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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

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:-