检查n ^ 2 – m ^ 2是否为质数

给定两个整数n和m。检查n ^ 2 – m ^ 2是否为质数。 n和m可能非常大。

例子：

Input : n = 6, m = 5
Output : YES

Input : n = 16, m = 13
Output : NO

一个简单的解决方案是先计算n ^ 2 – m ^ 2，然后检查它是否为素数。 n ^ 2 – m ^ 2可能非常大–它甚至可能不适合64位整数。检查素数当然不能天真地执行。

更好的解决方案是将n ^ 2- – m ^ 2表示为(nm)(n + m)。当且仅当nm = 1且n + m是素数时，才是素数。

C++

// CPP program to find n^2 - m^2
// is prime or not.
#include 
using namespace std;
  
// Check a number is prime or not
bool isprime(int x)
{
    // run a loop upto square of given number
    for (int i = 2; i * i <= x; i++)
        if (x % i == 0)
            return false;
    return true;
}
  
// Check if n^2 - m^2 is prime
bool isNSqMinusnMSqPrime(int m, int n)
{
    if (n - m == 1 and isprime(m + n))
        return true;
    else
        return false;
}
  
// Driver code
int main()
{
    int m = 13, n = 16;
    if (isNSqMinusnMSqPrime(m, n))
        cout << "YES";
    else
        cout << "NO";
  
    return 0;
}

Java

// Java program to find n^2 - m^2
// is prime or not.
  
class GFG
{
        // Check if a number is prime or not
        static boolean isprime(int x)
        {
            // run a loop upto square of given number
            for (int i = 2; i * i <= x; i++)
                if (x % i == 0)
                    return false;
            return true;
        }
          
        // Check if n^2 - m^2 is prime
        static boolean isNSqMinusnMSqPrime(int m, int n)
        {
            if (n - m == 1 && isprime(m + n))
                return true;
            else
                return false;
        }
          
        // Driver code
        public static void  main(String [] args)
        {
            int m = 13, n = 16;
            if (isNSqMinusnMSqPrime(m, n))
                System.out.println("YES");
            else
                System.out.println("NO");
          
        }
}
  
// This code is contributed
// by ihritik

Python3

# Python program to find n^2 - m^2 
# is prime or not. 
  
# Check a number is prime or not 
def isprime(x): 
  
    # run a loop upto square 
    # of given number 
    for i in range(2, math.sqrt(x)): 
        if (x % i == 0) :
            return False; 
    return True; 
  
# Check if n^2 - m^2 is prime 
def isNSqMinusnMSqPrime( m, n): 
  
    if (n - m == 1 and isprime(m + n)): 
        return True; 
    else:
        return False; 
  
# Driver code 
m = 13;
n = 16; 
if (isNSqMinusnMSqPrime(m, n)) :
    print ( "YES"); 
else:
    print ("NO"); 
  
# This code is contributed 
# by Shivi_Aggarwal

C#

// C# program to find n^2 - m^2 
// is prime or not. 
using System;
  
class GFG 
{ 
// Check if a number is prime or not 
static bool isprime(int x) 
{ 
    // run a loop upto square
    // of given number 
    for (int i = 2; i * i <= x; i++) 
        if (x % i == 0) 
            return false; 
    return true; 
} 
  
// Check if n^2 - m^2 is prime 
static bool isNSqMinusnMSqPrime(int m, 
                                int n) 
{ 
    if (n - m == 1 && isprime(m + n)) 
        return true; 
    else
        return false; 
} 
  
// Driver code 
public static void Main() 
{ 
    int m = 13, n = 16; 
    if (isNSqMinusnMSqPrime(m, n)) 
        Console.Write("YES"); 
    else
        Console.Write("NO"); 
} 
} 
  
// This code is contributed 
// by Smitha

输出：

NO

时间复杂度： O(sqrt(n + m))