现已上线！2025 年 Telegram 研究 — 年度关键洞察 
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
85 523
![int main() { ll N; cin >> N; vector w(N); for(ll i = 0; i < N; ++i) { cin >> w[i]; } vector s(N, 15); for(ll i](https://cdn.tlmtr.io/_next/static/media/empty-channel.35ed232f.svg)
int main() {
ll N;
cin >> N;
vector w(N);
for(ll i = 0; i < N; ++i) {
cin >> w[i];
}
vector s(N, 15);
for(ll i = N-2; i >= 0; --i) {
for(ll j = i+1; j < N; ++j) {
if(w[i] < w[j]) {
s[i] = 10;
for(ll k = j+1; k < N; ++k) {
if(w[j] > w[k]) {
s[i] = 5;
break;
}
}
break;
}
}
}
ll g = 0;
for(ll sc : s) {
g += sc;
}
cout << g << endl;
return 0;
}
chair manufacture
85 523
class TreeNode:
def init(self, val):
self.val = val
self.children = []
def build_tree(units, relationships):
nodes = {i: TreeNode(units[i]) for i in range(len(units))}
for rel in relationships:
parent, child = rel
nodes[parent].children.append(nodes[child])
return nodes[0]
def max_electricity_per_floor(root):
max_electricity = 0
floor_electricity = {}
def dfs(node, level):
nonlocal max_electricity
if level not in floor_electricity:
floor_electricity[level] = 0
floor_electricity[level] += node.val
max_electricity = max(max_electricity, floor_electricity[level])
for child in node.children:
dfs(child, level + 1)
dfs(root, 0)
return max_electricity
num = int(input())
units = list(map(int, input().split()))
numRel, memConnect = map(int, input().split())
relationships = [tuple(map(int, input().split())) for _ in range(numRel)]
root = build_tree(units, relationships)
result = max_electricity_per_floor(root)
print(result)
Apartment one
85 523
X(1), (2)
int count;
// Line 1
Antint i) { count 1)
int getcount () {return count)
// Line 5
public class Book
public static void main(String[] args)
System.out.println(AB.X.count + AB.Y.count); // Line 10
Ans)Compile time error at "// Line 10"
@allcoding1_official
85 523
Which of the statement(s) given below is/are correct?
1)Routers can be static or dynamic
2)Routers connect two or more separate networks,
Ans) both 1 and 2
@allcoding1_official
85 523
37
38
In MS Word, how can a document be split into multiple parts such that each part can have different formats and layouts
Ans)By creating sections
@allcoding1_official
85 523
How many pairs of letters are there in the word "DYNASTY" which has as many letters between them (from both the sides) as in the alphabets from A to Z?
A) 2
@allcoding1_official
85 523
Mostly, it is students from English mediun schools who appear in this competition
