检查给定的号码是否为 Emirp 号码
原文:https://www . geesforgeks . org/check-given-number-emirp-number-not/
一个 Emirp 数(质数向后拼)是一个质数，当它的十进制数字颠倒时会产生一个不同的质数。这个定义排除了相关的回文素数。 例:
Input : n = 13
Output : 13 is Emirp!
Explanation :
13 and 31 are both prime numbers.
Thus, 13 is an Emirp number.
Input : n = 27
Output : 27 is not Emirp.
目的:输入一个数字，找出这个数字是否是 emirp 数。
方法:输入一个数，首先检查它是否是素数。如果数是质数，那么我们找到原数的反数，检查反数是否为质数。如果反数也是质数，那么原数就是 Emirp 数，否则不是。 以下是上述办法的实施情况:。
C++
// C++ program to check if given
// number is Emirp or not.
#include <iostream>
using namespace std;
// Returns true if n is prime.
// Else false.
bool isPrime(int n)
{
// Corner case
if (n <= 1)
return false;
// Check from 2 to n-1
for (int i = 2; i < n; i++)
if (n % i == 0)
return false;
return true;
}
// Function will check whether number
// is Emirp or not
bool isEmirp(int n)
{
// Check if n is prime
if (isPrime(n) == false)
return false;
// Find reverse of n
int rev = 0;
while (n != 0) {
int d = n % 10;
rev = rev * 10 + d;
n /= 10;
}
// If both Original and Reverse are Prime,
// then it is an Emirp number
return isPrime(rev);
}
// Driver code
int main()
{
int n = 13; // Input number
if (isEmirp(n) == true)
cout << "Yes";
else
cout << "No";
}
// This code is contributed by Anant Agarwal.
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to check if given number is
// Emirp or not.
import java.io.*;
class Emirp {
// Returns true if n is prime. Else
// false.
public static boolean isPrime(int n)
{
// Corner case
if (n <= 1)
return false;
// Check from 2 to n-1
for (int i = 2; i < n; i++)
if (n % i == 0)
return false;
return true;
}
// Function will check whether number
// is Emirp or not
public static boolean isEmirp(int n)
{
// Check if n is prime
if (isPrime(n) == false)
return false;
// Find reverse of n
int rev = 0;
while (n != 0) {
int d = n % 10;
rev = rev * 10 + d;
n /= 10;
}
// If both Original and Reverse are Prime,
// then it is an Emirp number
return isPrime(rev);
}
// Driver Function
public static void main(String args[]) throws IOException
{
int n = 13; // Input number
if (isEmirp(n) == true)
System.out.println("Yes");
else
System.out.println("No");
}
}
Python 3
# Python3 code to check if
# given number is Emirp or not.
# Returns true if n is prime.
# Else false.
def isPrime( n ):
# Corner case
if n <= 1:
return False
# Check from 2 to n-1
for i in range(2, n):
if n % i == 0:
return False
return True
# Function will check whether
# number is Emirp or not
def isEmirp( n):
# Check if n is prime
n = int(n)
if isPrime(n) == False:
return False
# Find reverse of n
rev = 0
while n != 0:
d = n % 10
rev = rev * 10 + d
n = int(n / 10)
# If both Original and Reverse
# are Prime, then it is an
# Emirp number
return isPrime(rev)
# Driver Function
n = 13 # Input number
if isEmirp(n):
print("Yes")
else:
print("No")
# This code is contributed by "Sharad_Bhardwaj".
C
// C# program to check if given
// number is Emirp or not.
using System;
class Emirp {
// Returns true if n is prime
// Else false.
public static bool isPrime(int n)
{
// Corner case
if (n <= 1)
return false;
// Check from 2 to n-1
for (int i = 2; i < n; i++)
if (n % i == 0)
return false;
return true;
}
// Function will check whether number
// is Emirp or not
public static bool isEmirp(int n)
{
// Check if n is prime
if (isPrime(n) == false)
return false;
// Find reverse of n
int rev = 0;
while (n != 0) {
int d = n % 10;
rev = rev * 10 + d;
n /= 10;
}
// If both Original and Reverse are Prime,
// then it is an Emirp number
return isPrime(rev);
}
// Driver Function
public static void Main()
{
int n = 13; // Input number
if (isEmirp(n) == true)
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by vt_m.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to check if given
// number is Emirp or not.
// Returns true if n
// is prime else false
function isPrime($n)
{
// Corner case
if ($n <= 1)
return -1;
// Check from 2 to n-1
for ($i = 2; $i < $n; $i++)
if ($n % $i == 0)
return -1;
return 1;
}
// Function will check
// whether number is
// Emirp or not
function isEmirp($n)
{
// Check if n is prime
if (isPrime($n) == -1)
return -1;
// Find reverse of n
$rev = 0;
while ($n != 0)
{
$d = $n % 10;
$rev = $rev * 10 + $d;
$n /= 10;
}
// If both Original and
// Reverse are Prime,
// then it is an Emirp number
return isPrime($rev);
}
// Driver code
$n = 13;
if (isEmirp($n) ==-1)
echo "Yes";
else
echo "No";
// This code is contributed by ajit
?>
java 描述语言
<script>
// javascript program to check if given number is
// Emirp or not.
// Returns true if n is prime. Else
// false.
function isPrime(n)
{
// Corner case
if (n <= 1)
return false;
// Check from 2 to n-1
for (i = 2; i < n; i++)
if (n % i == 0)
return false;
return true;
}
// Function will check whether number
// is Emirp or not
function isEmirp(n)
{
// Check if n is prime
if (isPrime(n) == false)
return false;
// Find reverse of n
var rev = 0;
while (n != 0) {
var d = n % 10;
rev = rev * 10 + d;
n = parseInt(n/10);
}
// If both Original and Reverse are Prime,
// then it is an Emirp number
return isPrime(rev);
}
// Driver Function
var n = 13; // Input number
if (isEmirp(n) == true)
document.write("Yes");
else
document.write("No");
// This code contributed by Princi Singh
</script>
输出:
Yes
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