所有子集素数的乘积

给定一个大小为N的数组a[] 。子集的值是该子集中素数的乘积。在寻找价值副产品时，非素数被认为是 1。任务是找到所有可能子集的值的乘积。
例子：

Input: a[] = {3, 7}
Output: 20
The subsets are: {3} {7} {3, 7}
{3, 7} = 3 * 7 = 21
{3} = 3
{7} = 7
21 * 3 * 7 = 441
Input: a[] = {10, 2, 14, 3}
Output: 1679616

编程需要懂一点英语

朴素方法：一种朴素的方法是使用幂集找到所有子集，然后通过将子集的所有值相乘来找到乘积。可以使用 Sieve 检查 Prime。
时间复杂度： O(2 ^N )
有效方法：一种有效的方法是使用观察来解决问题。如果我们写出所有子序列，一个共同的观察点是每个数字在一个子集中出现2 ^(N-1)次，因此将导致2 ^(N-1)作为对乘积的贡献。遍历数组并检查数组中的元素是否为素数。如果它是素数，那么它对答案的贡献是arr[i] ^2(N-1)次。
下面是上述方法的实现：

C++

// C++ program to find the product of
// the multiplication of
// prime numbers in all possible subsets.
#include 
using namespace std;
 
// Sieve method to check prime or not
void sieve(int n, vector& prime)
{
    // Initially mark all primes
    for (int i = 2; i <= n; i++)
        prime[i] = true;
    prime[0] = prime[1] = false;
 
    // Iterate and mark all the
    // non primes as false
    for (int i = 2; i <= n; i++) {
        if (prime[i]) {
            // Multiples of prime marked as false
            for (int j = i * i; j <= n; j += i) {
                prime[j] = false;
            }
        }
    }
}
 
// Function to find the sum
// of sum of all the subset
int sumOfSubset(int a[], int n)
{
 
    // Get the maximum element
    int maxi = *max_element(a, a + n);
 
    // Declare a sieve array
    vector prime(maxi + 1);
 
    // Sieve function called
    sieve(maxi, prime);
 
    // Number of times an element
    // contributes to the answer
    int times = pow(2, n - 1);
 
    int sum = 1;
 
    // Iterate and check
    for (int i = 0; i < n; i++) {
        // If prime
        if (prime[a[i]])
        sum = sum * (pow(a[i], times)); // Contribution
    }
 
    return sum;
}
 
// Driver Code
int main()
{
    int a[] = { 3, 7 };
    int n = sizeof(a) / sizeof(a[0]);
    cout << sumOfSubset(a, n);
}

Java

// Java program to find the product of
// the multiplication of
// prime numbers in all possible subsets.
import java.util.*;
 
class GFG
{
 
// Sieve method to check prime or not
static void sieve(int n, boolean []prime)
{
    // Initially mark all primes
    for (int i = 2; i <= n; i++)
        prime[i] = true;
    prime[0] = prime[1] = false;
 
    // Iterate and mark all the
    // non primes as false
    for (int i = 2; i <= n; i++)
    {
        if (prime[i])
        {
            // Multiples of prime marked as false
            for (int j = i * i; j <= n; j += i)
            {
                prime[j] = false;
            }
        }
    }
}
 
// Function to find the sum
// of sum of all the subset
static int sumOfSubset(int a[], int n)
{
 
    // Get the maximum element
    int maxi = Arrays.stream(a).max().getAsInt();
 
    // Declare a sieve array
    boolean []prime = new boolean[maxi + 1];
 
    // Sieve function called
    sieve(maxi, prime);
 
    // Number of times an element
    // contributes to the answer
    int times = (int) Math.pow(2, n - 1);
 
    int sum = 1;
 
    // Iterate and check
    for (int i = 0; i < n; i++)
    {
        // If prime
        if (prime[a[i]])
        sum = (int) (sum * (Math.pow(a[i], times)));
    }
 
    return sum;
}
 
// Driver Code
public static void main(String[] args)
{
    int a[] = { 3, 7 };
    int n = a.length;
    System.out.println(sumOfSubset(a, n));
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 program to find the product of
# the multiplication of
# prime numbers in all possible subsets.
prime = [True for i in range(100)]
 
# Sieve method to check prime or not
def sieve(n, prime):
     
    # Initially mark all primes
    for i in range(1, n + 1):
        prime[i] = True
    prime[0] = prime[1] = False
 
    # Iterate and mark all the
    # non primes as false
    for i in range(2, n + 1):
        if (prime[i]):
             
            # Multiples of prime marked as false
            for j in range(2 * i, n + 1, i):
                prime[j] = False
 
# Function to find the Sum
# of Sum of all the subset
def SumOfSubset(a, n):
 
    # Get the maximum element
    maxi = max(a)
 
    # Declare a sieve array
 
    # Sieve function called
    sieve(maxi, prime)
 
    # Number of times an element
    # contributes to the answer
    times = pow(2, n - 1)
 
    Sum = 1
 
    # Iterate and check
    for i in range(n):
         
        # If prime
        if (prime[a[i]]):
            Sum = Sum * (pow(a[i], times)) # Contribution
 
    return Sum
 
# Driver Code
a = [3, 7]
n = len(a)
print(SumOfSubset(a, n))
 
# This code is contributed
# by Mohit Kumar

C#

// C# program to find the product of
// the multiplication of
// prime numbers in all possible subsets.
using System;
using System.Linq;
 
class GFG
{
 
// Sieve method to check prime or not
static void sieve(int n, Boolean []prime)
{
    // Initially mark all primes
    for (int i = 2; i <= n; i++)
        prime[i] = true;
    prime[0] = prime[1] = false;
 
    // Iterate and mark all the
    // non primes as false
    for (int i = 2; i <= n; i++)
    {
        if (prime[i])
        {
            // Multiples of prime marked as false
            for (int j = i * i; j <= n; j += i)
            {
                prime[j] = false;
            }
        }
    }
}
 
// Function to find the sum
// of sum of all the subset
static int sumOfSubset(int []a, int n)
{
 
    // Get the maximum element
    int maxi = a.Max();
 
    // Declare a sieve array
    Boolean []prime = new Boolean[maxi + 1];
 
    // Sieve function called
    sieve(maxi, prime);
 
    // Number of times an element
    // contributes to the answer
    int times = (int) Math.Pow(2, n - 1);
 
    int sum = 1;
 
    // Iterate and check
    for (int i = 0; i < n; i++)
    {
        // If prime
        if (prime[a[i]])
        sum = (int) (sum * (Math.Pow(a[i], times)));
    }
    return sum;
}
 
// Driver Code
public static void Main(String[] args)
{
    int []a = { 3, 7 };
    int n = a.Length;
    Console.WriteLine(sumOfSubset(a, n));
}
}
 
// This code is contributed by PrinciRaj1992

Javascript

输出：

时间复杂度：O(M log M) 用于预计算，其中 M 是最大元素，O(N) 用于迭代。
空间复杂度：O(M)
注意：由于arr[i] ^2(N-1)可能非常大，答案可能会溢出，最好使用更大的数据类型和 mod 操作来保存答案。