检查素数总和是否可被数组中的任何素数整除

给定数组arr [] ，任务是检查数组中素数的和是否可被数组中的任何素数整除。如果是，则打印“是” ，否则打印“否” 。

例子：

Input: arr[] = {2, 3}
Output: NO
Primes: 2, 3
Sum = 2 + 3 = 5 which is neither divisible by 2 nor 3

Input: arr[] = {1, 2, 3, 4, 5}
Output: YES
2 + 3 + 5 = 10 is divisible by 2 as well as 5

方法：这个想法是使用Eratosthenes的Sieve从数组生成所有素数到最大元素。

• 遍历数组并检查当前元素是否为质数。如果是素数，则更新sum = sum + arr [i] 。
• 再次遍历数组，并检查sum％arr [i] = 0 ，其中arr [i]是质数。如果是，则打印YES 。否则，最后打印NO 。

下面是上述方法的实现：

C++

// C++ program to check if sum of primes from an array
// is divisible by any of the primes from the same array
#include 
using namespace std;
  
// Function to print "YES" if sum of primes from an array
// is divisible by any of the primes from the same array
void SumDivPrime(int A[], int n)
{
    int max_val = *(std::max_element(A, A + n)) + 1;
  
    // 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;
        }
    }
  
    int sum = 0;
  
    // Traverse through the array
    for (int i = 0; i < n; ++i) {
        if (prime[A[i]])
            sum += A[i];
    }
  
    for (int i = 0; i < n; ++i) {
        if (prime[A[i]] && sum % A[i] == 0) {
            cout << "YES";
            return;
        }
    }
  
    cout << "NO";
}
  
// Driver program
int main()
{
    int A[] = { 1, 2, 3, 4, 5 };
    int n = sizeof(A) / sizeof(A[0]);
  
    SumDivPrime(A, n);
  
    return 0;
}

Java

// Java program to check if sum of primes from an array 
// is divisible by any of the primes from the same array 
class Solution
{
    //returns the maximum value
static int max_element(int A[])
{
    int max=Integer.MIN_VALUE;
      
    for(int i=0;i

---

Python3

# Python3 program to check if sum of 
# primes from an array is divisible
# by any of the primes from the same array
import math
  
# Function to print "YES" if sum of primes 
# from an array is divisible by any of the 
# primes from the same array
def SumDivPrime(A, n):
  
    max_val = max(A) + 1
  
    # 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] * (max_val + 1)
  
    # Remaining part of SIEVE
    prime[0] = False
    prime[1] = False
    for p in range(2, int(math.sqrt(max_val)) + 1):
  
        # If prime[p] is not changed, 
        # then it is a prime
        if prime[p] == True :
  
            # Update all multiples of p
            for i in range(2 * p, max_val + 1, p):
                prime[i] = False
  
    sum = 0
  
    # Traverse through the array
    for i in range(0, n):
        if prime[A[i]]:
            sum += A[i]
      
    for i in range(0, n):
        if prime[A[i]] and sum % A[i] == 0:
            print("YES")
            return
          
    print("NO")
  
# Driver Code
A = [ 1, 2, 3, 4, 5 ]
n = len(A)
  
SumDivPrime(A, n)
  
# This code is contributed 
# by saurabh_shukla

C#

// C# program to check if sum of primes
// from an array is divisible by any of 
// the primes from the same array 
class GFG 
{ 
  
//returns the maximum value 
static int max_element(int[] A) 
{ 
    int max = System.Int32.MinValue; 
      
    for(int i = 0; i < A.Length; i++) 
        if(max < A[i]) 
            max = A[i]; 
      
    return max; 
} 
  
  
// Function to print "YES" if sum of 
// primes from an array  is divisible 
// by any of the primes from the same array 
static void SumDivPrime(int[] A, int n) 
{ 
    int max_val = (max_element(A)) + 1; 
  
    // 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. 
    bool[] prime=new bool[max_val+1]; 
      
    //initilize the array 
    for(int i = 0; i <= max_val; i++) 
        prime[i] = 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; 
        } 
    } 
    int sum = 0; 
  
    // Traverse through the array 
    for (int i = 0; i < n; ++i)
    { 
        if (prime[A[i]]) 
            sum += A[i]; 
    } 
  
    for (int i = 0; i < n; ++i) 
    { 
        if (prime[A[i]] && sum % A[i] == 0) 
        { 
            System.Console.WriteLine( "YES"); 
            return; 
        } 
    } 
    System.Console.WriteLine("NO"); 
} 
  
// Driver code
public static void Main() 
{ 
    int []A = { 1, 2, 3, 4, 5 }; 
    int n = A.Length; 
    SumDivPrime(A, n); 
} 
} 
  
// This code is contributed by mits

输出：

YES