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allcoding1

allcoding1

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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 22 575 名订阅者,在 教育 类别中位列第 8 822,并在 印度 地区排名第 19 518

📊 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 22 575 名订阅者。

根据 12 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 -437,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 5.99%。内容发布后 24 小时内通常能获得 1.25% 的反应,占订阅者总量。
  • 帖子覆盖: 每篇帖子平均可获得 1 353 次浏览,首日通常累积 283 次浏览。
  • 互动与反馈: 受众积极参与,单帖平均反应数为 2
  • 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 13 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。

22 575
订阅者
-624 小时
-967
-43730
帖子存档
import java.util.Scanner; public class Main {     public static void main(String[] args) {         Scanner scanner = new Scanner(System.in);                 int n = scanner.nextInt();         int[] left = new int[n];         int[] right = new int[n];                 for (int i = 0; i < n; i++) {             left[i] = scanner.nextInt();             right[i] = scanner.nextInt();         }                 int sumLeft = 0;         int sumRight = 0;                 for (int i = 0; i < n; i++) {             sumLeft += left[i];             sumRight += right[i];         }                 if (sumLeft % 2 == 0 && sumRight % 2 == 0) {             System.out.println(0);             return;         }                 boolean possible = false;         for (int i = 0; i < n; i++) {             int newSumLeft = sumLeft - left[i] + right[i];             int newSumRight = sumRight - right[i] + left[i];             if (newSumLeft % 2 == 0 && newSumRight % 2 == 0) {                 possible = true;                 break;             }         }                 if (possible) {             System.out.println(1);         } else {             System.out.println(-1);         }     } }

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Count children
Count children

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Prime partition

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def count_arrays(N, K, X):     MOD = 1000000007         if N == 1:         return 1 if X == 1 else 0     a = [0] * (N + 1)     b = [0] * (N + 1)         a[1] = 1     b[1] = 0         for i in range(2, N + 1):         a[i] = (a[i - 1] * (K - 2) + b[i - 1] * (K - 1)) % MOD         b[i] = a[i - 1] % MOD         return b[N] N = int(input()) K = int(input()) X = int(input()) print(count_arrays(N, K, X)) Number Of Array code Thoughtwork exam Python3 Telegram:- @allcoding1

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allcoding1 - Telegram 频道 @allcoding1 的统计与分析