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ACCENTURE EXAM SOLUTIONS

ACCENTURE EXAM SOLUTIONS

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ACCENTURE EXAM SOLUTIONS (@coding_are) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 14 136 obunachidan iborat bo'lib, Taʼlim toifasida 14 103-o'rinni va Hindiston mintaqasida 28 083-o'rinni egallagan.

📊 Auditoriya ko‘rsatkichlari va dinamika

невідомо sanasidan buyon loyiha tez o‘sib, 14 136 obunachiga ega bo‘ldi.

22 Sentabr, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -130 ga, so‘nggi 24 soatda esa 2 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.

  • Tasdiqlash holati: Tasdiqlanmagan
  • Jalb etish (ER): Auditoriya o‘rtacha 4.14% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 1.53% ini tashkil etuvchi reaksiyalarni to‘playdi.
  • Post qamrovi: Har bir post o‘rtacha 586 marta ko‘riladi; birinchi sutkada odatda 217 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 Sentabr, 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.

14 136
Obunachilar
+224 soatlar
-337 kun
-13030 kun
Postlar arxiv
import sys def solve(N, K, A): n = N limit = 1 << n valid = [False] * limit for mask in range(limit): s = 0 for i in range(n): if (mask >> i) & 1: s += A[i] if s <= K: valid[mask] = True dp = [n+1] * limit dp[0] = 0 full = limit - 1 for mask in range(1, limit): sub = mask while sub: if valid[sub]: other = mask ^ sub if dp[other] + 1 < dp[mask]: dp[mask] = dp[other] + 1 sub = (sub - 1) & mask return dp[full] * K def main(): data = sys.stdin.read().split() it = iter(data) N = int(next(it)); K = int(next(it)) A = [int(next(it)) for _ in range(N)] print(solve(N, K, A)) if name == "main": main()

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; }