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allcoding1_official (@allcoding1_official) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 85 483 obunachidan iborat bo'lib, Texnologiyalar & Aralashmalar toifasida 1 507-o'rinni va Hindiston mintaqasida 3 480-o'rinni egallagan.

📊 Auditoriya ko‘rsatkichlari va dinamika

невідомо sanasidan buyon loyiha tez o‘sib, 85 483 obunachiga ega bo‘ldi.

24 Iyun, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -1 369 ga, so‘nggi 24 soatda esa -35 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.

  • Tasdiqlash holati: Tasdiqlanmagan
  • Jalb etish (ER): Auditoriya o‘rtacha 2.60% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 0.55% ini tashkil etuvchi reaksiyalarni to‘playdi.
  • Post qamrovi: Har bir post o‘rtacha 2 222 marta ko‘riladi; birinchi sutkada odatda 474 ta ko‘rish yig‘iladi.
  • Reaksiyalar va o‘zaro ta’sir: Auditoriya faol: har bir postga o‘rtacha 3 ta reaksiya keladi.
  • Tematik yo‘nalishlar: Kontent dsa, stack, namaste, javascript, dev kabi asosiy mavzularga jamlangan.

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Yuqori yangilanish chastotasi (oxirgi ma’lumot 25 Iyun, 2026 da olingan) sababli kanal doimo dolzarb va katta qamrovli bo‘lib qoladi. Analitika auditoriya kontent bilan faol hamkorlik qilishini, uni Texnologiyalar & Aralashmalar toifasidagi muhim ta’sir nuqtasiga aylantirishini ko‘rsatadi.

85 483
Obunachilar
-3524 soatlar
-2687 kunlar
-1 36930 kunlar
Postlar arxiv
#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 &amp;a, const Client &amp;b) { return a.total_invested_in_bonds &gt; 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 &lt;&lt; std::left &lt;&lt; std::setw(15) &lt;&lt; "Client Name" &lt;&lt; std::setw(25) &lt;&lt; "Total Invested in Bonds" &lt;&lt; "Total Invested in Stocks" &lt;&lt; std::endl; for (const auto &amp;client : clients) { if (client.total_invested_in_bonds &gt; 5000.00) { std::cout &lt;&lt; std::left &lt;&lt; std::setw(15) &lt;&lt; client.name &lt;&lt; std::fixed &lt;&lt; std::setprecision(2) &lt;&lt; std::setw(25) &lt;&lt; client.total_invested_in_bonds &lt;&lt; client.total_invested_in_stocks &lt;&lt; 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

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Candy shop - Mitsogo
Candy shop - Mitsogo

191
191

8:15
8:15

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
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)

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+2

#include <iostream>
#include <vector>
using namespace std;

bool isValidPermutation(const vector<int>&amp; permutation) {
    for (int i = 0; i &lt; permutation.size(); i++) {
        if (permutation[i] % (i + 1) != 0 || (i + 1) % permutation[i] != 0) {
            return false;
        }
    }
    return true;
}

void generatePermutations(vector<int>&amp; nums, int start, vector<vector<int>&gt;&amp; result) {
    if (start == nums.size()) {
        result.push_back(nums);
        return;
    }
    for (int i = start; i &lt; 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 &lt; N; i++) {
        nums[i] = i + 1;
    }
    vector<vector<int>&gt; permutations;
    generatePermutations(nums, 0, permutations);

    int count = 0;
    for (const auto&amp; perm : permutations) {
        if (isValidPermutation(perm)) {
            count++;
        }
    }
    return count;
}

int main() {
    int N = 2;
    cout &lt;&lt; "Number of valid permutations: " &lt;&lt; countValidPermutations(N) &lt;&lt; endl;
    return 0;
}

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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

#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 &lt;&lt; "The circles are touching at a single point." &lt;&lt; std::endl; } else if (distance &lt; sumOfRadii) { std::cout &lt;&lt; "The circles are intersecting." &lt;&lt; std::endl; } else if (distance == differenceOfRadii) { std::cout &lt;&lt; "The circles are touching from within or without." &lt;&lt; std::endl; } else { std::cout &lt;&lt; "The circles are not intersecting." &lt;&lt; std::endl; } if ((A.x == B.x) &amp;&amp; (A.y == B.y) &amp;&amp; (A.r == B.r)) { std::cout &lt;&lt; "The circles are concentric." &lt;&lt; std::endl; } return 0; } C++ Telegram:- @allcoding1_official

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Special String
+1
Special String

Solve the Equation Telegram:-
+1
Solve the Equation Telegram:-

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+1
➡️ Deal Price : ₹280 Buy Here :-

#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

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class Solution { public: &nbsp; int minOperations(vector&amp; nums, int x, int y) { &nbsp;&nbsp;&nbsp; int l = 0; &nbsp;&nbsp
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 ✅

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