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MTHREE exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer

MTHREE exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer

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إظهار المزيد

📈 نظرة تحليلية على قناة تيليجرام MTHREE exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer

تُعد قناة MTHREE exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer (@coding_are) في القطاع اللغوي الإنكليزية لاعباً نشطاً. يضم المجتمع حالياً 13 241 مشتركاً، محتلاً المرتبة 15 362 في فئة التعليم والمرتبة 32 092 في منطقة الهند.

📊 مؤشرات الجمهور والحراك

منذ تأسيسه في невідомо، حقق المشروع نمواً سريعاً وجمع 13 241 مشتركاً.

بحسب آخر البيانات بتاريخ 18 يونيو, 2026، تحافظ القناة على نشاط مستقر. خلال آخر 30 يوماً تغيّر عدد الأعضاء بمقدار -138، وفي آخر 24 ساعة بمقدار -2، مع بقاء الوصول العام مرتفعاً.

  • حالة التحقق: غير موثّقة
  • معدل التفاعل (ER): يبلغ متوسط تفاعل الجمهور 2.93‎%. وخلال أول 24 ساعة من النشر يحصد المحتوى عادةً 1.11‎% من ردود الفعل نسبةً إلى إجمالي المشتركين.
  • وصول المنشورات: يحصل كل منشور على متوسط 388 مشاهدة. وخلال اليوم الأول يجمع عادةً 147 مشاهدة.
  • التفاعلات والاستجابة: يتفاعل الجمهور بانتظام؛ متوسط التفاعلات لكل منشور يبلغ 2.
  • الاهتمامات الموضوعية: يركز المحتوى على مواضيع رئيسية مثل placement, gaurntee, suree, capgemini, infosy.

📝 الوصف وسياسة المحتوى

يصف المؤلف القناة بأنها مساحة للتعبير عن الآراء الذاتية:
🔥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...

بفضل وتيرة التحديث المرتفعة (أحدث البيانات بتاريخ 19 يونيو, 2026) تحافظ القناة على حداثتها ومستوى وصول مرتفع. وتُظهر التحليلات تفاعلاً نشطاً من الجمهور، ما يجعلها نقطة تأثير مهمة ضمن فئة التعليم.

13 241
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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; }

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; }