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 85 523 名订阅者,在 技术与应用 类别中位列第 1 507,并在 印度 地区排名第 3 480 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 85 523 名订阅者。
根据 24 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 -1 369,过去 24 小时变化为 -35,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 2.60%。内容发布后 24 小时内通常能获得 0.55% 的反应,占订阅者总量。
- 帖子覆盖: 每篇帖子平均可获得 2 222 次浏览,首日通常累积 474 次浏览。
- 互动与反馈: 受众积极参与,单帖平均反应数为 3。
- 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 25 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 技术与应用 类别中的关键影响点。
85 523
订阅者
-3524 小时
-2687 天
-1 36930 天
帖子存档
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def find_different_evenness_index(n, numbers):
odd_count = even_count = 0
odd_index = even_index = 0
for i, num in enumerate(numbers):
if num % 2 == 0:
even_count += 1
even_index = i + 1
else:
odd_count += 1
odd_index = i + 1
return odd_index if even_count > odd_count else even_index
n = int(input())
numbers = list(map(int, input().split()))
result = find_different_evenness_index(n, numbers)
print(result)
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n=int(input())
temp=list(map(int,list(bin(n)[2:])))
print(sum(temp))
Python3
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