原文:
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
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）
java 描述语言
output:
2 11 7
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