现已上线！2025 年 Telegram 研究 — 年度关键洞察 
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 85 483 名订阅者，在 技术与应用 类别中位列第 1 507，并在 印度 地区排名第 3 480 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 85 483 名订阅者。
根据 24 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 -1 369，过去 24 小时变化为 -35，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 2.60%。内容发布后 24 小时内通常能获得 0.55% 的反应，占订阅者总量。
- 帖子覆盖： 每篇帖子平均可获得 2 222 次浏览，首日通常累积 474 次浏览。
- 互动与反馈： 受众积极参与，单帖平均反应数为 3。
- 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 25 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。
85 483
订阅者
-3524 小时
-2687 天
-1 36930 天
帖子存档
85 467
#include <iostream>
#include <string>
#include <iomanip>
#include <vector>
#include <algorithm>
struct Client {
std::string name;
double total_invested_in_bonds;
double total_invested_in_stocks;
};
bool compareByBonds(const Client &a, const Client &b) {
return a.total_invested_in_bonds > b.total_invested_in_bonds;
}
int main() {
std::vector<client> clients = {
{"Client1", 7500.0, 3000.0},
{"Client2", 6000.0, 5000.0},
{"Client3", 4000.0, 6000.0},
// Add more clients as needed
};
// Sort clients based on total_invested_in_bonds in descending order
std::sort(clients.begin(), clients.end(), compareByBonds);
// Display the result
std::cout << std::left << std::setw(15) << "Client Name"
<< std::setw(25) << "Total Invested in Bonds"
<< "Total Invested in Stocks" << std::endl;
for (const auto &client : clients) {
if (client.total_invested_in_bonds > 5000.00) {
std::cout << std::left << std::setw(15) << client.name
<< std::fixed << std::setprecision(2)
<< std::setw(25) << client.total_invested_in_bonds
<< client.total_invested_in_stocks << std::endl;
}
}
return 0;
}
You can modify the clients vector to include more clients with their respective investments in bonds and stocks. When you run this program, it will display the result according to the specified requirements.</client></algorithm></vector></iomanip></string></iostream>
Telegram:- @allcoding1_official
85 467
To find the ratio of A to C, we can use the given ratios for AB and BC.
Given:
AB = 2:3
BC = 4:5
We want to find the ratio of A to C.
One way to solve this is to express both AB and BC in terms of a common variable and then use that to find the ratio of A to C.
Let's express AB and BC in terms of a common multiple of B.
Let's assume the common multiple is x. Then:
AB = 2x : 3x
BC = 4x : 5x
Now, to find the ratio of A to C, we can multiply the two ratios together:
A:B * B:C = A:C
So:
(2x : 3x) * (4x : 5x)
= (2x * 4x : 3x * 5x)
= (8x^2 : 15x^2)
Therefore, the ratio of (A to C is 8:15)
85 467
#include <iostream>
#include <vector>
using namespace std;
bool isValidPermutation(const vector<int>& permutation) {
for (int i = 0; i < permutation.size(); i++) {
if (permutation[i] % (i + 1) != 0 || (i + 1) % permutation[i] != 0) {
return false;
}
}
return true;
}
void generatePermutations(vector<int>& nums, int start, vector<vector<int>>& result) {
if (start == nums.size()) {
result.push_back(nums);
return;
}
for (int i = start; i < nums.size(); i++) {
swap(nums[start], nums[i]);
generatePermutations(nums, start + 1, result);
swap(nums[start], nums[i]);
}
}
int countValidPermutations(int N) {
vector<int> nums(N);
for (int i = 0; i < N; i++) {
nums[i] = i + 1;
}
vector<vector<int>> permutations;
generatePermutations(nums, 0, permutations);
int count = 0;
for (const auto& perm : permutations) {
if (isValidPermutation(perm)) {
count++;
}
}
return count;
}
int main() {
int N = 2;
cout << "Number of valid permutations: " << countValidPermutations(N) << endl;
return 0;
}85 467
int min(string str){
unordered_map<char,int>mp;
for(char :str){
mp[ch]++;
}
unodered_set<int>d;
for(auto it:mp){
d.insert(it.second);
}
return d.size();
}
Music Teacher
Only 4 test cases pass
Telegram:- @allcoding1_official
85 467
#include <iostream>
#include <cmath>
struct Circle {
double x; // x-coordinate of the center
double y; // y-coordinate of the center
double r; // radius
};
double distanceBetweenCenters(Circle A, Circle B) {
return sqrt(pow((B.x - A.x), 2) + pow((B.y - A.y), 2));
}
int main() {
Circle A, B;
// Example values
A.x = 0;
A.y = 0;
A.r = 5;
B.x = 10;
B.y = 0;
B.r = 7;
double distance = distanceBetweenCenters(A, B);
double sumOfRadii = A.r + B.r;
double differenceOfRadii = abs(A.r - B.r);
if (distance == sumOfRadii) {
std::cout << "The circles are touching at a single point." << std::endl;
} else if (distance < sumOfRadii) {
std::cout << "The circles are intersecting." << std::endl;
} else if (distance == differenceOfRadii) {
std::cout << "The circles are touching from within or without." << std::endl;
} else {
std::cout << "The circles are not intersecting." << std::endl;
}
if ((A.x == B.x) && (A.y == B.y) && (A.r == B.r)) {
std::cout << "The circles are concentric." << std::endl;
}
return 0;
}
C++
Telegram:- @allcoding1_official
85 467
#include<bits/stdc++.h>
using namespace std;
int main(){
int n,m;
cin>>n>>m;
vector<int>ans(n);
if(n<=m){
for(int i=0;i<n;i++){
ans[i]=i+1;
}
}
else{
int k=n/m;
// int r=n%m;
int j=0;
for(int i=0;i<n;i++){
ans[i]=j+1;
j++;
j%=m;
}
}
for(int i=0;i<n;i++){
cout<<ans[i]<<" ";
}
cout<<endl;
}
King Dreams ✅
Intuit
Telegram:- @meterials_available
85 467
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85 467
class Solution {
public:
int minOperations(vector& nums, int x, int y) {
int l = 0;
int r = ranges::max(nums);
while (l < r) {
const int m = (l + r) / 2;
if (isPossible(nums, x, y, m))
r = m;
else
l = m + 1;
}
return l;
}
private:
bool isPossible(const vector& nums, int x, int y, int m) {
long long additionalOps = 0;
for (const int num : nums)
additionalOps += max(0LL, (num - 1LL * y * m + x - y - 1) / (x - y));
return additionalOps <= m;
}
};
Job Execution ✅
