检查N是否为强质数

给定正整数N ，任务是检查N是否为强质数。
在数论中，强素数是大于最接近素数(即下一个和上一个素数)的算术平均值的素数。
前几个强质数是11、17、29、37、41、59、67、71，…
强素数P _n可以表示为-

其中n是其在有素数的有序集合中的索引。

例子：

Input: N = 11 
Output: Yes 
11 is 5th prime number, the arithmetic mean of 4th and 6th prime number i.e. 7 and 13 is 10. 
11 is greater than 10 so 11 is a strong prime. 

Input: N = 13 
Output: No 
13 is 6th prime number, the arithmetic mean of 5th (11) and 7th (17) is (11 + 17) / 2 = 14. 
13 is smaller than 14 so 13 is not a strong prime. 
 

方法：

• 如果N不是素数或它是第一个素数，即2，则打印No。
• 否则，找到最接近N的素数(左边一个，右边一个)，并将它们的算术平均值存储在mean中。
  • 如果N>等于，则打印是。
  • 其他打印号。

下面是上述方法的实现：

C++

// C++ program to check if given number is strong prime
#include 
using namespace std;
 
// Utility function to check
// if a number is prime or not
bool isPrime(int n)
{
    // Corner cases
    if (n <= 1)
        return false;
    if (n <= 3)
        return true;
 
    // This is checked so that we can skip
    // middle five numbers in below loop
    if (n % 2 == 0 || n % 3 == 0)
        return false;
 
    for (int i = 5; i * i <= n; i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return false;
 
    return true;
}
 
// Function that returns true if n is a strong prime
static bool isStrongPrime(int n)
{
    // If n is not a prime number or
    // n is the first prime then return false
    if (!isPrime(n) || n == 2)
        return false;
 
    // Initialize previous_prime to n - 1
    // and next_prime to n + 1
    int previous_prime = n - 1;
    int next_prime = n + 1;
 
    // Find next prime number
    while (!isPrime(next_prime))
        next_prime++;
 
    // Find previous prime number
    while (!isPrime(previous_prime))
        previous_prime--;
 
    // Arithmetic mean
    int mean = (previous_prime + next_prime) / 2;
 
    // If n is a strong prime
    if (n > mean)
        return true;
    else
        return false;
}
 
// Driver code
int main()
{
 
    int n = 11;
 
    if (isStrongPrime(n))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

Java

// Java program to check if given number is strong prime
class GFG {
 
    // Utility function to check
    // if a number is prime or not
    static boolean isPrime(int n)
    {
        // Corner cases
        if (n <= 1)
            return false;
        if (n <= 3)
            return true;
 
        // This is checked so that we can skip
        // middle five numbers in below loop
        if (n % 2 == 0 || n % 3 == 0)
            return false;
 
        for (int i = 5; i * i <= n; i = i + 6)
            if (n % i == 0 || n % (i + 2) == 0)
                return false;
 
        return true;
    }
 
    // Function that returns true if n is a strong prime
    static boolean isStrongPrime(int n)
    {
        // If n is not a prime number or
        // n is the first prime then return false
        if (!isPrime(n) || n == 2)
            return false;
 
        // Initialize previous_prime to n - 1
        // and next_prime to n + 1
        int previous_prime = n - 1;
        int next_prime = n + 1;
 
        // Find next prime number
        while (!isPrime(next_prime))
            next_prime++;
 
        // Find previous prime number
        while (!isPrime(previous_prime))
            previous_prime--;
 
        // Arithmetic mean
        int mean = (previous_prime + next_prime) / 2;
 
        // If n is a strong prime
        if (n > mean)
            return true;
        else
            return false;
    }
 
    // Driver code
    public static void main(String args[])
    {
 
        int n = 11;
 
        if (isStrongPrime(n))
            System.out.println("Yes");
 
        else
            System.out.println("No");
    }
}

Python3

# Python 3 program to check if given
# number is strong prime
from math import sqrt
 
# Utility function to check if a
# number is prime or not
def isPrime(n):
     
    # Corner cases
    if (n <= 1):
        return False
    if (n <= 3):
        return True
 
    # This is checked so that we can skip
    # middle five numbers in below loop
    if (n % 2 == 0 or n % 3 == 0):
        return False
     
    k = int(sqrt(n)) + 1
    for i in range(5, k, 6):
        if (n % i == 0 or n % (i + 2) == 0):
            return False
 
    return True
 
# Function that returns true if
# n is a strong prime
def isStrongPrime(n):
     
    # If n is not a prime number or
    # n is the first prime then return false
    if (isPrime(n) == False or n == 2):
        return False
 
    # Initialize previous_prime to n - 1
    # and next_prime to n + 1
    previous_prime = n - 1
    next_prime = n + 1
 
    # Find next prime number
    while (isPrime(next_prime) == False):
        next_prime += 1
 
    # Find previous prime number
    while (isPrime(previous_prime) == False):
        previous_prime -= 1
 
    # Arithmetic mean
    mean = (previous_prime + next_prime) / 2
 
    # If n is a strong prime
    if (n > mean):
        return True
    else:
        return False
 
# Driver code
if __name__ == '__main__':
    n = 11
 
    if (isStrongPrime(n)):
        print("Yes")
    else:
        print("No")
 
# This code is contributed by
# Sanjit_prasad

C#

// C# program to check if a given number is strong prime
using System;
class GFG {
 
    // Utility function to check
    // if a number is prime or not
    static bool isPrime(int n)
    {
        // Corner cases
        if (n <= 1)
            return false;
        if (n <= 3)
            return true;
 
        // This is checked so that we can skip
        // middle five numbers in below loop
        if (n % 2 == 0 || n % 3 == 0)
            return false;
 
        for (int i = 5; i * i <= n; i = i + 6)
            if (n % i == 0 || n % (i + 2) == 0)
                return false;
 
        return true;
    }
 
    // Function that returns true if n is a strong prime
    static bool isStrongPrime(int n)
    {
        // If n is not a prime number or
        // n is the first prime then return false
        if (!isPrime(n) || n == 2)
            return false;
 
        // Initialize previous_prime to n - 1
        // and next_prime to n + 1
        int previous_prime = n - 1;
        int next_prime = n + 1;
 
        // Find next prime number
        while (!isPrime(next_prime))
            next_prime++;
 
        // Find previous prime number
        while (!isPrime(previous_prime))
            previous_prime--;
 
        // Arithmetic mean
        int mean = (previous_prime + next_prime) / 2;
 
        // If n is a strong prime
        if (n > mean)
            return true;
        else
            return false;
    }
 
    // Driver code
    public static void Main()
    {
        int n = 11;
 
        if (isStrongPrime(n))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}

PHP

 $mean)
        return true;
    else
        return false;
}
 
// Driver code
$n = 11;
 
if (isStrongPrime($n))
    echo ("Yes");
else
    echo("No");
 
// This code is contributed
// by Shivi_Aggarwal
?>

Javascript

 
// Javascript program to check if
// given number is strong prime
 
// Utility function to check
// if a number is prime or not
function isPrime(n)
{
     
    // Corner cases
    if (n <= 1)
        return false;
    if (n <= 3)
        return true;
 
    // This is checked so that we can skip
    // middle five numbers in below loop
    if (n % 2 == 0 || n % 3 == 0)
        return false;
 
    for(let i = 5; i * i <= n; i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return false;
 
    return true;
}
 
// Function that returns true if
// n is a strong prime
function isStrongPrime(n)
{
     
    // If n is not a prime number or
    // n is the first prime then return false
    if (!isPrime(n) || n == 2)
        return false;
 
    // Initialize previous_prime to n - 1
    // and next_prime to n + 1
    let previous_prime = n - 1;
    let next_prime = n + 1;
 
    // Find next prime number
    while (!isPrime(next_prime))
        next_prime++;
 
    // Find previous prime number
    while (!isPrime(previous_prime))
        previous_prime--;
 
    // Arithmetic mean
    let mean = parseInt((previous_prime +
                         next_prime) / 2);
 
    // If n is a strong prime
    if (n > mean)
        return true;
    else
        return false;
}
 
// Driver code
let n = 11;
 
if (isStrongPrime(n))
    document.write("Yes");
else
    document.write("No");
     
// This code is contributed by souravmahato348
 

输出：

Yes