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LTM EXAM SOLUTIONS

LTM EXAM SOLUTIONS

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🔥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

إظهار المزيد

📈 نظرة تحليلية على قناة تيليجرام LTM EXAM SOLUTIONS

تُعد قناة LTM EXAM SOLUTIONS (@coding_are) في القطاع اللغوي الإنكليزية لاعباً نشطاً. يضم المجتمع حالياً 14 110 مشتركاً، محتلاً المرتبة 14 118 في فئة التعليم والمرتبة 28 107 في منطقة الهند.

📊 مؤشرات الجمهور والحراك

منذ تأسيسه في невідомо، حقق المشروع نمواً سريعاً وجمع 14 110 مشتركاً.

بحسب آخر البيانات بتاريخ 03 أكتوبر, 2026، تحافظ القناة على نشاط مستقر. خلال آخر 30 يوماً تغيّر عدد الأعضاء بمقدار -103، وفي آخر 24 ساعة بمقدار 2، مع بقاء الوصول العام مرتفعاً.

  • حالة التحقق: غير موثّقة
  • معدل التفاعل (ER): يبلغ متوسط تفاعل الجمهور 3.70‎%. وخلال أول 24 ساعة من النشر يحصد المحتوى عادةً 1.68‎% من ردود الفعل نسبةً إلى إجمالي المشتركين.
  • وصول المنشورات: يحصل كل منشور على متوسط 522 مشاهدة. وخلال اليوم الأول يجمع عادةً 237 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 04 أكتوبر, 2026) تحافظ القناة على حداثتها ومستوى وصول مرتفع. وتُظهر التحليلات تفاعلاً نشطاً من الجمهور، ما يجعلها نقطة تأثير مهمة ضمن فئة التعليم.

14 110
المشتركون
+224 ساعات
-177 أيام
-10330 أيام
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#include <iostream> #include <vector> #include <unordered_map> #include <algorithm> using namespace std; const int MOD = 1e9 + 7; vector<int> a; vector<vector<int>> g; vector<vector<int>> dp; vector<int> subtree_sum; int k; void dfs(int u, int parent) { dp[u][0] = 1; subtree_sum[u] = a[u]; for (int v : g[u]) { if (v != parent) { dfs(v, u); vector<int> tmp(k + 1, 0); for (int i = 0; i <= k; ++i) { if (dp[u][i]) { for (int j = 0; j <= k - i; ++j) { if (dp[v][j]) { tmp[i + j] = (tmp[i + j] + (long long)dp[u][i] * dp[v][j]) % MOD; } } } } for (int i = 0; i <= k; ++i) { dp[u][i] = (dp[u][i] + tmp[i]) % MOD; } } } for (int i = k; i >= a[u]; --i) { dp[u][i] = (dp[u][i] + dp[u][i - a[u]]) % MOD; } } int main() { int n, m; cin >> n >> m >> k; a.resize(n + 1); g.resize(n + 1); dp.resize(n + 1, vector<int>(k + 1, 0)); subtree_sum.resize(n + 1, 0); for (int i = 1; i <= n; ++i) { cin >> a[i]; } for (int i = 0; i < m; ++i) { int x, y; cin >> x >> y; g[x].push_back(y); g[y].push_back(x); } long long result = 0; dfs(1, -1); for (int i = 1; i <= n; ++i) { result = (result + dp[i][k]) % MOD; } cout << result << endl; return 0; } Tree Function 🌳🌲🌴

class SegmentTree: def init(self, data): self.n = len(data) self.tree = [0] * (2 * self.n) self.build(data) def build(self, data): for i in range(self.n): self.tree[self.n + i] = data[i] for i in range(self.n - 1, 0, -1): self.tree[i] = max(self.tree[2 * i], self.tree[2 * i + 1]) def range_max(self, left, right): left += self.n right += self.n max_val = 0 while left < right: if left % 2: max_val = max(max_val, self.tree[left]) left += 1 if right % 2: right -= 1 max_val = max(max_val, self.tree[right]) left //= 2 right //= 2 return max_val def calculate_total_xp(N, A, Bonus): seg_tree = SegmentTree() seg_tree.init(Bonus) last_seen = [-1] * (N // 2 + 1) right_multiple = [-1] * N for i in range(N - 1, -1, -1): power = A[i] next_pos = float('inf') for multiple in range(power, N // 2 + 1, power): if last_seen[multiple] != -1: next_pos = min(next_pos, last_seen[multiple]) if next_pos != float('inf'): right_multiple[i] = next_pos last_seen[power] = i total_xp = 0 for i in range(N): if right_multiple[i] != -1: max_bonus = seg_tree.range_max(i, right_multiple[i] + 1) total_xp += max_bonus return total_xp N = int(input().strip()) A = [int(input().strip()) for _ in range(N)] Bonus = [int(input().strip()) for _ in range(N)] result = calculate_total_xp(N, A, Bonus) print(result) Some help

#include <bits/stdc++.h> using namespace std; vector<int> sieve(int max_number) { vector<bool> is_prime(max_number + 1, true); vector<int> primes; for (int number = 2; number <= max_number; ++number) { if (is_prime[number]) { primes.push_back(number); for (int multiple = number * number; multiple <= max_number; multiple += number) { is_prime[multiple] = false; } } } return primes; } int count_divisors(int number, const vector<int>& primes) { int divisor_count = 0; int prime_count = primes.size(); for (int subset = 1; subset < (1 << prime_count); ++subset) { long long least_common_multiple = 1; int bit_count = 0; for (int bit = 0; bit < prime_count; ++bit) { if (subset & (1 << bit)) { least_common_multiple *= primes[bit]; bit_count++; if (least_common_multiple > number) break; } } if (least_common_multiple > number) continue; if (bit_count % 2 == 1) divisor_count += number / least_common_multiple; else divisor_count -= number / least_common_multiple; } return divisor_count; } int count_non_divisors(int number, const vector<int>& primes) { if (number == 0) return 0; return number - count_divisors(number, primes); } int count_non_divisors_range(int max_prime, int left, int right) { vector<int> primes = sieve(max_prime); int right_count = count_non_divisors(right, primes); int left_count = count_non_divisors(left - 1, primes); return right_count - left_count; } int main() { int max_prime; string left_str, right_str; cin >> max_prime >> left_str >> right_str; cout << count_non_divisors_range(max_prime, stoi(left_str), stoi(right_str)) << endl; return 0; } Divisible string

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