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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 14 136 подписчиков, занимая 14 103 место в категории Образование и 28 083 место в регионе Индия.

📊 Показатели аудитории и динамика

С момента создания невідомо проект демонстрирует стремительный рост, собрав аудиторию из 14 136 подписчиков.

Согласно последним данным от 21 сентября, 2026, канал показывает стабильную активность. За последние 30 дней изменение числа участников составило -136, а за последние 24 часа — 1, при этом общий охват остаётся высоким.

  • Статус верификации: Не верифицирован
  • Уровень вовлечённости (ER): Средний показатель вовлечённости аудитории составляет 4.15%. В первые 24 часа после публикации контент обычно набирает 1.53% реакций от общего числа подписчиков.
  • Охват публикаций: В среднем каждый пост получает 586 просмотров. В течение первых суток публикация набирает 216 просмотров.
  • Реакции и взаимодействия: Аудитория активно поддерживает контент: среднее количество реакций на один пост — 1.
  • 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 23 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.

14 139
Подписчики
+124 часа
-367 дней
-13630 дней
Архив постов
include using namespace std; int N, M; vector> A; map, bool> dp; vector> dirs = {{1, 0}, {0, 1}}; bool dfs(int ax, int ay, int bx, int by, bool turn) { auto key = make_tuple(ax, ay, bx, by, turn); if (dp.count(key)) return dp[key]; if (turn) { for (auto [dx, dy] : dirs) { int nx = ax + dx, ny = ay + dy; if (nx < N && ny < M && A[nx][ny] >= A[bx][by]) { if (dfs(nx, ny, bx, by, !turn)) return dp[key] = true; } } return dp[key] = false; } else { for (auto [dx, dy] : dirs) { int nx = bx + dx, ny = by + dy; if (nx < N && ny < M && A[nx][ny] >= A[ax][ay]) { if (!dfs(ax, ay, nx, ny, !turn)) return dp[key] = false; } } return dp[key] = true; } } int solve(int n, int m, int q, vector> grid, vector> queries) { N = n; M = m; A = grid; int res = 0; for (int i = 0; i < q; ++i) { dp.clear(); int x1 = queries[i][0] - 1; int y1 = queries[i][1] - 1; int x2 = queries[i][2] - 1; int y2 = queries[i][3] - 1; if (dfs(x1, y1, x2, y2, true)) res |= (1 << i); } return res; }

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def GetAnswer(N, A): MOD = 10**9 + 7 table = [[0] * (N+1) for _ in range(N+1)] table[0][0] = 1 nums = [0] + A for i in range(1, N+1): next_table = [[0] * (N+1) for _ in range(N+1)] for x in range(i): for y in range(i): next_table[x][y] = table[x][y] for col in range(i): total = 0 for row in range(i): if row == 0 or nums[row] < nums[i]: total = (total + table[row][col]) % MOD next_table[i][col] = total for row in range(i): total = 0 for col in range(i): if col == 0 or nums[col] > nums[i]: total = (total + table[row][col]) % MOD next_table[row][i] = total table = next_table result = 0 for row in table: result = (result + sum(row)) % MOD return result Share to All Groups 📣 Follow the Codeing_area( Srksvk) channel on WhatsApp: https://whatsapp.com/channel/0029VaicY2a65yD2YNehWP2j

#include <bits/stdc++.h> using namespace std; string trim(string); long calc(int N, int M, int X, vector<int> A, vector<int> B) { const int MAX = 100005; vector<bool> visited(MAX, false); unordered_set<int> corrupted(B.begin(), B.end()); queue<pair<int, int>> q; q.push({0, 0}); visited[0] = true; while (!q.empty()) { auto [step, jumps] = q.front(); q.pop(); for (int jump : A) { int next = step + jump; if (next == X) return jumps + 1; if (next <= X && !visited[next] && corrupted.find(next) == corrupted.end()) { visited[next] = true; q.push({next, jumps + 1}); } } } return -1; } int main() { cout << calc(1, 1, 5, {1}, {6}) << endl; cout << calc(2, 2, 5, {1, 4}, {1, 3}) << endl; cout << calc(4, 2, 4, {2, 3, 4, 1}, {1, 3}) << endl; return 0; } Share To Clg Group, public channel groups ✅ Follow the Codeing_area( Srksvk) channel on WhatsApp: https://whatsapp.com/channel/0029VaicY2a65yD2YNehWP2j

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import sys import heapq from collections import defaultdict def Diff_LCM(N, A): MOD = 10**9 + 7 M = max(abs(x) for x in A) spf = [0] * (M + 1) if M >= 1: spf[1] = 1 for i in range(2, M + 1): if spf[i] == 0: spf[i] = i ii = i * i if ii <= M: for j in range(ii, M + 1, i): if spf[j] == 0: spf[j] = i factors = [[] for _ in range(N)] global_exp = {} for i, v in enumerate(A): x = abs(v) while x > 1: p = spf[x] cnt = 0 while x % p == 0: x //= p cnt += 1 factors[i].append((p, cnt)) if cnt > global_exp.get(p, 0): global_exp[p] = cnt if not global_exp: return 0 contrib = {p: [] for p in global_exp} at_pos = [[] for _ in range(N)] for i, fl in enumerate(factors): for p, cnt in fl: if global_exp[p] == cnt: contrib[p].append(i) at_pos[i].append(p) ptr = {p: 0 for p in global_exp} next_pos = {} heap = [] for p, lst in contrib.items(): pos0 = lst[0] next_pos[p] = pos0 heap.append((-pos0, p)) heapq.heapify(heap) S = [0] * (N + 1) for i in range(N): S[i+1] = S[i] + A[i] size = 1 while size < N + 1: size <<= 1 INF = 10**30 tree = [-INF] * (2 * size) for i in range(N + 1): tree[size + i] = S[i] for i in range(size - 1, 0, -1): tree[i] = max(tree[2*i], tree[2*i + 1]) def rmq(l, r): l += size r += size m = -INF while l <= r: if l & 1: m = max(m, tree[l]) l += 1 if not r & 1: m = max(m, tree[r]) r -= 1 l //= 2; r //= 2 return m ans = 0 for l in range(N): if l > 0: for p in at_pos[l-1]: ptr[p] += 1 idx = ptr[p] lst = contrib[p] newpos = lst[idx] if idx < len(lst) else N next_pos[p] = newpos heapq.heappush(heap, (-newpos, p)) while True: negpos, p = heap[0] pos = -negpos if next_pos[p] != pos: heapq.heappop(heap) else: r_end = pos break lo = l + 1 hi = r_end if lo <= hi: best_prefix = rmq(lo, hi) ans = max(ans, best_prefix - S[l]) return ans % MOD def main(): data = sys.stdin.read().split() N = int(data[0]) A = list(map(int, data[1:])) print(Diff_LCM(N, A)) if name == "main": main()

long Pastry_Contrast(int N,int Q,vector>Pastry,vector>Queries){ vectorF(N),D(N); for(int i=0;i>e; vectorst; for(int i=0;i=D[i]){ int u=st.back();st.pop_back(); e.push_back({i,u,(F[i]-F[u])*(D[i]+D[u])}); } if(!st.empty()){ int u=st.back(); e.push_back({i,u,(F[i]-F[u])*(D[i]+D[u])}); } st.push_back(i); } sort(e.begin(),e.end(),[](auto&a,auto&b){return a[0]t(2*p,LLONG_MAX); auto upd=&{ int x=u+p; t[x]=min(t[x],v); for(x>>=1;x;x>>=1) t[x]=min(t[2*x],t[2*x+1]); }; auto qmn=&{ long s=LLONG_MAX; for(l+=p,r+=p;l<=r;l>>=1,r>>=1){ if(l&1) s=min(s,t[l++]); if(!(r&1)) s=min(s,t[r--]); } return s; }; vector>q(Q); for(int i=0;i

int Tom_Jerry(int N, vector<int> Tom, vector<int> Jerry) { int L = 2*N; vector<vector<int>> G(N+1); for(int i=0;i<N;i++) G[Tom[i]].push_back(Jerry[i]); int B = floor(sqrt(2.0*N)); vector<int> small, large; for(int v=1;v<=N;v++) if(!G[v].empty()) { if(v<=B) small.push_back(v); else large.push_back(v); } vector<vector<pair<int,int>>> comp(B+1); for(int v: small){ auto &g = G[v]; sort(g.begin(),g.end()); auto &c = comp[v]; for(int x: g){ if(c.empty()||c.back().first!=x) c.emplace_back(x,1); else c.back().second++; } } long long ans=0; int M = small.size(); for(int i=0;i<M;i++){ int v = small[i]; auto &cv = comp[v]; long long P = 1LL*v*v; if(P<=L){ int l=0, r=cv.size()-1; while(l<=r){ long long s=cv[l].first+cv[r].first; if(s<P) l++; else if(s>P) r--; else{ if(l==r) ans += 1LL*cv[l].second*(cv[l].second-1)/2; else ans += 1LL*cv[l].second*cv[r].second; l++; r--; } } } for(int j=i+1;j<M;j++){ int u = small[j]; long long P2 = 1LL*v*u; if(P2> L) break; auto &cu = comp[u]; int l=0, r=cu.size()-1; while(l<cv.size() && r>=0){ long long s=cv[l].first+cu[r].first; if(s<P2) l++; else if(s>P2) r--; else{ ans += 1LL*cv[l].second*cu[r].second; l++; r--; } } } } for(int v: large){ auto &gv = G[v]; for(int x: gv){ for(int u: small){ long long P = 1LL*v*u; if(P> L) break; int need = P - x; auto &cu = comp[u]; int lo=0, hi=cu.size(); while(lo<hi){ int m=(lo+hi)/2; if(cu[m].first<need) lo=m+1; else hi=m; } if(lo<cu.size() && cu[lo].first==need) ans += cu[lo].second; } } } return (int)ans; }

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