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allcoding1_official (@allcoding1_official) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 78 961 obunachidan iborat bo'lib, Texnologiyalar & Aralashmalar toifasida 1 613-o'rinni va Hindiston mintaqasida 3 828-o'rinni egallagan.
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невідомо sanasidan buyon loyiha tez o‘sib, 78 961 obunachiga ega bo‘ldi.
25 Oktabr, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -1 654 ga, so‘nggi 24 soatda esa -52 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.
- Tasdiqlash holati: Tasdiqlanmagan
- Jalb etish (ER): Auditoriya o‘rtacha 1.79% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 0.62% ini tashkil etuvchi reaksiyalarni to‘playdi.
- Post qamrovi: Har bir post o‘rtacha 1 414 marta ko‘riladi; birinchi sutkada odatda 489 ta ko‘rish yig‘iladi.
- Reaksiyalar va o‘zaro ta’sir: Auditoriya faol: har bir postga o‘rtacha 0 ta reaksiya keladi.
- Tematik yo‘nalishlar: Kontent dsa, stack, namaste, javascript, dev kabi asosiy mavzularga jamlangan.
📝 Tavsif va kontent siyosati
Kanal uchun tavsif kiritilmagan.
Yuqori yangilanish chastotasi (oxirgi ma’lumot 26 Oktabr, 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.
78 961
Obunachilar
-5224 soatlar
-3677 kun
-1 65430 kun
Postlar arxiv
78 949
import math
AC
def is_nprime(x): if x <2: return False for i in range(2, int(math.sqrt(x)) + 1): if x % i == 0: return x == i*i return True
def count_nearly_primes(L, R): nPrimes = 0 for x in range(L+1, R): if is_nprime(x): nPrimes += 1 return nPrimes
L = int(input()) R = int(input()) result = count_nearly..primes(L, R) print(result)
Count of Nearly Primes
Python3
78 949
def count_arrays(N, K, X):
MOD = 1000000007
if N == 1:
return 1 if X == 1 else 0
a = [0] * (N + 1)
b = [0] * (N + 1)
a[1] = 1
b[1] = 0
for i in range(2, N + 1):
a[i] = (a[i - 1] * (K - 2) + b[i - 1] * (K - 1)) % MOD
b[i] = a[i - 1] % MOD
return b[N]
N = int(input())
K = int(input())
X = int(input())
print(count_arrays(N, K, X))
Number Of Array code
Thoughtwork exam
Python3
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All courses (100 rupees)
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Repost from allcoding1
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📌IT learning courses
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Repost from allcoding1
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All courses (100 rupees)
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#include <iostream>
#include <vector>
#include <climits>
#include <algorithm>
using namespace std;
int minimum_energy_cost_recursive(int index,int idxy, int left_consecutive, int right_consecutive, int N, int X, int Y, int El, int Er, vector<int>& weights) {
// Base case: if we have reached the end of bags
if (index > idxy) {
return 0;
}
// Calculate the cost of picking from the left and the right
int cost_left = weights[index] * X + (left_consecutive ? El : 0) +
minimum_energy_cost_recursive(index + 1,idxy, 1, 0, N, X, Y, El, Er, weights);
int cost_right = weights[idxy] * Y + (right_consecutive ? Er : 0) +
minimum_energy_cost_recursive(index ,idxy-1, 0, 1, N, X, Y, El, Er, weights);
// Return the minimum of both choices
return min(cost_left, cost_right);
}
int main() {
int N, X, Y, El, Er;
cin >> N >> X >> Y >> El >> Er;
vector<int> weights(N);
for (int i = 0; i < N; ++i) {
cin >> weights[i];
}
cout << minimum_energy_cost_recursive(0,N-1, 0, 0, N, X, Y, El, Er, weights) << endl;
return 0;
}
Amazon Hackon exam
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