📜  数组中所有素数在K可整除位置上的XOR

📅  最后修改于: 2021-05-04 23:34:09             🧑  作者: Mango

给定大小为N的整数和整数K的数组arr ,任务是查找所有素数且在K可整除的位置上的所有数字的XOR

例子:

天真的方法:遍历数组,对于每个可被k整除的位置i ,检查其是否为素数。如果是质数,则计算该数字与先前答案的XOR。

高效方法:一种有效方法是使用Eratosthenes筛。筛子阵列可用于检查O(1)时间内是否为质数。

下面是上述方法的实现:

C++
// C++ program to find XOR of
// all Prime numbers in an Array
// at positions divisible by K
  
#include 
using namespace std;
#define MAX 1000005
  
void SieveOfEratosthenes(vector& prime)
{
    // 0 and 1 are not prime numbers
    prime[1] = false;
    prime[0] = false;
  
    for (int p = 2; p * p < MAX; p++) {
  
        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p] == true) {
  
            // Update all multiples of p
            for (int i = p * 2;
                 i < MAX; i += p)
                prime[i] = false;
        }
    }
}
  
// Function to find the required XOR
void prime_xor(int arr[], int n, int k)
{
  
    vector prime(MAX, true);
  
    SieveOfEratosthenes(prime);
  
    // To store XOR of the primes
    long long int ans = 0;
  
    // Traverse the array
    for (int i = 0; i < n; i++) {
  
        // If the number is a prime
        if (prime[arr[i]]) {
  
            // If index is divisible by k
            if ((i + 1) % k == 0) {
                ans ^= arr[i];
            }
        }
    }
  
    // Print the xor
    cout << ans << endl;
}
  
// Driver code
int main()
{
    int arr[] = { 2, 3, 5, 7, 11, 8 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int K = 2;
  
    // Function call
    prime_xor(arr, n, K);
  
    return 0;
}


Java
// Java program to find XOR of
// all Prime numbers in an Array
// at positions divisible by K
class GFG {
  
    static int MAX = 1000005;
    static boolean prime[] = new boolean[MAX];
      
    static void SieveOfEratosthenes(boolean []prime)
    {
        // 0 and 1 are not prime numbers
        prime[1] = false;
        prime[0] = false;
      
        for (int p = 2; p * p < MAX; p++) {
      
            // If prime[p] is not changed,
            // then it is a prime
            if (prime[p] == true) {
      
                // Update all multiples of p
                for (int i = p * 2;i < MAX; i += p)
                    prime[i] = false;
            }
        }
    }
      
    // Function to find the required XOR
    static void prime_xor(int arr[], int n, int k)
    {
          
        for(int i = 0; i < MAX; i++)
            prime[i] = true;
      
        SieveOfEratosthenes(prime);
      
        // To store XOR of the primes
        int ans = 0;
      
        // Traverse the array
        for (int i = 0; i < n; i++) {
      
            // If the number is a prime
            if (prime[arr[i]]) {
      
                // If index is divisible by k
                if ((i + 1) % k == 0) {
                    ans ^= arr[i];
                }
            }
        }
      
        // Print the xor
        System.out.println(ans);
    }
      
    // Driver code
    public static void main (String[] args) 
    {
        int arr[] = { 2, 3, 5, 7, 11, 8 };
        int n = arr.length;
        int K = 2;
      
        // Function call
        prime_xor(arr, n, K);
      
    }
}
  
// This code is contributed by Yash_R


Python3
# Python3 program to find XOR of 
# all Prime numbers in an Array 
# at positions divisible by K 
MAX = 1000005 
  
def SieveOfEratosthenes(prime) : 
  
    # 0 and 1 are not prime numbers 
    prime[1] = False; 
    prime[0] = False; 
  
    for p in range(2, int(MAX ** (1/2))) :
  
        # If prime[p] is not changed, 
        # then it is a prime 
        if (prime[p] == True) :
  
            # Update all multiples of p 
            for i in range(p * 2, MAX, p) :
                prime[i] = False; 
  
# Function to find the required XOR 
def prime_xor(arr, n, k) : 
  
    prime = [True]*MAX ; 
  
    SieveOfEratosthenes(prime); 
  
    # To store XOR of the primes 
    ans = 0; 
  
    # Traverse the array 
    for i in range(n) :
  
        # If the number is a prime 
        if (prime[arr[i]]) :
  
            # If index is divisible by k 
            if ((i + 1) % k == 0) :
                ans ^= arr[i]; 
  
    # Print the xor 
    print(ans); 
  
# Driver code 
if __name__ == "__main__" : 
  
    arr = [ 2, 3, 5, 7, 11, 8 ]; 
    n = len(arr); 
    K = 2; 
  
    # Function call 
    prime_xor(arr, n, K); 
  
# This code is contributed by Yash_R


C#
// C# program to find XOR of
// all Prime numbers in an Array
// at positions divisible by K
using System;
  
class GFG {
  
    static int MAX = 1000005;
    static bool []prime = new bool[MAX];
      
    static void SieveOfEratosthenes(bool []prime)
    {
        // 0 and 1 are not prime numbers
        prime[1] = false;
        prime[0] = false;
      
        for (int p = 2; p * p < MAX; p++) {
      
            // If prime[p] is not changed,
            // then it is a prime
            if (prime[p] == true) {
      
                // Update all multiples of p
                for (int i = p * 2;i < MAX; i += p)
                    prime[i] = false;
            }
        }
    }
      
    // Function to find the required XOR
    static void prime_xor(int []arr, int n, int k)
    {
          
        for(int i = 0; i < MAX; i++)
            prime[i] = true;
      
        SieveOfEratosthenes(prime);
      
        // To store XOR of the primes
        int ans = 0;
      
        // Traverse the array
        for (int i = 0; i < n; i++) {
      
            // If the number is a prime
            if (prime[arr[i]]) {
      
                // If index is divisible by k
                if ((i + 1) % k == 0) {
                    ans ^= arr[i];
                }
            }
        }
      
        // Print the xor
        Console.WriteLine(ans);
    }
      
    // Driver code
    public static void Main (string[] args) 
    {
        int []arr = { 2, 3, 5, 7, 11, 8 };
        int n = arr.Length;
        int K = 2;
      
        // Function call
        prime_xor(arr, n, K);
      
    }
}
  
// This code is contributed by Yash_R


输出:
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