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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

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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 Telegram:- @allcoding1_official

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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 Telegram:-

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