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allcoding1_official

allcoding1_official

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📈 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 84 444 名订阅者,在 技术与应用 类别中位列第 1 495,并在 印度 地区排名第 3 530

📊 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 84 444 名订阅者。

根据 12 七月, 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 -1 602,过去 24 小时变化为 -68,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 2.00%。内容发布后 24 小时内通常能获得 0.84% 的反应,占订阅者总量。
  • 帖子覆盖: 每篇帖子平均可获得 1 688 次浏览,首日通常累积 712 次浏览。
  • 互动与反馈: 受众积极参与,单帖平均反应数为 1
  • 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 13 七月, 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。

84 444
订阅者
-6824 小时
-3987
-1 60230
帖子存档
#include <bits/stdc++.h> using namespace std; int main() { int n; cin>>n; set<int> s; for(int i=0;i<=n;i++) { int x; cin>>x; s.insert(x); } cout<<s.size(); return 0; } Primes with a twist Code 🔗 Telegram: @allcoding1 🔔Unmute this channel to never miss any updates

#include <stdio.h> int nth_ap(int a, int d, int n) { // using formula to find the // Nth term t(n) = a(1) + (n-1)*d return (a + (n - 1) * d); } //main function int main() { // starting number int a = 2; // Common difference int d = 1; // N th term to be find int n = 5; printf("The %dth term of AP :%d\n", n, nth_ap(a,d,n)); return 0; } arithmetic progression first term C language

#include <iostream> #include <string> using namespace std; int main() { string str,str1,str2; cin>>str; int a,b; for(int i=0;i<str.size();i++) { if(str[i]=='+') str1=str.substr(0, i); if(str[i]=='=') str2=str.substr(i+1,str.size()-1); } a=std::atoi(str1.c_str()); b=std::atoi(str2.c_str()); cout<<b-a; return 0; } Arithmetic Code TCS exam Ans 🔗 Telegram: @allcoding1 🔔Unmute this channel to never miss any updates

🎯Company name: Apple Inc Job Position: Security Engineer Location: Hyderabad Job Type: Full time Experience: Freshers Qualification: BE/ B.Tech/ ME/ M.Tech Batch: 2017/ 2018/ 2019/ 2020/ 2021 Salary: Min 10 LPA (Expected) Apply:- https://www.joboffersadda.com/2022/04/apple-recruitment-be-btech-me-mtech.html 🔗 Telegram: @allcoding1 🔔Unmute this channel to never miss any updates

3 TCS exam Ans 🔗 Telegram: @allcoding1 🔔Unmute this channel to never miss any updates
3 TCS exam Ans 🔗 Telegram: @allcoding1 🔔Unmute this channel to never miss any updates

1 TCS exam Ans 🔗 Telegram: @allcoding1 🔔Unmute this channel to never miss any updates
1 TCS exam Ans 🔗 Telegram: @allcoding1 🔔Unmute this channel to never miss any updates

25 TCS exam Ans 🔗 Telegram: @allcoding1 🔔Unmute this channel to never miss any updates
25 TCS exam Ans 🔗 Telegram: @allcoding1 🔔Unmute this channel to never miss any updates

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n=int(input()) a=list(map(int,input().split())) e=[i for i in a if i%2==0 ] b=[i for i in a if i%2!=0 ] e.extend(b) print(*e) Play online game Python Accenture exam Ans 🔗 Telegram: @allcoding1 🔔Unmute this channel to never miss any updates

A = input() B = input() if A == B: examcell print(len(A)*2) else: print(0) GUESS-ME Python Accenture exam Ans 🔗 Telegram: @allcoding1 🔔Unmute this channel to never miss any updates

B TCS exam Ans 🔗 Telegram: @allcoding1 🔔Unmute this channel to never miss any updates
B TCS exam Ans 🔗 Telegram: @allcoding1 🔔Unmute this channel to never miss any updates

C TCS exam Ans 🔗 Telegram: @allcoding1 🔔Unmute this channel to never miss any updates
C TCS exam Ans 🔗 Telegram: @allcoding1 🔔Unmute this channel to never miss any updates

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C TCS exam Ans 🔗 Telegram: @allcoding1 🔔Unmute this channel to never miss any updates
C TCS exam Ans 🔗 Telegram: @allcoding1 🔔Unmute this channel to never miss any updates

B TCS exam Ans 🔗 Telegram: @allcoding1 🔔Unmute this channel to never miss any updates
B TCS exam Ans 🔗 Telegram: @allcoding1 🔔Unmute this channel to never miss any updates

B TCS exam Ans 🔗 Telegram: @allcoding1 🔔Unmute this channel to never miss any updates
B TCS exam Ans 🔗 Telegram: @allcoding1 🔔Unmute this channel to never miss any updates

RP TCS exam Ans 🔗 Telegram: @allcoding1 🔔Unmute this channel to never miss any updates
RP TCS exam Ans 🔗 Telegram: @allcoding1 🔔Unmute this channel to never miss any updates

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