Prime Subset Product Problem

Given an array arr[] of N integers. Value of a subset of array A is defined as the product of all prime numbers in that subset. If there are no primes in the subset then the value of that subset is 1. The task is to calculate the product of values of all possible non-empty subsets of the given array modulus 100000007.

Examples:

Input: arr[] = {3, 7}
Output: 441
val({3}) = 3
val({7}) = 7
val({3, 7}) = 3 * 7 = 21
3 * 7 * 21 = 441



Input: arr[] = {1, 1, 1}
Output: 1

Approach: Since it is known that a number occurs 2N – 1 times in all the subset of the given array of size N. So, if a number X is prime then the contibution of X will be X * X * X * ….. * 2N – 1 times i.e.

Since 2N – 1 will also be a large number, it cannot be calculated directly. Fermat’s Theorem will be used to calculate the power here.

After that, the value of each element can be calculated easily.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include<bits/stdc++.h>
using namespace std;
  
int power(int a, int b, int mod)
{
    int aa = 1;
    while(b)
    {
        if(b & 1)
        
            aa = aa * a;
            aa %= mod;
        }
        a = a * a;
        a %= mod;
        b /= 2;
    }
    return aa;
}
  
// Function to return the prime subset 
// product of the given array
int product(int A[], int n)
{
  
    // Create Sieve to check whether a
    // number is prime or not
    int N = 100010;
    int mod = 1000000007;
    vector<int> prime(N, 1);
    prime[0] = prime[1] = 0;
    int i = 2;
    while (i * i < N)
    {
        if (prime[i])
            for (int j = 2 * i;
                     j <= N;j += i)
                prime[j] = 0;
  
        i += 1;
    }
  
    // Length of the array
    // Calculating 2^(n-1) % mod
    int t = power(2, n - 1, mod - 1);
  
    int ans = 1;
  
    for (int j = 0; j < n; j++)
    {
        int i = A[j];
  
        // If element is prime then add
        // its contribution in the result
        if( prime[i])
        {
            ans *= power(i, t, mod);
            ans %= mod;
        }
    }
    return ans;
}
  
// Driver code
int main()
{
    int A[] = {3, 7};
      
    int n = sizeof(A) / sizeof(A[0]);
      
    printf("%d", product(A, n));
}
  
// This code is contributed by Mohit Kumar 

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Java

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// Java implementation of the approach
class GFG
{
static int power(int a, int b, int mod)
{
    int aa = 1;
    while(b > 0)
    {
        if(b % 2 == 1)
        
            aa = aa * a;
            aa %= mod;
        }
        a = a * a;
        a %= mod;
        b /= 2;
    }
    return aa;
}
  
// Function to return the prime subset 
// product of the given array
static int product(int A[], int n)
{
  
    // Create Sieve to check whether a
    // number is prime or not
    int N = 100010;
    int mod = 1000000007;
    int []prime = new int[N];
    for (int j = 0; j < N; j++)
    {
        prime[j] = 1;
    }
      
    prime[0] = prime[1] = 0;
    int i = 2;
    while (i * i < N)
    {
        if (prime[i] == 1)
            for (int j = 2 * i;
                    j < N;j += i)
                prime[j] = 0;
  
        i += 1;
    }
  
    // Length of the array
    // Calculating 2^(n-1) % mod
    int t = power(2, n - 1, mod - 1);
  
    int ans = 1;
  
    for (int j = 0; j < n; j++)
    {
        i = A[j];
  
        // If element is prime then add
        // its contribution in the result
        if( prime[i] == 1)
        {
            ans *= power(i, t, mod);
            ans %= mod;
        }
    }
    return ans;
}
  
// Driver code
public static void main (String[] args)
{
    int A[] = {3, 7};
      
    int n = A.length;
      
    System.out.printf("%d", product(A, n));
}
}
  
// This code is contributed by Rajput-Ji

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Python3

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# Python3 implementation of the approach
  
# Function to return the prime subset 
# product of the given array
def product(A):
      
    # Create Sieve to check whether a 
    # number is prime or not
    N = 100010
    mod = 1000000007
    prime = [1] * N
    prime[0] = prime[1] = 0
    i = 2
    while i * i < N:
        if prime[i]:
            for j in range(i * i, N, i):
                prime[j] = 0
          
        i += 1
      
    # Length of the array
    n = len(A)
      
    # Calculating 2^(n-1) % mod
    t = pow(2, n-1, mod-1)
      
    ans = 1
      
    for i in A:
          
        # If element is prime then add
        # its contribution in the result
        if prime[i]:
            ans *= pow(i, t, mod)
            ans %= mod
              
    return ans
      
# Driver code
A = [3, 7]
print(product(A))

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C#

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// C# implementation of the approach
using System;
  
class GFG
{
static int power(int a, int b, int mod)
{
    int aa = 1;
    while(b > 0)
    {
        if(b % 2 == 1)
        
            aa = aa * a;
            aa %= mod;
        }
        a = a * a;
        a %= mod;
        b /= 2;
    }
    return aa;
}
  
// Function to return the prime subset 
// product of the given array
static int product(int []A, int n)
{
  
    // Create Sieve to check whether a
    // number is prime or not
    int N = 100010;
    int mod = 1000000007;
    int []prime = new int[N];
    for (int j = 0; j < N; j++)
    {
        prime[j] = 1;
    }
      
    prime[0] = prime[1] = 0;
    int i = 2;
    while (i * i < N)
    {
        if (prime[i] == 1)
            for (int j = 2 * i;
                     j < N; j += i)
                prime[j] = 0;
  
        i += 1;
    }
  
    // Length of the array
    // Calculating 2^(n-1) % mod
    int t = power(2, n - 1, mod - 1);
  
    int ans = 1;
  
    for (int j = 0; j < n; j++)
    {
        i = A[j];
  
        // If element is prime then add
        // its contribution in the result
        if( prime[i] == 1)
        {
            ans *= power(i, t, mod);
            ans %= mod;
        }
    }
    return ans;
}
  
// Driver code
public static void Main(String[] args)
{
    int []A = {3, 7};
      
    int n = A.Length;
      
    Console.Write("{0}", product(A, n));
}
}
      
// This code is contributed by Rajput-Ji

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Output:

441


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Improved By : mohit kumar 29, Rajput-Ji



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