现已上线！2025 年 Telegram 研究 — 年度关键洞察 
ACCENTURE EXAM HELP ! CISCO EXAM !
前往频道在 Telegram
🔥Guys plz Stop fearing for daily exams 📝 👨💻 @srksvk is here to help you all at lowest cost possible.💪 🌀 ” Our Only Aim Is To Let Get Placed To You In A Reputed Company 🔥Effort from our side = 💯 📱Main Channel: @coding_are 📱Tel I'd : @srksvk
显示更多📈 Telegram 频道 ACCENTURE EXAM HELP ! CISCO EXAM ! 的分析概览
频道 ACCENTURE EXAM HELP ! CISCO EXAM ! (@coding_are) 英语 语言赛道中的 是活跃参与者。目前社区聚集了 13 208 名订阅者，在 教育 类别中位列第 15 294，并在 印度 地区排名第 31 490 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 13 208 名订阅者。
根据 30 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 -144，过去 24 小时变化为 6，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 3.08%。内容发布后 24 小时内通常能获得 1.29% 的反应，占订阅者总量。
- 帖子覆盖： 每篇帖子平均可获得 407 次浏览，首日通常累积 170 次浏览。
- 互动与反馈： 受众积极参与，单帖平均反应数为 2。
- 主题关注点： 内容集中在 placement, gaurntee, suree, capgemini, infosy 等核心主题上。
📝 描述与内容策略
作者将该频道定位为表达主观观点的平台：
“🔥Guys plz Stop fearing for daily exams 📝
👨💻 @srksvk is here to help you all at lowest cost possible.💪
🌀 ” Our Only Aim Is To Let Get Placed To You In A Reputed Company
🔥Effort from our side = 💯
📱Main Channel: @coding_are
📱Tel I'd : @srks...”
凭借高频更新（最新数据采集于 01 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 教育 类别中的关键影响点。
13 208
订阅者
+624 小时
-127 天
-14430 天
帖子存档
For BNY mellon answers join the group
Join ..👇👇👇👇
https://t.me/ibmsolution
+1IBM exam successfully done by laptop access ✅✅✅✅
Both code fully passed with all test case passed ✅✅
Contact for placement exams @srksvk
IBM Hiring
Role :- Application Developer
Location - Banglore
Full time
You will get the Exam link in 1-2 Hours.
Apply Link
https://careers.ibm.com/job/19372309/application-developer-content-courseware-design-bangalore-in/
Apply soon link may be expire anytime.
U got Exam mail also Immediately
Those who want clear IBM exam so contact fast and book your slotts
Contact @srksvk
+2Lumens exam successfully done by laptop access bro ✅✅✅✅✅✅✅
All critical mcq are done with 100% correct answer ✅✅
Contact for placement exams @srksvk
Virtusa and ibm for answer solution.
Join ..
https://t.me/ibmsolution
Virtusa Exam
All Codes available
Help available
Contact @VickySir12
+1IBM exam successfully done by remote access ✅✅✅
Both code fully paased with all tets case passed ✅✅
Contact for exams @srksvk
E Litmus. , Virtusa Exam , ibm exam
Help available
Book your slot
Limited Slots Available.
Contact @srksvk
100% Clearance guarantee.and genuine help
Remote access available ✅
+1IBM exam successfully done by remote access ✅✅✅
Both code fully paased with all tets case passed ✅✅
Contact for exams @srksvk
+5Amazon sde exam successfully done by remote access ✅✅✅
Bith code fully passed with all tets case passed ✅✅
All MCQ 💯% correct ✅✅✅
Contact for placement exam @srksvk
E Litmus. , Virtusa Exam , ibm exam
Help available
Book your slot
Limited Slots Available.
Contact @srksvk
100% Clearance guarantee.and genuine help
Remote access available ✅
Hello Everyone Welcome Guys
We tried our best to help you in free in TCS CODEVITAA 💯
Mostly code successfully ✅✅💯💯
Due to lot of rush we are unable to provide personal support to each candidate
But we posted codes here to help you.
So support Us
And we will help u in any placement drive
Thank You.
Guys Submit your exam
Only 9 mins left
Otherwise at end moment server issue you will face
Shared everyonee fasttt
I will uploded all code ...full working
Shared ✅✅✅✅
----- PICK UP SERVICE -----
from collections import defaultdict
def pick_up_service(N, start, connections):
graph = defaultdict(list)
taxes = defaultdict(int)
for i in range(N - 1):
city1, city2, goods, tax = connections[i]
# graph[city1].update({city2: (goods, tax)})
# graph[city2].update({city1: (goods, tax)})
graph[city1].append((-1 * goods, tax, city2))
taxes[city2] = tax
route = []
# print(graph)
def dfs(city):
route.append(city)
for n in sorted(graph[city]):
dfs(n[2])
route.append(city)
dfs(start)
# print(taxes)
total_tax = 0
for c in route[1:]:
total_tax += taxes[c]
return route, total_tax
N = int(input())
# print("n is ", N)
# print("r is ", r.split('\n'))
cons = []
for _ in range(N-1):
l = input()
ls = l.split()
cons.append((ls[0], ls[1], int(ls[2]), int(ls[3])))
ans, t = pick_up_service(N, cons[0][0], cons)
print("-".join(ans))
print(t, end="")
Fully working ✅😁😁😁😁
Passes ✅✅✅✅
Share @codeing_area
