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Tcs exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer

Tcs exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer

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📈 Telegram 频道 Tcs exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer 的分析概览

频道 Tcs exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer (@coding_are) 英语 语言赛道中的 是活跃参与者。目前社区聚集了 13 273 名订阅者,在 教育 类别中位列第 15 343,并在 印度 地区排名第 32 337

📊 受众指标与增长动态

невідомо 创建以来,项目保持高速增长,吸引了 13 273 名订阅者。

根据 13 六月, 2026 的最新数据,频道保持稳定运转。过去 30 天订阅人数变化为 114,过去 24 小时变化为 -6,整体触达仍然可观。

  • 认证状态: 未认证
  • 互动率 (ER): 平均受众互动率为 2.87%。内容发布后 24 小时内通常能获得 1.19% 的反应,占订阅者总量。
  • 帖子覆盖: 每篇帖子平均可获得 381 次浏览,首日通常累积 158 次浏览。
  • 互动与反馈: 受众积极参与,单帖平均反应数为 1
  • 主题关注点: 内容集中在 placement, gaurntee, suree, capgemini, infosy 等核心主题上。

📝 描述与内容策略

作者将该频道定位为表达主观观点的平台:
🔥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...

凭借高频更新(最新数据采集于 14 六月, 2026),频道始终保持新鲜度与高覆盖。分析显示受众积极互动,使其成为 教育 类别中的关键影响点。

13 273
订阅者
-624 小时
-347
+11430
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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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Taimur and Shadiur Code : ALL PASSED ✅ MOD = 10**9+7 def get_ans(N, Tpos, Shpos): a0 = min(Tpos, Shpos) - 1 b0 = N - max(Tpos, Shpos) dp = [[0]*(b0+1) for _ in range(a0+1)] dp2 = [[0]*(b0+1) for _ in range(a0+1)] dp[a0][b0] = 1 for _ in range(N): for b in range(b0+1): run = 0 for a in range(a0, -1, -1): run = (run + dp[a][b]) % MOD dp2[a][b] = run for a in range(a0+1): run = 0 for b in range(b0, -1, -1): run = (run + dp[a][b]) % MOD dp2[a][b] = (dp2[a][b] + run) % MOD for a in range(a0+1): for b in range(b0+1): dp2[a][b] = (dp2[a][b] - dp[a][b]) % MOD dp, dp2 = dp2, [[0]*(b0+1) for _ in range(a0+1)] return sum(sum(row) for row in dp) % MOD N = int(input()) Tpos = int(input()) Shpos = int(input()) print(get_ans(N, Tpos, Shpos)) *

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