Honaker 质数
Honaker 质数为质数 P ，使得 P 的位数和 P 的指数位数之和为质数。 很少有 Honaker 质数是:
131, 263, 457, 1039, 1049, 1091, 1301, 1361, 1433, 1571, 1913, 1933, 2141, 2221,…
检查 N 是否是 Honaker 质数
给定一个整数 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”。
示例:
输入: N = 131 输出:是 说明: 131 位数之和= 1+3+1 = 5 32 位数之和= 3 + 2 = 5
输入:N = 161 T3】输出:否
方法:思路是找到给定数字的索引，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 <bits/stdc++.h>
#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 语言(一种计算机语言，尤用于创建网站)
// 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
Python 3
# 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
java 描述语言
<script>
// 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
</script>
Output:
No
参考:T2https://oeis.org/A033548
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