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 85 759 名订阅者,在 技术与应用 类别中位列第 1 509,并在 印度 地区排名第 3 538 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 85 759 名订阅者。
根据 18 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 -1 515,过去 24 小时变化为 -32,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 3.54%。内容发布后 24 小时内通常能获得 0.74% 的反应,占订阅者总量。
- 帖子覆盖: 每篇帖子平均可获得 3 034 次浏览,首日通常累积 637 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 19 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。
85 759
订阅者
-3224 小时
-3877 天
-1 51530 天
帖子存档
85 744
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85 744
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85 744
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85 744
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using System;
using System.Collections.Generic;
class Program
{
static void Main()
{
int N = int.Parse(Console.ReadLine());
var graph = new Dictionary>();
var indegrees = new Dictionary();
for (int i = 0; i < N; i++)
{
var edge = Console.ReadLine().Split();
string from = edge[0];
string to = edge[1];
if (!graph.ContainsKey(from))
graph[from] = new List();
graph[from].Add(to);
if (!indegrees.ContainsKey(from))
indegrees[from] = 0;
if (!indegrees.ContainsKey(to))
indegrees[to] = 0;
indegrees[to]++;
}
var words = Console.ReadLine().Split();
var levels = new Dictionary();
var queue = new Queue();
foreach (var node in indegrees.Keys)
{
if (indegrees[node] == 0)
{
levels[node] = 1; // Root level is 1
queue.Enqueue(node);
}
}
while (queue.Count > 0)
{
var current = queue.Dequeue();
int currentLevel = levels[current];
if (graph.ContainsKey(current))
{
foreach (var neighbor in graph[current])
{
if (!levels.ContainsKey(neighbor))
{
levels[neighbor] = currentLevel + 1;
queue.Enqueue(neighbor);
}
indegrees[neighbor]--;
if (indegrees[neighbor] == 0 && !levels.ContainsKey(neighbor))
{
queue.Enqueue(neighbor);
}
}
}
}
int totalValue = 0;
foreach (var word in words)
{
if (levels.ContainsKey(word))
{
totalValue += levels[word];
}
else
{
totalValue += -1;
}
}
Console.Write(totalValue);
}
}
String Puzzle
C+
85 744
int main() {
string s;
cin >> s;
int n = s.length(), res = 0;
vector v(n);
for (int i = 0; i < n; ++i) cin >> v[i];
int lw = s[0] - '0', lwv = v[0];
for (int i = 1; i < n; ++i) {
if (s[i] - '0' == lw) {
res += min(lwv, v[i]);
lwv = max(lwv, v[i]);
} else {
lw = s[i] - '0';
lwv = v[i];
}
}
cout << res;
return 0;
}
Form alternating string
Change accordingly before submitting to avoid plagiarism
85 744
#include<stdio.h>
int main()
{
int n;
scanf("%d",&n);
int arr[n];
for(int i=0;i<n;i++)
{
scanf("%d",&arr[i]);
}
int count=0;
int num=arr[0];
for(int i=1;i<n;i++)
{
if(num!=arr[i])
count++;
}
printf("%d",count);
}
Count press
85 744
int main() {
string s;
cin >> s;
int n = s.length(), res = 0;
vector v(n);
for (int i = 0; i < n; ++i) cin >> v[i];
int lw = s[0] - '0', lwv = v[0];
for (int i = 1; i < n; ++i) {
if (s[i] - '0' == lw) {
res += min(lwv, v[i]);
lwv = max(lwv, v[i]);
} else {
lw = s[i] - '0';
lwv = v[i];
}
}
cout << res;
return 0;
}
85 744
#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <cmath>
#include <algorithm>
using namespace std;
struct Line {
int x1, y1, x2, y2;
};
int countCells(Line line, pair<int, int> star, bool split) {
if (line.x1 == line.x2) {
if (split) {
return min(abs(star.second - line.y1), abs(star.second - line.y2)) + 1;
}
else {
return abs(line.y1 - line.y2) + 1;
}
}
else {
if (split) {
return min(abs(star.first - line.x1), abs(star.first - line.x2)) + 1;
}
else {
return abs(line.x1 - line.x2) + 1;
}
}
}
bool intersects(Line a, Line b, pair<int, int>& intersection) {
if (a.x1 == a.x2 && b.y1 == b.y2) {
if (b.x1 <= a.x1 && a.x1 <= b.x2 && a.y1 <= b.y1 && b.y1 <= a.y2) {
intersection = {a.x1, b.y1};
return true;
}
}
if (a.y1 == a.y2 && b.x1 == b.x2) {
if (a.x1 <= b.x1 && b.x1 <= a.x2 && b.y1 <= a.y1 && a.y1 <= b.y2) {
intersection = {b.x1, a.y1};
return true;
}
}
return false;
}
int main() {
int N, K;
cin >> N;
vector<Line> lines(N);
for (int i = 0; i < N; ++i) {
cin >> lines[i].x1 >> lines[i].y1 >> lines[i].x2 >> lines[i].y2;
if (lines[i].x1 > lines[i].x2 || (lines[i].x1 == lines[i].x2 && lines[i].y1 > lines[i].y2)) {
swap(lines[i].x1, lines[i].x2);
swap(lines[i].y1, lines[i].y2);
}
}
cin >> K;
map<pair<int, int>, vector<Line>> stars;
for (int i = 0; i < N; ++i) {
for (int j = i + 1; j < N; ++j) {
pair<int, int> intersection;
if (intersects(lines[i], lines[j], intersection)) {
stars[intersection].push_back(lines[i]);
stars[intersection].push_back(lines[j]);
}
}
}
int asiylam = 0;
for (auto& star : stars) {
if (star.second.size() / 2 == K) {
vector<int> intensities;
for (auto& line : star.second) {
intensities.push_back(countCells(line, star.first, true));
}
asiylam += *min_element(intensities.begin(), intensities.end());
}
}
cout << asiylam << endl;
return 0;
}
Magic Star Intensity Code
C++
TCS Exam
85 744
int main() {
int n, m, k, d = 1, rv = 0;
cin >> n >> m;
vector> f(n);
for (int i = 0, u, v; i < m; ++i) {
cin >> u >> v;
f[u].insert(v);
f[v].insert(u);
}
cin >> k;
vector w(n, true);
rv = n;
while (rv < k) {
vector nw(n, false);
for (int i = 0; i < n; ++i) {
int cnt = 0;
for (int fr : f[i]) cnt += w[fr];
if (w[i] && cnt == 3) nw[i] = true;
else if (!w[i] && cnt < 3) nw[i] = true;
}
w = nw;
rv += count(w.begin(), w.end(), true);
++d;
}
cout << d;
return 0;
}
Office Rostering
C+
TCS exam
85 744
#include<stdio.h>
int main()
{
int n;
scanf("%d",&n);
int arr[n];
for(int i=0;i<n;i++)
{
scanf("%d",&arr[i]);
}
int count=0;
int num=arr[0];
for(int i=1;i<n;i++)
{
if(num!=arr[i])
count++;
}
printf("%d",count);
}
C Language
TCS 1st Qsn
---------------------------------------------------------
N=int(input())
K=int(input())
price=list(map(int,input().split()))
vol=list(map(int,input().split()))
maxvol=0
volu=0
maxvol=max(vol)
for i in range(0,N):
if (maxvol==vol[i] and price[i]<=K):
K=K-price[i]
volu=maxvol
for i in range(0,N):
for j in range(i+1,N+1):
if (price[i]<=K and price[i]==price[j]):
if (vol[i]>vol[j]):
volu=volu+vol[i]
K=K-price[i]
else:
volu=volu+vol[j]
K=K-price[j]
elif (price[i]<=K and price[i]!=price[j]):
K=K-price[i]
-------
include<stdio.h>
int main()
{
int n;
scanf("%d",&n);
int arr[n];
for(int i=0;i<n;i++)
{
scanf("%d",&arr[i]);
}
int count=0;
int num=arr[0];
for(int i=1;i<n;i++)
{
if(num!=arr[i])
count++;
}
printf("%d",count);
}
Array Code in C language
85 744
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85 744
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85 744
MIUM RESOURCES
➙Data science
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➙Artificial Intelligence .
➙AWS Certified .
➙Cloud
➙BIG DATA
➙Data Analytics
➙BI
➙Google Cloud Platform
➙IT Training
➙MBA
➙Machine Learning
➙Deep Learning
➙Ethical Hacking
➙SPSS
➙Statistics
➙Data Base
➙Learning language resources ( English , French , German )
All courses at 30 rupees
Contact:- @meterials_available
85 744
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现已上线!2025 年 Telegram 研究 — 年度关键洞察 
