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📈 Telegram 频道 allcoding1 的分析概览

频道 allcoding1 (@allcoding1) 英语 语言赛道中的 是活跃参与者。目前社区聚集了 21 578 名订阅者,在 教育 类别中位列第 9 062,并在 印度 地区排名第 19 025

📊 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 21 578 名订阅者。

根据 29 八月, 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 -360,过去 24 小时变化为 -8,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 5.78%。内容发布后 24 小时内通常能获得 1.10% 的反应,占订阅者总量。
  • 帖子覆盖: 每篇帖子平均可获得 1 247 次浏览,首日通常累积 238 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, learning 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 30 八月, 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。

21 578
订阅者
-824 小时
-787
-36030
帖子存档
#include <iostream> #include <vector> #include <climits> #include <algorithm> using namespace std; int minimum_energy_cost_recursive(int index,int idxy, int left_consecutive, int right_consecutive, int N, int X, int Y, int El, int Er, vector<int>& weights) {     // Base case: if we have reached the end of bags     if (index > idxy) {         return 0;     }         // Calculate the cost of picking from the left and the right     int cost_left = weights[index] * X + (left_consecutive ? El : 0) +                     minimum_energy_cost_recursive(index + 1,idxy, 1, 0, N, X, Y, El, Er, weights);     int cost_right = weights[idxy] * Y + (right_consecutive ? Er : 0) +                      minimum_energy_cost_recursive(index ,idxy-1, 0, 1, N, X, Y, El, Er, weights);         // Return the minimum of both choices     return min(cost_left, cost_right); } int main() {     int N, X, Y, El, Er;     cin >> N >> X >> Y >> El >> Er;         vector<int> weights(N);     for (int i = 0; i < N; ++i) {         cin >> weights[i];     }         cout << minimum_energy_cost_recursive(0,N-1, 0, 0, N, X, Y, El, Er, weights) << endl;         return 0; } Amazon Hackon exam

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