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不要被骗子欺骗!Telemetrio 会找到并标记这些频道 👉 如果想查看标记,请订阅

allcoding1_official

allcoding1_official

前往频道在 Telegram

📈 Telegram 频道 allcoding1_official 的分析概览

频道 allcoding1_official (@allcoding1_official) 英语 语言赛道中的 是活跃参与者。目前社区聚集了 79 517 名订阅者,在 技术与应用 类别中位列第 1 587,并在 印度 地区排名第 3 849 位。

📊 受众指标与增长动态

自 невідомо 创建以来,项目保持高速增长,吸引了 79 517 名订阅者。

根据 16 十月, 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 -1 729,过去 24 小时变化为 -20,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 0.77%。内容发布后 24 小时内通常能获得 0.65% 的反应,占订阅者总量。
  • 帖子覆盖: 每篇帖子平均可获得 616 次浏览,首日通常累积 517 次浏览。
  • 互动与反馈: 受众积极参与,单帖平均反应数为 0。
  • 主题关注点: 内容集中在 dsa, stack, namaste, javascript, dev 等核心主题上。

📝 描述与内容策略

尚未提供频道描述。

凭借高频更新(最新数据采集于 17 十月, 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 技术与应用 类别中的关键影响点。

79 517
订阅者
-2024 小时
-3397 天
-1 72930 天
帖子存档
Company: Accentur Job Description Skill required: Membership - Life Sciences Regulatory Operations Designation: Health Operat
Company: Accentur Job Description Skill required: Membership - Life Sciences Regulatory Operations Designation: Health Operations New Associate Qualifications:Any Graduation Years of Experience:0 to 1 years Apply Now: https://www.accenture.com/in-en/careers/jobdetails?id=AIOC-S01645412_en

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Infosys Exam codes

def solve(S: str) -> int: def cost(ch): pos=[i for c in enumerate(S) if c==ch] k=len(pos) if k<=1: return 0 q=[pos[i]-i for i in range(k)] m=q[k//2] return sum(abs(x-m) for x in q) return min(cost('0'),cost('1'))

import sys input = sys.stdin.readline def solve(N: int, V: list) -> int: if N == 1: return V[0] A = [[0] * N for _ in range(N)] B = [[0] * N for _ in range(N)] for gap in range(1, N): for l in range(N - gap): r = l + gap best = -10**30 for step in (1, 2): nl = l + step if nl >= r: best = max(best, 0) else: best = max(best, V[nl] + B[nl][r]) A[l][r] = best worst = 10**30 for step in (1, 2): nr = r - step if nr <= l: worst = min(worst, 0) else: worst = min(worst, A[l][nr]) B[l][r] = worst return V[0] + A[0][N - 1] if name == "main": data = list(map(int, sys.stdin.read().split())) N = data[0] V = data[1:1 - N] print(solve(N, V))

import sys def solve(): data = list(map(int, sys.stdin.read().split())) n = data[0] W = data[1] items = data[2:] g0 = [] g1 = [] g2 = [] idx = 0 for _ in range(n): w = items[idx] v = items[idx + 1] g = items[idx + 2] idx += 3 if g == 0: g0.append((w, v)) elif g == 1: g1.append((w, v)) else: g2.append((w, v)) dp = [0] * (W + 1) for w, v in g0: for cap in range(W, w - 1, -1): dp[cap] = max(dp[cap], dp[cap - w] + v) best1 = [(0, 0)] for w, v in g1: best1.append((w, v)) best2 = [(0, 0)] for w, v in g2: best2.append((w, v)) ans = 0 for w1, v1 in best1: for w2, v2 in best2: total_w = w1 - w2 if total_w <= W: ans = max(ans, v1 + v2 + dp[W - total_w]) print(ans) if name == "main": solve()

import sys, math def solve(): data = sys.stdin.read().strip().split() n = int(data[0]) x = int(data[1]) ans = 1 prev = 1 for _ in range(2, n + 1): need = x // math.gcd(x, prev) cur = (prev // need + 1) * need ans ^= cur prev = cur print(ans) if name == "main": solve()

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PWC is Hiring Role:Specialist(Apprenticeship Model) Eligibility: MBA/PGDM (2 Years) - 2026 batch CTC: Rs. 3,60,000 per annum Application Deadline: 16 May 2026 11:00 PM Note: The candidate will be under the Apprenticeship model for 12 months and conversion to FTE will be based on the performance of the candidate Apply Link: https://app.joinsuperset.com/join/#/signup/student/jobprofiles/f00fe55c-ee4d-418d-b72f-65ef64e5cd27

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