数组中所有素数的总和

给定一个包含 N 个正整数的数组 arr[]。任务是编写一个程序来查找给定数组中所有素数元素的总和。
例子：

Input: arr[] = {1, 3, 4, 5, 7} 
Output: 15 
There are three primes, 3, 5 and 7 whose sum =15.
Input: arr[] = {1, 2, 3, 4, 5, 6, 7} 
Output: 17 
 

朴素方法：一个简单的解决方案是遍历数组并继续检查每个元素是否为素数，并同时添加素数元素。
有效方法：使用 Eratosthenes 的筛子生成直到数组最大元素的所有素数，并将它们存储在哈希中。现在遍历数组并使用筛子找到那些为素数的元素的总和。
下面是高效方法的实现：

C++

// CPP program to find sum of
// primes in given array.
#include 
using namespace std;
 
// Function to find count of prime
int primeSum(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;
        }
    }
 
    // Sum all primes in arr[]
    int sum = 0;
    for (int i = 0; i < n; i++)
        if (prime[arr[i]])
            sum += arr[i];
 
    return sum;
}
 
// Driver code
int main()
{
 
    int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << primeSum(arr, n);
 
    return 0;
}

Java

// Java program to find sum of
// primes in given array.
import java.util.*;
 
class GFG
{
 
// Function to find count of prime
static int primeSum(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.
    Vector prime = new Vector<>(max_val + 1);
    for(int i = 0; i < max_val + 1; i++)
        prime.add(i,Boolean.TRUE);
 
    // Remaining part of SIEVE
    prime.add(0,Boolean.FALSE);
    prime.add(1,Boolean.FALSE);
    for (int p = 2; p * p <= max_val; p++)
    {
 
        // If prime[p] is not changed, then
        // it is a prime
        if (prime.get(p) == true)
        {
 
            // Update all multiples of p
            for (int i = p * 2; i <= max_val; i += p)
                prime.add(i,Boolean.FALSE);
        }
    }
 
    // Sum all primes in arr[]
    int sum = 0;
    for (int i = 0; i < n; i++)
        if (prime.get(arr[i]))
            sum += arr[i];
 
    return sum;
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
    int n = arr.length;
    System.out.print(primeSum(arr, n));
}
}
 
/* This code contributed by PrinciRaj1992 */

Python

# Python3 program to find sum of
# primes in given array.
 
# Function to find count of prime
def primeSum( 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=[True for i in range(max_val + 1)]
 
    # Remaining part of SIEVE
    prime[0] = False
    prime[1] = False
    for p in range(2, max_val + 1):
        if(p * p > max_val):
            break
 
        # 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
 
    # Sum all primes in arr[]
    sum = 0
 
    for i in range(n):
        if (prime[arr[i]]):
            sum += arr[i]
 
    return sum
 
# Driver code
arr =[1, 2, 3, 4, 5, 6, 7]
 
n = len(arr)
 
print(primeSum(arr, n))
 
# This code is contributed by mohit kumar 29

C#

// C# program to find sum of
// primes in given array.
using System;
using System.Linq;
using System.Collections.Generic;
 
class GFG
{
 
// Function to find count of prime
static int primeSum(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.
    List prime = new List(max_val + 1);
    for(int i = 0; i < max_val + 1; i++)
        prime.Insert(i,true);
 
    // Remaining part of SIEVE
    prime.Insert(0, false);
    prime.Insert(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.Insert(i,false);
        }
    }
 
    // Sum all primes in arr[]
    int sum = 0;
    for (int i = 0; i < n; i++)
        if (prime[arr[i]])
            sum += arr[i];
 
    return sum;
}
 
// Driver code
public static void Main(String[] args)
{
    int []arr = { 1, 2, 3, 4, 5, 6, 7 };
    int n = arr.Length;
    Console.WriteLine(primeSum(arr, n));
}
}
 
// This code contributed by Rajput-Ji

Javascript

// Javascript program to find sum of
// primes in given array.
 
// Function to find count of prime
function primeSum(arr, n) {
    // Find maximum value in the array
    let max_val = arr.sort((a, b) => b - a)[0];
 
    // 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.
    let prime = new Array(max_val + 1).fill(true);
 
    // Remaining part of SIEVE
    prime[0] = false;
    prime[1] = false;
    for (let 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 (let i = p * 2; i <= max_val; i += p)
                prime[i] = false;
        }
    }
 
    // Sum all primes in arr[]
    let sum = 0;
    for (let i = 0; i < n; i++)
        if (prime[arr[i]])
            sum += arr[i];
 
    return sum;
}
 
// Driver code
 
let arr = [1, 2, 3, 4, 5, 6, 7];
let n = arr.length;
 
document.write(primeSum(arr, n));
 
// This code is contributed by _saurabh_jaiswal.

输出：

17

时间复杂度： O(n*loglogn)