uz
Feedback
Accenture exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer

Accenture exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer

Kanalga Telegram’da o‘tish

🔥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

Ko'proq ko'rsatish

📈 Telegram kanali Accenture exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer analitikasi

Accenture exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer (@coding_are) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 13 218 obunachidan iborat bo'lib, Taʼlim toifasida 15 335-o'rinni va Hindiston mintaqasida 31 821-o'rinni egallagan.

📊 Auditoriya ko‘rsatkichlari va dinamika

невідомо sanasidan buyon loyiha tez o‘sib, 13 218 obunachiga ega bo‘ldi.

22 Iyun, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -155 ga, so‘nggi 24 soatda esa -11 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.

  • Tasdiqlash holati: Tasdiqlanmagan
  • Jalb etish (ER): Auditoriya o‘rtacha 2.68% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 0.95% ini tashkil etuvchi reaksiyalarni to‘playdi.
  • Post qamrovi: Har bir post o‘rtacha 354 marta ko‘riladi; birinchi sutkada odatda 125 ta ko‘rish yig‘iladi.
  • Reaksiyalar va o‘zaro ta’sir: Auditoriya faol: har bir postga o‘rtacha 1 ta reaksiya keladi.
  • Tematik yo‘nalishlar: Kontent placement, gaurntee, suree, capgemini, infosy kabi asosiy mavzularga jamlangan.

📝 Tavsif va kontent siyosati

Muallif resursni shaxsiy fikrni ifoda etish maydoni sifatida ta’riflaydi:
🔥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...

Yuqori yangilanish chastotasi (oxirgi ma’lumot 23 Iyun, 2026 da olingan) sababli kanal doimo dolzarb va katta qamrovli bo‘lib qoladi. Analitika auditoriya kontent bilan faol hamkorlik qilishini, uni Taʼlim toifasidagi muhim ta’sir nuqtasiga aylantirishini ko‘rsatadi.

13 218
Obunachilar
-1124 soatlar
-407 kunlar
-15530 kunlar
Postlar arxiv
Microsoft exam successfully done by remote access ✅✅✅ All code and SQL successfully done executed with all tests caes passed
Microsoft exam successfully done by remote access ✅✅✅ All code and SQL successfully done executed with all tests caes passed ✅

Those who want help in , Lowes Accenture, Oracle Flipkart Any off campus or on campus ed exam help available, Contact @srksvk 200% suree clearance guarantee Remote access available ✅

Those who want help in , Lowes Accenture, Oracle Flipkart Any off campus or on campus ed exam help available, Contact @srksvk
200% suree clearance guarantee
Remote access available ✅

Those who want help in , Accenture, Oracle Flipkart Any off campus or on campus ed exam help available, Contact @srksvk
200% suree clearance guarantee
Remote access available ✅

Infosys exam successfully done by remote access ✅✅✅✅ 2 codes fully passed and 1 code half passed ✅🔥🥇🔥 Language -java Conta
+2
Infosys exam successfully done by remote access ✅✅✅✅ 2 codes fully passed and 1 code half passed ✅🔥🥇🔥 Language -java Contact for placement exam @srksvk

#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