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📜  检查N是否为二面体质数

📅  最后修改于: 2021-04-22 01:35:09             🧑  作者: Mango

给定整数N ，任务是检查N是否为二面质数。二面体质数是素数，当在七段显示器中读取时，无论其方向和表面如何，都可以将其本身或另一个质数读取。
例子：

方法：为素数测试预先计算素数筛。 Eratosthenes的筛子可以n * logn * logn时间计算。对数字及其不同方向进行素性测试。如果该数字通过了素数测试，请检查是否有任何数字属于排除集[3、4、6、7、9]。如果数字通过两个测试，则返回true。
下面是上述方法的实现：

C++
// C++ implementation of the above approach
#include 
using namespace std;
 
bool isPrime[int(1e5) + 5];
 
// Function to return the reverse
// of a number
int reverse(int n)
{
    int temp = n;
    int sum = 0;
    while (temp > 0) {
        int rem = temp % 10;
        sum = sum * 10 + rem;
        temp /= 10;
    }
    return sum;
}
 
// Function to generate mirror reflection
// of a number
int mirror(int n)
{
    int temp = n;
    int sum = 0;
    while (temp > 0) {
        int rem = temp % 10;
        if (rem == 2)
            rem = 5;
        else if (rem == 5)
            rem = 2;
        sum = sum * 10 + rem;
        temp /= 10;
    }
    return sum;
}
 
// Function to initialize prime number sieve
bool sieve()
{
    memset(isPrime, true, sizeof isPrime);
 
    isPrime[0] = isPrime[1] = false;
 
    for (int i = 2; i <= int(1e5); i++) {
        for (int j = 2; i * j <= int(1e5); j++) {
            isPrime[i * j] = false;
        }
    }
}
 
// Function that returns true if n is
// Dihedral Prime number
bool isDihedralPrime(int n)
{
    // Check if the orientations of n
    // is also prime
    if (!isPrime[n]
        || !isPrime[mirror(n)]
        || !isPrime[reverse(n)]
        || !isPrime[reverse(mirror(n))])
        return false;
 
    int temp = n;
 
    while (temp > 0) {
        int rem = temp % 10;
        if (rem == 3 || rem == 4 || rem == 6
            || rem == 7 || rem == 9)
            return false;
        temp /= 10;
    }
 
    return true;
}
 
// Driver code
int main()
{
    sieve();
 
    int n = 18181;
    if (isDihedralPrime(n))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

Java
// Java implementation of the above approach
import java.util.*;
 
class GFG
{
 
    static boolean[] isPrime = new boolean[(int) (1e5) + 5];
 
    // Function to return the reverse
    // of a number
    static int reverse(int n)
    {
        int temp = n;
        int sum = 0;
        while (temp > 0)
        {
            int rem = temp % 10;
            sum = sum * 10 + rem;
            temp /= 10;
        }
        return sum;
    }
 
    // Function to generate mirror reflection
    // of a number
    static int mirror(int n)
    {
        int temp = n;
        int sum = 0;
        while (temp > 0)
        {
            int rem = temp % 10;
            if (rem == 2)
            {
                rem = 5;
            }
             
            else if (rem == 5)
            {
                rem = 2;
            }
            sum = sum * 10 + rem;
            temp /= 10;
        }
        return sum;
    }
 
    // Function to initialize
    // prime number sieve
    static void sieve()
    {
        Arrays.fill(isPrime, true);
 
        isPrime[0] = isPrime[1] = false;
 
        for (int i = 2;
                 i <= (int) 1e5; i++)
        {
            for (int j = 2;
                     i * j <= (int) 1e5; j++)
            {
                isPrime[i * j] = false;
            }
        }
    }
 
    // Function that returns true if n is
    // Dihedral Prime number
    static boolean isDihedralPrime(int n)
    {
         
        // Check if the orientations of n
        // is also prime
        if (!isPrime[n] ||
            !isPrime[mirror(n)] ||
            !isPrime[reverse(n)] ||
            !isPrime[reverse(mirror(n))])
        {
            return false;
        }
 
        int temp = n;
 
        while (temp > 0)
        {
            int rem = temp % 10;
            if (rem == 3 || rem == 4 ||
                rem == 6 || rem == 7 ||
                rem == 9)
            {
                return false;
            }
            temp /= 10;
        }
 
        return true;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        sieve();
 
        int n = 18181;
        if (isDihedralPrime(n))
        {
            System.out.println("Yes");
        }
        else
        {
            System.out.println("No");
        }
    }
}
 
// This code is contributed by Rajput-Ji

Python3
# Python implementation of the above approach
isPrime = (int(1e5)+5)*[True]
 
# Function to return the reverse
# of a number
def reverse(n):
    temp = n
    sum = 0
    while temp>0:
        rem = temp % 10
        sum = sum * 10 + rem
        temp//= 10
 
    return sum
 
# Function to generate mirror reflection
# of a number
def mirror(n):
    temp = n
    sum = 0
    while temp>0:
        rem = temp % 10
        if rem == 2:
            rem = 5
        elif rem == 5:
            rem = 2
        sum = sum * 10 + rem
        temp//= 10
 
    return sum
 
# Function to initialize prime number sieve
def sieve():
 
    isPrime[0] = isPrime[1] = False
 
    for i in range(2, int(1e5)+1):
        j = 2
        while i * j<= int(1e5):
            isPrime[i * j] = False
            j+= 1
 
 
# Function that returns true if n is
# Dihedral Prime number
def isDihedralPrime(n):
     
    # Check if the orientations of n is also prime
    if (not isPrime[n]) or (not isPrime[mirror(n)]) \
        or (not isPrime[reverse(n)]) \
        or (not isPrime[reverse(mirror(n))]):
        return False
 
    temp = n
 
    while temp>0:
        rem = temp % 10;
        if rem == 3 or rem == 4 or \
            rem == 6 or rem == 7 or rem == 9:
            return False
        temp//= 10
 
    return True
 
# Driver code
if __name__ == '__main__':
 
    sieve()
     
    n = 18181
    if isDihedralPrime(n):
        print("Yes")
    else :
        print("No")

C#
// C# implementation of the above approach
using System;
     
class GFG
{
    static Boolean[] isPrime = new Boolean[(int) (1e5) + 5];
 
    // Function to return the reverse
    // of a number
    static int reverse(int n)
    {
        int temp = n;
        int sum = 0;
        while (temp > 0)
        {
            int rem = temp % 10;
            sum = sum * 10 + rem;
            temp /= 10;
        }
        return sum;
    }
 
    // Function to generate mirror reflection
    // of a number
    static int mirror(int n)
    {
        int temp = n;
        int sum = 0;
        while (temp > 0)
        {
            int rem = temp % 10;
            if (rem == 2)
            {
                rem = 5;
            }
             
            else if (rem == 5)
            {
                rem = 2;
            }
            sum = sum * 10 + rem;
            temp /= 10;
        }
        return sum;
    }
 
    // Function to initialize
    // prime number sieve
    static void sieve()
    {
        for(int k = 0; k < isPrime.Length; k++)
            isPrime[k] = true;
 
        isPrime[0] = isPrime[1] = false;
 
        for (int i = 2;
                 i <= (int) 1e5; i++)
        {
            for (int j = 2;
                     i * j <= (int) 1e5; j++)
            {
                isPrime[i * j] = false;
            }
        }
    }
 
    // Function that returns true if n is
    // Dihedral Prime number
    static Boolean isDihedralPrime(int n)
    {
         
        // Check if the orientations of n
        // is also prime
        if (!isPrime[n] ||
            !isPrime[mirror(n)] ||
            !isPrime[reverse(n)] ||
            !isPrime[reverse(mirror(n))])
        {
            return false;
        }
 
        int temp = n;
 
        while (temp > 0)
        {
            int rem = temp % 10;
            if (rem == 3 || rem == 4 ||
                rem == 6 || rem == 7 ||
                rem == 9)
            {
                return false;
            }
            temp /= 10;
        }
 
        return true;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        sieve();
 
        int n = 18181;
        if (isDihedralPrime(n))
        {
            Console.WriteLine("Yes");
        }
        else
        {
            Console.WriteLine("No");
        }
    }
}
 
// This code is contributed by Rajput-Ji

PHP
 0)
    {
        $rem = $temp % 10;
        $sum = $sum * 10 + $rem;
        $temp = floor($temp / 10);
    }
    return $sum;
}
 
// Function to generate mirror reflection
// of a number
function mirror($n)
{
    $temp = $n;
    $sum = 0;
    while ($temp > 0)
    {
        $rem = $temp % 10;
        if ($rem == 2)
            $rem = 5;
        else if ($rem == 5)
            $rem = 2;
        $sum = $sum * 10 + $rem;
        $temp = floor($temp / 10);
    }
    return $sum;
}
 
// Function to initialize prime number sieve
function sieve()
{
    $GLOBALS['isPrime'][0] = $GLOBALS['isPrime'][1] = false;
 
    for ($i = 2; $i <= floor(1e4); $i++)
    {
        for ($j = 2; $i * $j <= floor(1e4); $j++)
        {
            $GLOBALS['isPrime'][$i * $j] = false;
        }
    }
}
 
// Function that returns true if n is
// Dihedral Prime number
function isDihedralPrime($n)
{
    // Check if the orientations of n
    // is also prime
    if (!$GLOBALS['isPrime'][$n] ||
        !$GLOBALS['isPrime'][mirror($n)] ||
        !$GLOBALS['isPrime'][reverse($n)] ||
        !$GLOBALS['isPrime'][reverse(mirror($n))])
        return false;
 
    $temp = $n;
 
    while ($temp > 0)
    {
        $rem = $temp % 10;
        if ($rem == 3 || $rem == 4 ||
            $rem == 6 || $rem == 7 || $rem == 9)
            return false;
             
        $temp = floor($temp / 10);
    }
 
    return true;
}
 
// Driver code
sieve();
 
$n = 18181;
if (isDihedralPrime($n))
    echo "Yes";
else
    echo "No";
 
// This code is contributed by Ryuga
?>

输出： 
Yes

时间复杂度： O(n * log(log n))

辅助空间： O(10 5 )