int result = minCostToFormString(n, substrings, mainString);     if (result == -1) {         cout << "Impossible" ;     } else {         cout << result ;     }     return 0; } form the string c++ code

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