用数组中数字的质数和打印质数
给定一个数组 arr[] ，任务是打印数组中的加法素数。 加性素数:这样的素数，它们的位数之和也是素数，如 2、3、7、11、23 是加性素数但不是 13、19、31 等。 举例:
Input: arr[] = {2, 4, 6, 11, 12, 18, 7}
Output: 2, 11, 7
Input: arr[] = {2, 3, 19, 13, 25, 7}
Output: 2, 3, 7
一种简单的方法是遍历所有的数组元素。对于每个元素，检查它是否是加法素数。 当数组很小或数组值很大时，上述方法可以。对于数值相对较小的大型数组，我们使用筛选来存储最大数组元素的素数。然后检查当前元素是否是质数。如果是，那么检查它的数字之和是否也是质数。如果是，则打印该号码。 以下是上述方法的实施:
C++
// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to store the primes
void sieve(int maxEle, int prime[])
{
prime[0] = prime[1] = 1;
for (int i = 2; i * i <= maxEle; i++) {
if (!prime[i]) {
for (int j = 2 * i; j <= maxEle; j += i)
prime[j] = 1;
}
}
}
// Function to return the sum of digits
int digitSum(int n)
{
int sum = 0;
while (n) {
sum += n % 10;
n = n / 10;
}
return sum;
}
// Function to print additive primes
void printAdditivePrime(int arr[], int n)
{
int maxEle = *max_element(arr, arr + n);
int prime[maxEle + 1];
memset(prime, 0, sizeof(prime));
sieve(maxEle, prime);
for (int i = 0; i < n; i++) {
// If the number is prime
if (prime[arr[i]] == 0) {
int sum = digitSum(arr[i]);
// Check if it's digit sum is prime
if (prime[sum] == 0)
cout << arr[i] << " ";
}
}
}
// Driver code
int main()
{
int a[] = { 2, 4, 6, 11, 12, 18, 7 };
int n = sizeof(a) / sizeof(a[0]);
printAdditivePrime(a, n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the above approach
import java.util.Arrays;
class GFG
{
// Function to store the primes
static void sieve(int maxEle, int prime[])
{
prime[0] = prime[1] = 1;
for (int i = 2; i * i <= maxEle; i++)
{
if (prime[i]==0)
{
for (int j = 2 * i; j <= maxEle; j += i)
prime[j] = 1;
}
}
}
// Function to return the sum of digits
static int digitSum(int n)
{
int sum = 0;
while (n > 0)
{
sum += n % 10;
n = n / 10;
}
return sum;
}
// Function to print additive primes
static void printAdditivePrime(int arr[], int n)
{
int maxEle = Arrays.stream(arr).max().getAsInt();
int prime[] = new int[maxEle + 1];
sieve(maxEle, prime);
for (int i = 0; i < n; i++)
{
// If the number is prime
if (prime[arr[i]] == 0)
{
int sum = digitSum(arr[i]);
// Check if it's digit sum is prime
if (prime[sum] == 0)
System.out.print(arr[i]+" ");
}
}
}
// Driver code
public static void main(String[] args)
{
int a[] = { 2, 4, 6, 11, 12, 18, 7 };
int n =a.length;
printAdditivePrime(a, n);
}
}
// This code is contributed by chandan_jnu
Python 3
# Python3 implementation of the
# above approach
# from math lib import sqrt
from math import sqrt
# Function to store the primes
def sieve(maxEle, prime) :
prime[0], prime[1] = 1 , 1
for i in range(2, int(sqrt(maxEle)) + 1) :
if (not prime[i]) :
for j in range(2 * i , maxEle + 1, i) :
prime[j] = 1
# Function to return the sum of digits
def digitSum(n) :
sum = 0
while (n) :
sum += n % 10
n = n // 10
return sum
# Function to print additive primes
def printAdditivePrime(arr, n):
maxEle = max(arr)
prime = [0] * (maxEle + 1)
sieve(maxEle, prime)
for i in range(n) :
# If the number is prime
if (prime[arr[i]] == 0):
sum = digitSum(arr[i])
# Check if it's digit sum is prime
if (prime[sum] == 0) :
print(arr[i], end = " ")
# Driver code
if __name__ == "__main__" :
a = [ 2, 4, 6, 11, 12, 18, 7 ]
n = len(a)
printAdditivePrime(a, n)
# This code is contributed by Ryuga
C
// C# implementation of the above approach
using System.Linq;
using System;
class GFG
{
// Function to store the primes
static void sieve(int maxEle, int[] prime)
{
prime[0] = prime[1] = 1;
for (int i = 2; i * i <= maxEle; i++)
{
if (prime[i] == 0)
{
for (int j = 2 * i; j <= maxEle; j += i)
prime[j] = 1;
}
}
}
// Function to return the sum of digits
static int digitSum(int n)
{
int sum = 0;
while (n > 0)
{
sum += n % 10;
n = n / 10;
}
return sum;
}
// Function to print additive primes
static void printAdditivePrime(int []arr, int n)
{
int maxEle = arr.Max();
int[] prime = new int[maxEle + 1];
sieve(maxEle, prime);
for (int i = 0; i < n; i++)
{
// If the number is prime
if (prime[arr[i]] == 0)
{
int sum = digitSum(arr[i]);
// Check if it's digit sum is prime
if (prime[sum] == 0)
Console.Write(arr[i] + " ");
}
}
}
// Driver code
static void Main()
{
int[] a = { 2, 4, 6, 11, 12, 18, 7 };
int n = a.Length;
printAdditivePrime(a, n);
}
}
// This code is contributed by chandan_jnu
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the above approach
// Function to store the primes
function sieve($maxEle, &$prime)
{
$prime[0] = $prime[1] = 1;
for ($i = 2; $i * $i <= $maxEle; $i++)
{
if (!$prime[$i])
{
for ($j = 2 * $i;
$j <= $maxEle; $j += $i)
$prime[$j] = 1;
}
}
}
// Function to return the sum of digits
function digitSum($n)
{
$sum = 0;
while ($n)
{
$sum += $n % 10;
$n = $n / 10;
}
return $sum;
}
// Function to print additive primes
function printAdditivePrime($arr, $n)
{
$maxEle = max($arr);
$prime = array_fill(0, $maxEle + 1, 0);
sieve($maxEle, $prime);
for ($i = 0; $i < $n; $i++)
{
// If the number is prime
if ($prime[$arr[$i]] == 0)
{
$sum = digitSum($arr[$i]);
// Check if it's digit sum is prime
if ($prime[$sum] == 0)
print($arr[$i] . " ");
}
}
}
// Driver code
$a = array(2, 4, 6, 11, 12, 18, 7);
$n = count($a);
printAdditivePrime($a, $n);
// This code is contributed by chandan_jnu
?>
java 描述语言
<script>
// Javascript implementation of the above approach
// Function to store the primes
function sieve(maxEle, prime)
{
prime[0] = prime[1] = 1;
for (var i = 2; i * i <= maxEle; i++) {
if (!prime[i]) {
for (var j = 2 * i; j <= maxEle; j += i)
prime[j] = 1;
}
}
}
// Function to return the sum of digits
function digitSum(n)
{
var sum = 0;
while (n) {
sum += n % 10;
n = parseInt(n / 10);
}
return sum;
}
// Function to print additive primes
function printAdditivePrime(arr, n)
{
var maxEle = arr.reduce((a,b)=> Math.max(a,b));
var prime = Array(maxEle + 1).fill(0);
sieve(maxEle, prime);
for (var i = 0; i < n; i++) {
// If the number is prime
if (prime[arr[i]] == 0) {
var sum = digitSum(arr[i]);
// Check if it's digit sum is prime
if (prime[sum] == 0)
document.write( arr[i] + " ");
}
}
}
// Driver code
var a = [ 2, 4, 6, 11, 12, 18, 7 ];
var n = a.length;
printAdditivePrime(a, n);
</script>
Output:
2 11 7
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