Honaker素数

Honaker素数是素数P，使得P的数字和总和P的指数的数字之和是一个素数。
很少有Honaker素数是：

131, 263, 457, 1039, 1049, 1091, 1301, 1361, 1433, 1571, 1913, 1933, 2141, 2221,… 

检查N是否为Honaker素数

给定整数N ，任务是检查N是否为Honaker质数。如果N是Honaker质数，则打印“是”，否则打印“否” 。

例子：

Input: N = 131 
Output: Yes 
Explanation: 
Sum of digits of 131 = 1 + 3 + 1 = 5 
Sum of digits of 32 = 3 + 2 = 5

Input: N = 161 
Output: No 

方法：想法是找到给定数字的索引，并检查索引和N的位数之和是否相同。如果相同，则N为Honaker质数，并打印“是”，否则打印“否” 。

C++

// C++ program for the above approach
#include 
#define limit 10000000
using namespace std;
 
int position[limit + 1];
 
// Function to precompute the position
// of every prime number using Sieve
void sieve()
{
    // 0 and 1 are not prime numbers
    position[0] = -1, position[1] = -1;
 
    // Variable to store the position
    int pos = 0;
 
    for (int i = 2; i <= limit; i++) {
 
        if (position[i] == 0) {
 
            // Incrementing the position for
            // every prime number
            position[i] = ++pos;
            for (int j = i * 2; j <= limit; j += i)
                position[j] = -1;
        }
    }
}
 
// Function to get sum of digits
int getSum(int n)
{
    int sum = 0;
    while (n != 0) {
        sum = sum + n % 10;
        n = n / 10;
    }
    return sum;
}
 
// Function to check whether the given number
// is Honaker Prime number or not
bool isHonakerPrime(int n)
{
    int pos = position[n];
    if (pos == -1)
        return false;
    return getSum(n) == getSum(pos);
}
 
// Driver Code
int main()
{
    // Precompute the prime numbers till 10^6
    sieve();
 
    // Given Number
    int N = 121;
 
    // Function Call
    if (isHonakerPrime(N))
        cout << "Yes";
    else
        cout << "No";
}

Java

// Java program for above approach
class GFG{
 
static final int limit = 10000000;
static int []position = new int[limit + 1];
     
// Function to precompute the position
// of every prime number using Sieve
static void sieve()
{
    // 0 and 1 are not prime numbers
    position[0] = -1;
    position[1] = -1;
     
    // Variable to store the position
    int pos = 0;
    for (int i = 2; i <= limit; i++)
    {
        if (position[i] == 0)
        {
     
            // Incrementing the position for
            // every prime number
            position[i] = ++pos;
            for (int j = i * 2; j <= limit; j += i)
                position[j] = -1;
        }
    }
}
 
// Function to get sum of digits
static int getSum(int n)
{
    int sum = 0;
    while (n != 0)
    {
        sum = sum + n % 10;
        n = n / 10;
    }
    return sum;
}
 
// Function to check whether the given number
// is Honaker Prime number or not
static boolean isHonakerPrime(int n)
{
    int pos = position[n];
    if (pos == -1)
        return false;
    return getSum(n) == getSum(pos);
}
 
// Driver code
public static void main(String[] args)
{
    // Precompute the prime numbers till 10^6
    sieve();
 
    // Given Number
    int N = 121;
 
    // Function Call
    if (isHonakerPrime(N))
        System.out.print("Yes\n");
    else
        System.out.print("No\n");
}
}
 
// This code is contributed by shubham

Python3

# Python3 program for the above approach
limit = 10000000
 
position = [0] * (limit + 1)
 
# Function to precompute the position
# of every prime number using Sieve
def sieve():
     
    # 0 and 1 are not prime numbers
    position[0] = -1
    position[1] = -1
 
    # Variable to store the position
    pos = 0
 
    for i in range(2, limit + 1):
        if (position[i] == 0):
             
            # Incrementing the position for
            # every prime number
            pos += 1
            position[i] = pos
             
            for j in range(i * 2, limit + 1, i):
                position[j] = -1
 
# Function to get sum of digits
def getSum(n):
 
    Sum = 0
     
    while (n != 0):
        Sum = Sum + n % 10
        n = n // 10
  
    return Sum
 
# Function to check whether the given
# number is Honaker Prime number or not
def isHonakerPrime(n):
 
    pos = position[n]
     
    if (pos == -1):
        return False
         
    return bool(getSum(n) == getSum(pos))
 
# Driver code
 
# Precompute the prime numbers till 10^6
sieve()
 
# Given Number
N = 121
 
# Function Call
if (isHonakerPrime(N)):
    print("Yes")
else:
    print("No")
 
# This code is contributed by divyeshrabadiya07

C#

// C# program for above approach
using System;
class GFG{
 
static readonly int limit = 10000000;
static int []position = new int[limit + 1];
     
// Function to precompute the position
// of every prime number using Sieve
static void sieve()
{
    // 0 and 1 are not prime numbers
    position[0] = -1;
    position[1] = -1;
     
    // Variable to store the position
    int pos = 0;
    for (int i = 2; i <= limit; i++)
    {
        if (position[i] == 0)
        {
     
            // Incrementing the position for
            // every prime number
            position[i] = ++pos;
            for (int j = i * 2; j <= limit; j += i)
                position[j] = -1;
        }
    }
}
 
// Function to get sum of digits
static int getSum(int n)
{
    int sum = 0;
    while (n != 0)
    {
        sum = sum + n % 10;
        n = n / 10;
    }
    return sum;
}
 
// Function to check whether the given number
// is Honaker Prime number or not
static bool isHonakerPrime(int n)
{
    int pos = position[n];
    if (pos == -1)
        return false;
    return getSum(n) == getSum(pos);
}
 
// Driver code
public static void Main(String[] args)
{
    // Precompute the prime numbers till 10^6
    sieve();
 
    // Given Number
    int N = 121;
 
    // Function Call
    if (isHonakerPrime(N))
        Console.Write("Yes\n");
    else
        Console.Write("No\n");
}
}
 
// This code is contributed by 29AjayKumar

Javascript

// Javascript program for above approach
     const  limit = 10000000;
    let position = Array(limit + 1).fill(0);
 
    // Function to precompute the position
    // of every prime number using Sieve
    function sieve()
    {
     
        // 0 and 1 are not prime numbers
        position[0] = -1;
        position[1] = -1;
 
        // Variable to store the position
        let pos = 0;
        for (let i = 2; i <= limit; i++)
        {
            if (position[i] == 0)
            {
 
                // Incrementing the position for
                // every prime number
                position[i] = ++pos;
                for (let j = i * 2; j <= limit; j += i)
                    position[j] = -1;
            }
        }
    }
 
    // Function to get sum of digits
    function getSum( n) {
        let sum = 0;
        while (n != 0) {
            sum = sum + n % 10;
            n = parseInt(n / 10);
        }
        return sum;
    }
 
    // Function to check whether the given number
    // is Honaker Prime number or not
    function isHonakerPrime( n) {
        let pos = position[n];
        if (pos == -1)
            return false;
        return getSum(n) == getSum(pos);
    }
 
    // Driver code
    // Precompute the prime numbers till 10^6
    sieve();
 
    // Given Number
    let N = 121;
 
    // Function Call
    if (isHonakerPrime(N))
        document.write("Yes\n");
    else
        document.write("No\n");
 
// This code is contributed by aashish1995

输出：

No

参考： https : //oeis.org/A033548