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 78 863 名订阅者，在 技术与应用 类别中位列第 1 625，并在 印度 地区排名第 3 815 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 78 863 名订阅者。
根据 28 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 -1 626，过去 24 小时变化为 -63，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 2.72%。内容发布后 24 小时内通常能获得 0.61% 的反应，占订阅者总量。
- 帖子覆盖： 每篇帖子平均可获得 2 148 次浏览，首日通常累积 478 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 29 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。
78 863
订阅者
-6324 小时
-3517 天
-1 62630 天
帖子存档
78 863
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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)
Accenture
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n=int(input())
temp=list(map(int,list(bin(n)[2:])))
print(sum(temp))
Python3
78 863
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