allcoding1
前往频道在 Telegram
📈 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 22 409 名订阅者,在 教育 类别中位列第 8 808,并在 印度 地区排名第 18 944 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 22 409 名订阅者。
根据 26 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 -372,过去 24 小时变化为 -9,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 4.76%。内容发布后 24 小时内通常能获得 1.46% 的反应,占订阅者总量。
- 帖子覆盖: 每篇帖子平均可获得 1 068 次浏览,首日通常累积 327 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, 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 28 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。
22 412
订阅者
-924 小时
-797 天
-37230 天
帖子存档
22 409
#include <iostream>
using namespace std;
int main() {
int t,n,rev=0;
cin>>n;
while(n--)
{
cin>>t;
rev=0;
int a;
a=t;
while(a!=0)
{
int r=a%10;
rev=rev*10+r;
a=a/10;
}
if(rev==t)
cout<<"wins"<<endl;
else
cout<<"losses"<<endl;
}
return 0;
}
The block game code
C++
22 409
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Batch 2018/2019/2020/2021/2022
Salary Rs.3.5 LPA
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22 409
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Batch - 2023
Degree - BE/ME/MCA/MSc
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Telegram - @allcoding1
22 409
➡️Private channel link in
🅾️Instagram available
Guys❣️ go and check
https://instagram.com/allcoding1_official?igshid=YmMyMTA2M2Y=
https://instagram.com/allcoding1_official?igshid=YmMyMTA2M2Y=
https://instagram.com/allcoding1_official?igshid=YmMyMTA2M2Y=
22 409
Private channel for Tomorrow TCS Exam Ans
Ask your 10 friends to follow Instagram And subscribe🛑 & LIKE👍 youtube channel and send screenshot to
He will add you to private channel for tomorrow Ans
YouTube channel:-
https://youtu.be/ofDvcsP3OWY
https://youtu.be/ofDvcsP3OWY
https://youtu.be/ofDvcsP3OWY
Instagram:-
https://instagram.com/allcoding1_official?igshid=YmMyMTA2M2Y=
https://instagram.com/allcoding1_official?igshid=YmMyMTA2M2Y=
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Send A screen shots in this group
https://t.me/+FQoo8kIBWww2Y2Fl
Share with your friends and college group's
22 409
def jogging_ways(n: int) -> int:
a, b = 1, 1
for i in range(n):
a, b = b, a+b
return a %10000007
Run Walk and Run
Python 3
Telegram:- @allcoding1
22 409
def santa(num,d):
if d.get(num):
return 1
rem = sum = 0;
while(num > 0):
rem = num%10;
sum = sum + (rem*rem);
num = num//10;
return sum
def gift(n):
pre = [0 for i in range(n + 1)]
d = {}
for i in range(1, n + 1):
result = i
while(result != 1 and result != 4):
result = santa(result,d);
if(result == 1):
d[i] = 1
pre[i] += i + pre[i-1]
else:
pre[i] = pre[i-1]
return pre
pre = findall(1000001)
t = int(input())
for _ in range(t):
l,r = map(int,input().split())
print(pre[r] - pre[l-1])
Python
Telegram :-- @allcoding1
22 409
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2022 Batch |11LPA
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Telegram - @allcoding1
现已上线!2025 年 Telegram 研究 — 年度关键洞察 
