📜  具有素数频率的数组中元素的总和

📅  最后修改于: 2021-10-28 01:33:00             🧑  作者: Mango

给定一个数组arr ,任务是找到数组中具有素数频率的元素的总和。
注意: 1 既不是质数也不是合数。
例子:

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。
  • 使用在上一步中计算的 Sieve 数组计算具有质数频率的元素的总和。

下面是上述方法的实现:

C++
// C++ program to find sum of elements
// in an array having prime frequency
#include 
using namespace std;
 
// 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 sum of elements
// in an array having prime frequency
int sumOfElements(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]]++;
 
    int sum = 0;
 
    // 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]) {
            sum += (it->first);
        }
    }
 
    return sum;
}
 
// Driver code
int main()
{
    int arr[] = { 5, 4, 6, 5, 4, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << sumOfElements(arr, n);
    return 0;
}


Java
// Java program to find sum of elements
// in an array having prime frequency
import java.util.*;
 
class GFG
{
 
    // 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 sum of elements
    // in an array having prime frequency
    static int sumOfElements(int arr[], int n)
    {
        boolean prime[] = new boolean[n + 1];
        Arrays.fill(prime, true);
     
        SieveOfEratosthenes(prime, n + 1);
     
        int i, j;
     
        // Map is used to store
        // element frequencies
        HashMap m = new HashMap<>();
        for (i = 0; i < n; i++)
        {
            if(m.containsKey(arr[i]))
                m.put(arr[i], m.get(arr[i]) + 1);
            else
                m.put(arr[i], 1);
        }
     
        int sum = 0;
     
        // Traverse the map
        for (Map.Entry entry : m.entrySet())
        {
            int key = entry.getKey();
            int value = entry.getValue();
             
            // Count the number of elements
            // having prime frequencies
            if (prime[value])
            {
                sum += (key);
            }
        }
     
        return sum;
    }
     
    // Driver code
    public static void main(String args[])
    {
        int arr[] = { 5, 4, 6, 5, 4, 6 };
        int n = arr.length;
     
        System.out.println(sumOfElements(arr, n));
    }
}
 
// This code is contributed by ghanshyampandey


Python3
# Python3 program to find Sum of elements
# in an array having prime frequency
import math as mt
 
# 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, mt.ceil(mt.sqrt(p_size + 1))):
 
        # 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(p * 2, p_size + 1, p):
                prime[i] = False
         
# Function to return the Sum of elements
# in an array having prime frequency
def SumOfElements(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):
        if arr[i] in m.keys():
            m[arr[i]] += 1
        else:
            m[arr[i]] = 1
             
    Sum = 0
 
    # Traverse the map using iterators
    for i in m:
         
        # Count the number of elements
        # having prime frequencies
        if (prime[m[i]]):
            Sum += (i)
     
    return Sum
 
# Driver code
arr = [5, 4, 6, 5, 4, 6 ]
n = len(arr)
print(SumOfElements(arr, n))
 
# This code is contributed
# by Mohit kumar 29


C#
// C# program to find sum of elements
// in an array having prime frequency
using System;
using System.Collections.Generic;
 
class GFG
{
 
    // 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 sum of elements
    // in an array having prime frequency
    static int sumOfElements(int []arr, int n)
    {
        bool []prime = new bool[n + 1];
        for(int i = 0; i < n+1; i++)
            prime[i] = true;
     
        SieveOfEratosthenes(prime, n + 1);
 
     
        // Map is used to store
        // element frequencies
        Dictionary m = new Dictionary();
        for (int i = 0 ; i < n; i++)
        {
            if(m.ContainsKey(arr[i]))
            {
                var val = m[arr[i]];
                m.Remove(arr[i]);
                m.Add(arr[i], val + 1);
            }
            else
            {
                m.Add(arr[i], 1);
            }
        }
     
        int sum = 0;
     
        // Traverse the map
        foreach(KeyValuePair entry in m)
        {
            int key = entry.Key;
            int value = entry.Value;
             
            // Count the number of elements
            // having prime frequencies
            if (prime[value])
            {
                sum += (key);
            }
        }
     
        return sum;
    }
     
    // Driver code
    public static void Main(String []args)
    {
        int []arr = { 5, 4, 6, 5, 4, 6 };
        int n = arr.Length;
     
        Console.WriteLine(sumOfElements(arr, n));
    }
}
 
// This code is contributed by 29AjayKumar


Javascript


输出:
15