具有素频的阵列中元素的乘积

给定一个包含N 个元素的数组arr[] ，任务是找到数组中具有素数频率的元素的乘积。由于产品可能很大，所以打印产品模10 ⁹ + 7 。请注意， 1既不是质数也不是合数。
例子：

Input: arr[] = {5, 4, 6, 5, 4, 6} 
Output: 120 
All the elements appear 2 times which is a prime 
So, 5 * 4 * 6 = 120
Input: arr[] = {1, 2, 3, 3, 2, 3, 2, 3, 3} 
Output: 6 
Only 2 and 3 appears prime number of times i.e. 3 and 5 respectively. 
So, 2 * 3 = 6 
 

方法：

• 遍历数组并将所有元素的频率存储在地图中。
• 构建 Eratosthenes 筛，用于在 O(1) 时间内测试数字的素性。
• 使用上一步中计算的 Sieve 数组计算具有素数频率的元素的乘积。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
#define MOD 1000000007
 
// Function to create Sieve to check primes
void SieveOfEratosthenes(bool prime[], int p_size)
{
    // False here indicates
    // that it is not prime
    prime[0] = false;
    prime[1] = false;
 
    for (int p = 2; p * p <= p_size; p++) {
 
        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p]) {
 
            // Update all multiples of p,
            // set them to non-prime
            for (int i = p * 2; i <= p_size; i += p)
                prime[i] = false;
        }
    }
}
 
// Function to return the product of elements
// in an array having prime frequency
int productPrimeFreq(int arr[], int n)
{
    bool prime[n + 1];
    memset(prime, true, sizeof(prime));
 
    SieveOfEratosthenes(prime, n + 1);
 
    int i, j;
 
    // Map is used to store
    // element frequencies
    unordered_map m;
    for (i = 0; i < n; i++)
        m[arr[i]]++;
 
    long product = 1;
 
    // Traverse the map using iterators
    for (auto it = m.begin(); it != m.end(); it++) {
 
        // Count the number of elements
        // having prime frequencies
        if (prime[it->second]) {
            product *= (it->first % MOD);
            product %= MOD;
        }
    }
 
    return (int)(product);
}
 
// Driver code
int main()
{
    int arr[] = { 5, 4, 6, 5, 4, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << productPrimeFreq(arr, n);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
     
class GFG
{
static int MOD = 1000000007;
 
// Function to create Sieve to check primes
static void SieveOfEratosthenes(boolean prime[],
                                int p_size)
{
    // False here indicates
    // that it is not prime
    prime[0] = false;
    prime[1] = false;
 
    for (int p = 2; p * p <= p_size; p++)
    {
 
        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p])
        {
 
            // Update all multiples of p,
            // set them to non-prime
            for (int i = p * 2;
                     i <= p_size; i += p)
                prime[i] = false;
        }
    }
}
 
// Function to return the product of elements
// in an array having prime frequency
static int productPrimeFreq(int arr[], int n)
{
    boolean []prime = new boolean[n + 1];
    for (int i = 0; i < n; i++)
        prime[i] = true;
 
    SieveOfEratosthenes(prime, n + 1);
 
    int i, j;
 
    // Map is used to store
    // element frequencies
    HashMap mp = new HashMap();
 
    for (i = 0 ; i < n; i++)
    {
        if(mp.containsKey(arr[i]))
        {
            mp.put(arr[i], mp.get(arr[i]) + 1);
        }
        else
        {
            mp.put(arr[i], 1);
        }
    }
    long product = 1;
 
    // Traverse the map using iterators
    for (Map.Entry it : mp.entrySet())
    {
 
        // Count the number of elements
        // having prime frequencies
        if (prime[it.getValue()])
        {
            product *= (it.getKey() % MOD);
            product %= MOD;
        }
    }
    return (int)(product);
}
 
// Driver code
static public void main (String []arg)
{
    int arr[] = { 5, 4, 6, 5, 4, 6 };
    int n = arr.length;
 
    System.out.println(productPrimeFreq(arr, n));
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 implementation of the approach
MOD = 1000000007
 
# Function to create Sieve to check primes
def SieveOfEratosthenes(prime, p_size):
     
    # False here indicates
    # that it is not prime
    prime[0] = False
    prime[1] = False
 
    for p in range(2, p_size):
 
        # If prime[p] is not changed,
        # then it is a prime
        if (prime[p]):
 
            # Update all multiples of p,
            # set them to non-prime
            for i in range(2 * p, p_size, p):
                prime[i] = False
 
# Function to return the product of elements
# in an array having prime frequency
def productPrimeFreq(arr, n):
    prime = [True for i in range(n + 1)]
 
    SieveOfEratosthenes(prime, n + 1)
 
    i, j = 0, 0
 
    # Map is used to store
    # element frequencies
    m = dict()
    for i in range(n):
        m[arr[i]] = m.get(arr[i], 0) + 1
 
    product = 1
 
    # Traverse the map using iterators
    for it in m:
 
        # Count the number of elements
        # having prime frequencies
        if (prime[m[it]]):
            product *= it % MOD
            product %= MOD
 
    return product
 
# Driver code
arr = [5, 4, 6, 5, 4, 6]
n = len(arr)
 
print(productPrimeFreq(arr, n))
 
# This code is contributed by Mohit Kumar

C#

// C# implementation for above approach
using System;
using System.Collections.Generic;
 
class GFG
{
static int MOD = 1000000007;
 
// Function to create Sieve to check primes
static void SieveOfEratosthenes(bool []prime,
                                int p_size)
{
    // False here indicates
    // that it is not prime
    prime[0] = false;
    prime[1] = false;
 
    for (int p = 2; p * p <= p_size; p++)
    {
 
        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p])
        {
 
            // Update all multiples of p,
            // set them to non-prime
            for (int i = p * 2;
                     i <= p_size; i += p)
                prime[i] = false;
        }
    }
}
 
// Function to return the product of elements
// in an array having prime frequency
static int productPrimeFreq(int []arr, int n)
{
    bool []prime = new bool[n + 1];
    int i;
    for (i = 0; i < n; i++)
        prime[i] = true;
 
    SieveOfEratosthenes(prime, n + 1);
 
    // Map is used to store
    // element frequencies
    Dictionary mp = new Dictionary();
    for (i = 0 ; i < n; i++)
    {
        if(mp.ContainsKey(arr[i]))
        {
            var val = mp[arr[i]];
            mp.Remove(arr[i]);
            mp.Add(arr[i], val + 1);
        }
        else
        {
            mp.Add(arr[i], 1);
        }
    }
    long product = 1;
 
    // Traverse the map using iterators
    foreach(KeyValuePair it in mp)
    {
 
        // Count the number of elements
        // having prime frequencies
        if (prime[it.Value])
        {
            product *= (it.Key % MOD);
            product %= MOD;
        }
    }
    return (int)(product);
}
 
// Driver code
static public void Main (String []arg)
{
    int []arr = { 5, 4, 6, 5, 4, 6 };
    int n = arr.Length;
 
    Console.WriteLine(productPrimeFreq(arr, n));
}
}
 
// This code is contributed by Princi Singh

Javascript

// JavaScript implementation of the approach
let MOD = 1000000007;
 
// Function to create Sieve to check primes
function SieveOfEratosthenes(prime, p_size){
    // False here indicates
    // that it is not prime
    prime[0] = false;
    prime[1] = false;
    for (let p = 2; p * p <= p_size; p++) {
        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p]) {
            // Update all multiples of p,
            // set them to non-prime
            for (let i = p * 2; i <= p_size; i += p)
                prime[i] = false;
        }
    }
    return prime;
}
 
// Function to return the product of elements
// in an array having prime frequency
function productPrimeFreq(arr, n){
    let prime = [];
    for(let i = 0;i<n+1;i++){
        prime.push(true);
    }
    prime = SieveOfEratosthenes(prime, n + 1);
    let i, j;
    // Map is used to store
    // element frequencies
    let m = new Map();
    for (i = 0; i < n; i++){
      if(m[arr[i]])
        m[arr[i]]++;
      else
        m[arr[i]] = 1;
    }
 
    let product = 1;
 
    // Traverse the map using iterators
    for (var it in m) {
        // Count the number of elements
        // having prime frequencies
        if (prime[m[it]]) {
            product *= (it % MOD);
            product %= MOD;
        }
    }
 
    return (product);
}
 
// Driver code
let a = [ 5, 4, 6, 5, 4, 6 ];
let len = a.length;
document.write(productPrimeFreq(a, len));

输出：

120

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