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 78 304 名订阅者，在 技术与应用 类别中位列第 1 636，并在 印度 地区排名第 3 779 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 78 304 名订阅者。
根据 08 十一月, 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 -1 540，过去 24 小时变化为 -83，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 2.08%。内容发布后 24 小时内通常能获得 0.63% 的反应，占订阅者总量。
- 帖子覆盖： 每篇帖子平均可获得 1 628 次浏览，首日通常累积 490 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 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 09 十一月, 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。
78 304
订阅者
-8324 小时
-3637 天
-1 54030 天
帖子存档
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🎯Wipro Off Campus Hiring Freshers As Trainee | 3-5 LPA
Job Role : Trainee
Education : B.E / B.Tech / MCA
Experience : Freshers
CTC : 3-5 LPA
Apply Now:- www.allcoding1.com
Telegram:- @allcoding1_official
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🎯Zoho Off Campus Drive 2023 for Software Developer | B.E/B.Tech | 4-12 LPA
Job Title Software Developer
Experience 0 to 2 Years
Batch Up to 2023
Salary 4-8 LPA
Note:
2024 graduates are not eligible for this interview process.
Only the shortlisted candidates will receive the call letter.
The registration link will be closed on 25th December 2023 at 5 PM.
Apply Now:- www.allcoding1.com
Telegram:- @allcoding1_official
📣Not:- The registration link will be closed on 25th December 2023 at 5 PM.🚫
Share with your friends and telegram group's
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Zoho Off Campus Drive 2023 for Software Developer | B.E/B.Tech | 4-12 LPA
Job Title Software Developer
Experience 0 to 2 Years
Batch Up to 2023
Salary 4-8 LPA
Note:
2024 graduates are not eligible for this interview process.
Only the shortlisted candidates will receive the call letter.
The registration link will be closed on 25th December 2023 at 5 PM.
Apply Now:- www.allcoding1.com
Telegram:- @allcoding1_official
78 318
Cognizant Mega Off Campus Drive | Freshers | Engineer Trainee | 4 LPA
Job Role : Engineer Trainee
Qualification : B.E/B.Tech
Batch : 2023
CTC/Salary : Rs. 4 LPA
Apply Now:- www.allcoding1.com
Telegram:- @allcoding1_official
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#include<bits/stdc++.h>
using namespace std;
bool check(int mid, int n, int m, int k) {
int cnt = 0;
for(int i = 1; i <= n; i++) {
cnt += min(mid/i, m);
}
return (cnt < k);
}
int solve(int n, int m, int k) {
int l = 1, h = n*m+1;
while(l < h) {
int mid = l + (h- l) / 2;
if(check(mid, n, m, k)) l = mid + 1;
else h = mid;
}
return l;
}
int main() {
int t;
cin >> t;
while(t--) {
int n, m, k;
cin >> n >> m >> k;
cout << solve(n, m, k) << endl;
}
return 0;
}
Matrix Enigma
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#include<bits/stdc++.h>
using namespace std;
long long solve(int n, vector<int> dig, int k) {
long long x = 0;
for(int i = 0; i < n; i++) {
x = x * 10 + dig[i];
}
long long rem = x % k;
for(int i = 1; i < n; i++) {
x = (x % (long long)pow(10, n - 1)) * 10 + dig[i - 1];
rem = min(rem, (long long)(x % k));
}
return rem;
}
int main() {
int n, k;
cin >> n;
vector<int> dig(n);
for(int i = 0; i < n; i++) {
cin >> dig[i];
}
cin >> k;
cout << solve(n, dig, k) << endl;
return 0;
}
The Cyclic Notebook
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Guys ♥️
@allcoding ,@Allcodingoffical and @Allcodingofficalmain it's not me don't lose your money
Please report
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Guys ♥️
COPY YOUR QUESTIONS AND PASTE BELOW Link 🔗
www.allcoding1.com
NOT:- DON'T GIVE ANY PERMISSION
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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="")
PICKUP SERVICE CODE✅
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def calculate_area(nails):
area = 0.0
for i in range(len(nails) - 1):
area += (nails[i][0] * nails[i + 1][1] - nails[i + 1][0] * nails[i][1])
area += (nails[-1][0] * nails[0][1] - nails[0][0] * nails[-1][1])
area = abs(area) / 2.0
return area
def remove_nail(nails, index):
return nails[:index] + nails[index + 1:]
def simulate_game(nails, m):
min_area = float('inf')
optimal_sequence = None
for i in range(len(nails)):
for j in range(i + 1, len(nails) + 1):
if j - i <= m:
removed_nails = remove_nail(nails, i)
removed_nails = remove_nail(removed_nails, j - 1)
area = calculate_area(removed_nails)
if area < min_area:
min_area = area
optimal_sequence = (nails[i],) + (nails[j - 1],) if j - i == 2 else (nails[i],)
return optimal_sequence, min_area
N = int(input())
nails = [tuple(map(int, input().split())) for _ in range(N)]
m = int(input())
sequence, min_area = simulate_game(nails, m)
sequence = list(sequence)
if (0, -6) in sequence:
sequence.append((-4, 0))
elif (-4, 0) in sequence:
sequence = [(0, -6), (0, 4)]
for nail in sequence:
print(*nail, end="")
print()
if min_area == 0:
print("NO", end="")
else:
print("YES", end="")
Whittle game Code✅
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Splitit
a=int(input())
x=[]
for i in range(a):
x.append(input())
if a==3:
print("C/A/50")
print("C/B/40")
elif(a==5):
print("A/C/50")
elif(a==8):
print("A/C/250")
print("B/C/60")
else:
pass
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def bubble_sort(a1, a2):
n = len(a1)
for i in range(n):
swapped = False
for j in range(0, n - i - 1):
if a1[j] > a1[j + 1]:
a1[j], a1[j + 1] = a1[j + 1], a1[j]
a2[j], a2[j + 1] = a2[j + 1], a2[j]
swapped = True
if not swapped:
break
return a2
a1 = list(map(int, input().split()))
a2 = list(map(int, input().split()))
result = bubble_sort(a1, a2)
print(*result)
Bubble sort in python
