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import java.util.Scanner; public class Main {     public static void main(String[] args) {         Scanner scanner = new Scanner(System.in);                 int n = scanner.nextInt();         int[] left = new int[n];         int[] right = new int[n];                 for (int i = 0; i < n; i++) {             left[i] = scanner.nextInt();             right[i] = scanner.nextInt();         }                 int sumLeft = 0;         int sumRight = 0;                 for (int i = 0; i < n; i++) {             sumLeft += left[i];             sumRight += right[i];         }                 if (sumLeft % 2 == 0 && sumRight % 2 == 0) {             System.out.println(0);             return;         }                 boolean possible = false;         for (int i = 0; i < n; i++) {             int newSumLeft = sumLeft - left[i] + right[i];             int newSumRight = sumRight - right[i] + left[i];             if (newSumLeft % 2 == 0 && newSumRight % 2 == 0) {                 possible = true;                 break;             }         }                 if (possible) {             System.out.println(1);         } else {             System.out.println(-1);         }     } }

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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