非素数与数组素数的乘积的绝对差

给定一个正数数组，任务是计算非素数与素数的乘积之间的绝对差。
注意： 1既不是素数也不是非素数。
例子：

Input : arr[] = {1, 3, 5, 10, 15, 7}
Output : 45
Explanation : Product of non-primes = 150
              Product of primes = 105

Input : arr[] = {3, 4, 6, 7} 
Output : 3

天真的方法：一个简单的解决方案是遍历数组并检查每个元素是否为素数。如果数字是素数，则将其乘以表示素数乘积的乘积P2，否则检查其是否不是1，然后将其乘以非素数的乘积，即P1。遍历整个数组后，取两者之间的绝对差(P1-P2)。
时间复杂度： O(N * sqrt(N))
高效方法：使用Eratosthenes筛子生成所有素数，直到数组的最大元素，并将它们存储在哈希中。现在，遍历数组并检查哈希图中是否存在该数字。然后，将这些数字乘以乘积P2，否则检查它是否不是1，然后将其乘以乘积P1。遍历整个数组后，显示两者之间的绝对差。
下面是上述方法的实现：

C++

// C++ program to find the Absolute Difference
// between the Product of Non-Prime numbers
// and Prime numbers of an Array
   
#include 
using namespace std;
   
// Function to find the difference between
// the product of non-primes and the
// product of primes of an array.
int calculateDifference(int arr[], int n)
{
    // Find maximum value in the array
    int max_val = *max_element(arr, arr + n);
   
    // USE SIEVE TO FIND ALL PRIME NUMBERS LESS
    // THAN OR EQUAL TO max_val
    // Create a boolean array "prime[0..n]". A
    // value in prime[i] will finally be false
    // if i is Not a prime, else true.
    vector prime(max_val + 1, true);
   
    // Remaining part of SIEVE
    prime[0] = false;
    prime[1] = false;
    for (int p = 2; p * p <= max_val; 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_val; i += p)
                prime[i] = false;
        }
    }
   
    // Store the product of primes in P1 and
    // the product of non primes in P2
    int P1 = 1, P2 = 1;
    for (int i = 0; i < n; i++) {
   
        if (prime[arr[i]]) {
   
            // the number is prime
            P1 *= arr[i];
        }
        else if (arr[i] != 1) {
   
            // the number is non-prime
            P2 *= arr[i];
        }
    }
   
    // Return the absolute difference
    return abs(P2 - P1);
}
   
// Driver Code
int main()
{
    int arr[] = { 1, 3, 5, 10, 15, 7 };
    int n     = sizeof(arr) / sizeof(arr[0]);
   
    // Find the absolute difference
    cout << calculateDifference(arr, n);
   
    return 0;
}

Java

// Java program to find the Absolute Difference
// between the Product of Non-Prime numbers
// and Prime numbers of an Array
import java.util.*; 
import java.util.Arrays;
import java.util.Collections;
 
   
class GFG{
 
    // Function to find the difference between
    // the product of non-primes and the
    // product of primes of an array.
    public static int calculateDifference(int []arr, int n)
    {
        // Find maximum value in the array
 
        int max_val = Arrays.stream(arr).max().getAsInt();
       
        // USE SIEVE TO FIND ALL PRIME NUMBERS LESS
        // THAN OR EQUAL TO max_val
        // Create a boolean array "prime[0..n]". A
        // value in prime[i] will finally be false
        // if i is Not a prime, else true.
        boolean[] prime = new boolean[max_val + 1];
        Arrays.fill(prime, true);
       
        // Remaining part of SIEVE
        prime[0] = false;
        prime[1] = false;
        for (int p = 2; p * p <= max_val; 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_val ;i += p)
                    prime[i] = false;
            }
        }
       
        // Store the product of primes in P1 and
        // the product of non primes in P2
        int P1 = 1, P2 = 1;
        for (int i = 0; i < n; i++) {
       
            if (prime[arr[i]]) {
       
                // the number is prime
                P1 *= arr[i];
            }
            else if (arr[i] != 1) {
       
                // the number is non-prime
                P2 *= arr[i];
            }
        }
       
        // Return the absolute difference
        return Math.abs(P2 - P1);
    }
       
    // Driver Code
    public static void main(String []args)
    {
        int[] arr = new int []{ 1, 3, 5, 10, 15, 7 };
        int n     = arr.length;
       
        // Find the absolute difference
        System.out.println(calculateDifference(arr, n));
       
        System.exit(0);
    }
}
// This code is contributed
// by Harshit Saini

Python3

# Python3 program to find the Absolute Difference
# between the Product of Non-Prime numbers
# and Prime numbers of an Array
 
   
# Function to find the difference between
# the product of non-primes and the
# product of primes of an array.
def calculateDifference(arr, n):
    # Find maximum value in the array
    max_val = max(arr)
   
    # USE SIEVE TO FIND ALL PRIME NUMBERS LESS
    # THAN OR EQUAL TO max_val
    # Create a boolean array "prime[0..n]". A
    # value in prime[i] will finally be false
    # if i is Not a prime, else true.
 
    prime    = (max_val + 1) * [True]
   
    # Remaining part of SIEVE
    prime[0] = False
    prime[1] = False
    p = 2
 
    while p * p <= max_val:
 
        # 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_val+1, p):
                prime[i] = False
        p += 1
   
    # Store the product of primes in P1 and
    # the product of non primes in P2
    P1 = 1 ; P2 = 1
    for i in range(n):
 
        if prime[arr[i]]:
            # the number is prime
            P1 *= arr[i]
 
        elif arr[i] != 1:
            # the number is non-prime
            P2 *= arr[i]
   
    # Return the absolute difference
    return abs(P2 - P1)
   
# Driver Code
if __name__ == '__main__':
    arr   = [ 1, 3, 5, 10, 15, 7 ]
    n     = len(arr)
   
    # Find the absolute difference
    print(calculateDifference(arr, n))
# This code is contributed
# by Harshit Saini

C#

// C# program to find the Absolute Difference
// between the Product of Non-Prime numbers
// and Prime numbers of an Array
using System;
using System.Linq;
   
class GFG{
 
    // Function to find the difference between
    // the product of non-primes and the
    // product of primes of an array.
    static int calculateDifference(int []arr, int n)
    {
        // Find maximum value in the array
        int max_val = arr.Max();
       
        // USE SIEVE TO FIND ALL PRIME NUMBERS LESS
        // THAN OR EQUAL TO max_val
        // Create a boolean array "prime[0..n]". A
        // value in prime[i] will finally be false
        // if i is Not a prime, else true.
        var prime   = Enumerable.Repeat(true,
                                    max_val+1).ToArray();
       
        // Remaining part of SIEVE
        prime[0] = false;
        prime[1] = false;
        for (int p = 2; p * p <= max_val; 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_val; i += p)
                    prime[i] = false;
            }
        }
       
        // Store the product of primes in P1 and
        // the product of non primes in P2
        int P1 = 1, P2 = 1;
        for (int i = 0; i < n; i++) {
       
            if (prime[arr[i]]) {
       
                // the number is prime
                P1 *= arr[i];
            }
            else if (arr[i] != 1) {
       
                // the number is non-prime
                P2 *= arr[i];
            }
        }
       
        // Return the absolute difference
        return Math.Abs(P2 - P1);
    }
       
    // Driver Code
    public static void Main()
    {
        int[] arr = new int []{ 1, 3, 5, 10, 15, 7 };
        int n     = arr.Length;
       
        // Find the absolute difference
        Console.WriteLine(calculateDifference(arr, n));
    }
}
// This code is contributed
// by Harshit Saini

Javascript

 
// Javascript program to find the Absolute Difference
// between the Product of Non-Prime numbers
// and Prime numbers of an Array
 
    // Function to find the difference between
    // the product of non-primes and the
    // product of primes of an array.
    function calculateDifference(arr , n)
    {
        // Find maximum value in the array
 
        var max_val = Math.max.apply(Math,arr);
 
        // USE SIEVE TO FIND ALL PRIME NUMBERS LESS
        // THAN OR EQUAL TO max_val
        // Create a boolean array "prime[0..n]". A
        // value in prime[i] will finally be false
        // if i is Not a prime, else true.
        var prime = Array(max_val + 1).fill(true);
     
 
        // Remaining part of SIEVE
        prime[0] = false;
        prime[1] = false;
        for (p = 2; p * p <= max_val; p++) {
 
            // If prime[p] is not changed, then
            // it is a prime
            if (prime[p] == true) {
 
                // Update all multiples of p
                for (i = p * 2; i <= max_val; i += p)
                    prime[i] = false;
            }
        }
 
        // Store the product of primes in P1 and
        // the product of non primes in P2
        var P1 = 1, P2 = 1;
        for (i = 0; i < n; i++) {
 
            if (prime[arr[i]]) {
 
                // the number is prime
                P1 *= arr[i];
            } else if (arr[i] != 1) {
 
                // the number is non-prime
                P2 *= arr[i];
            }
        }
 
        // Return the absolute difference
        return Math.abs(P2 - P1);
    }
 
    // Driver Code
     
        var arr = [ 1, 3, 5, 10, 15, 7 ];
        var n = arr.length;
 
        // Find the absolute difference
        document.write(calculateDifference(arr, n));
 
// This code contributed by gauravrajput1
 

输出：

45

时间复杂度： O(N * log(log(N))
空间复杂度： O(MAX(N，max_val))，其中max_val是给定数组中元素的最大值。