计算其差也是素数的素数对

给定一个整数N ，任务是计算范围[1, N]内的素数对的数量，使得每对元素之间的差也是一个素数。

例子：

Input: N = 5 
Output: 2 
Explanations: 
Pair of prime numbers in the range [1, 5] whose difference between elements is also a prime number are: 
(2, 5) = 3 (Prime number) 
(3, 5) = 2 (Prime number) 
Therefore, the count of pairs of the prime numbers whose difference is also a prime number is 2. 

Input: N = 11 
Output: 4

朴素的方法：解决这个问题的最简单的方法是生成范围[1, N] 中所有可能的元素对，对于每一对，检查元素和元素之间的差是否是素数数与否。如果发现为真，则增加计数。最后，打印计数。

时间复杂度： O(N ² * √N)
辅助空间： O(1)

高效的方法：为了优化上述方法，该想法基于以下观察：

Odd number – Even Number = Odd Number 
Odd number – Odd number = Even number 
2 is the only even prime number. 
Therefore, the problem reduces to check only for those pairs of prime numbers whose difference between the elements of the pair is equal to 2. 
 

请按照以下步骤解决问题：

• 初始化一个变量，比如cntPairs来存储素数对的计数，这样每对元素之间的差异也是一个素数。
• 初始化一个数组，比如sieve[]来检查[1, N]范围内的数字是否是素数。
• 使用埃拉托色尼筛法找出[1, N]范围内的所有素数。
• 迭代范围[2, N] ，对于给定范围内的每个元素，检查筛 [i]和筛 [i – 2]是否为真。如果发现为真，则将 cntPairs的值增加2 。
• 最后，打印cntPairs的值。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
 
#include 
using namespace std;
 
// Function to find all prime
// numbers in the range [1, N]
vector SieveOfEratosthenes(
    int N)
{
   
    // isPrime[i]: Stores if i is
    // a prime number or not
    vector isPrime(N, true);
   
    isPrime[0] = false;
    isPrime[1] = false;
   
    // Calculate all prime numbers up to
    // Max using Sieve of Eratosthenes
    for (int p = 2; p * p <= N; p++) {
   
        // If P is a prime number
        if (isPrime[p]) {
   
            // Set all multiple of P
            // as non-prime
            for (int i = p * p; i <= N;
                 i += p) {
   
                // Update isPrime
                isPrime[i] = false;
            }
        }
    }
    return isPrime;
}
 
// Function to count pairs of
// prime numbers in the range [1, N]
// whose difference is prime
int cntPairsdiffOfPrimeisPrime(int N)
{
     
    // Function to count pairs of
    // prime numbers whose difference  
    // is also a prime number
    int cntPairs = 0;
     
     
    // isPrime[i]: Stores if i is
    // a prime number or not
    vector isPrime
          = SieveOfEratosthenes(N);
     
    // Iterate over the range [2, N]
    for (int i = 2; i <= N; i++) {
         
         
        // If i and i - 2 is
        // a prime number
        if (isPrime[i] &&
            isPrime[i - 2]) {
               
               
            // Update cntPairs
            cntPairs += 2;
        }
    }
    return cntPairs;
}
 
// Driver Code
int main()
{
    int N = 5;
    cout << cntPairsdiffOfPrimeisPrime(N);
    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
     
// Function to find all prime
// numbers in the range [1, N]
public static boolean[] SieveOfEratosthenes(int N)
{
     
    // isPrime[i]: Stores if i is
    // a prime number or not
    boolean[] isPrime = new boolean[N + 1];
    Arrays.fill(isPrime, true);
 
    isPrime[0] = false;
    isPrime[1] = false;
 
    // Calculate all prime numbers up to
    // Max using Sieve of Eratosthenes
    for(int p = 2; p * p <= N; p++)
    {
         
        // If P is a prime number
        if (isPrime[p])
        {
             
            // Set all multiple of P
            // as non-prime
            for(int i = p * p; i <= N; i += p)
            {
                 
                // Update isPrime
                isPrime[i] = false;
            }
        }
    }
    return isPrime;
}
 
// Function to count pairs of
// prime numbers in the range [1, N]
// whose difference is prime
public static int cntPairsdiffOfPrimeisPrime(int N)
{
     
    // Function to count pairs of
    // prime numbers whose difference
    // is also a prime number
    int cntPairs = 0;
 
    // isPrime[i]: Stores if i is
    // a prime number or not
    boolean[] isPrime = SieveOfEratosthenes(N);
 
    // Iterate over the range [2, N]
    for(int i = 2; i <= N; i++)
    {
         
        // If i and i - 2 is
        // a prime number
        if (isPrime[i] && isPrime[i - 2])
        {
             
            // Update cntPairs
            cntPairs += 2;
        }
    }
    return cntPairs;
}
 
// Driver Code
public static void main(String args[])
{
    int N = 5;
     
    System.out.println(cntPairsdiffOfPrimeisPrime(N));
}
}
 
// This code is contributed by hemanth gadarla

Python3

# Python3 program to implement
# the above approach
from math import sqrt
 
# Function to find all prime
# numbers in the range [1, N]
def SieveOfEratosthenes(N):
     
    # isPrime[i]: Stores if i is
    # a prime number or not
    isPrime = [True for i in range(N + 1)]
   
    isPrime[0] = False
    isPrime[1] = False
   
    # Calculate all prime numbers up to
    # Max using Sieve of Eratosthenes
    for p in range(2, int(sqrt(N)) + 1, 1):
         
        # If P is a prime number
        if (isPrime[p]):
             
            # Set all multiple of P
            # as non-prime
            for i in range(p * p, N + 1, p):
                 
                # Update isPrime
                isPrime[i] = False
                 
    return isPrime
 
# Function to count pairs of
# prime numbers in the range [1, N]
# whose difference is prime
def cntPairsdiffOfPrimeisPrime(N):
     
    # Function to count pairs of
    # prime numbers whose difference  
    # is also a prime number
    cntPairs = 0
     
    # isPrime[i]: Stores if i is
    # a prime number or not
    isPrime = SieveOfEratosthenes(N)
     
    # Iterate over the range [2, N]
    for i in range(2, N + 1, 1):
         
        # If i and i - 2 is
        # a prime number
        if (isPrime[i] and isPrime[i - 2]):
             
            # Update cntPairs
            cntPairs += 2
 
    return cntPairs
 
# Driver Code
if __name__ == '__main__':
     
    N = 5
     
    print(cntPairsdiffOfPrimeisPrime(N))
 
# This code is contributed by ipg2016107

C#

// C# program to implement
// the above approach 
using System;
  
class GFG{
     
// Function to find all prime
// numbers in the range [1, N]
public static bool[] SieveOfEratosthenes(int N)
{
     
    // isPrime[i]: Stores if i is
    // a prime number or not
    bool[] isPrime = new bool[N + 1];
    for(int i = 0; i < N + 1; i++)
    {
        isPrime[i] = true;
    }
  
    isPrime[0] = false;
    isPrime[1] = false;
  
    // Calculate all prime numbers up to
    // Max using Sieve of Eratosthenes
    for(int p = 2; p * p <= N; p++)
    {
         
        // If P is a prime number
        if (isPrime[p])
        {
             
            // Set all multiple of P
            // as non-prime
            for(int i = p * p; i <= N; i += p)
            {
                 
                // Update isPrime
                isPrime[i] = false;
            }
        }
    }
    return isPrime;
}
  
// Function to count pairs of
// prime numbers in the range [1, N]
// whose difference is prime
public static int cntPairsdiffOfPrimeisPrime(int N)
{
     
    // Function to count pairs of
    // prime numbers whose difference
    // is also a prime number
    int cntPairs = 0;
  
    // isPrime[i]: Stores if i is
    // a prime number or not
    bool[] isPrime = SieveOfEratosthenes(N);
  
    // Iterate over the range [2, N]
    for(int i = 2; i <= N; i++)
    {
         
        // If i and i - 2 is
        // a prime number
        if (isPrime[i] && isPrime[i - 2])
        {
             
            // Update cntPairs
            cntPairs += 2;
        }
    }
    return cntPairs;
}
  
// Driver Code
public static void Main()
{
    int N = 5;
      
    Console.WriteLine(cntPairsdiffOfPrimeisPrime(N));
}
}
 
// This code is contributed by susmitakundugoaldanga

Javascript

 
// JavaScript program to implement
// the above approach
 
// Function to find all prime
// numbers in the range [1, N]
function SieveOfEratosthenes(N)
{
      
    // isPrime[i]: Stores if i is
    // a prime number or not
    let isPrime = [];
    for(let i = 0; i < N + 1; i++)
    {
        isPrime[i] = true;
    }
  
    isPrime[0] = false;
    isPrime[1] = false;
  
    // Calculate all prime numbers up to
    // Max using Sieve of Eratosthenes
    for(let p = 2; p * p <= N; p++)
    {
          
        // If P is a prime number
        if (isPrime[p])
        {
              
            // Set all multiple of P
            // as non-prime
            for(let i = p * p; i <= N; i += p)
            {
                  
                // Update isPrime
                isPrime[i] = false;
            }
        }
    }
    return isPrime;
}
  
// Function to count pairs of
// prime numbers in the range [1, N]
// whose difference is prime
function cntPairsdiffOfPrimeisPrime(N)
{
      
    // Function to count pairs of
    // prime numbers whose difference
    // is also a prime number
    let cntPairs = 0;
  
    // isPrime[i]: Stores if i is
    // a prime number or not
    let isPrime = SieveOfEratosthenes(N);
  
    // Iterate over the range [2, N]
    for(let i = 2; i <= N; i++)
    {
          
        // If i and i - 2 is
        // a prime number
        if (isPrime[i] && isPrime[i - 2])
        {
              
            // Update cntPairs
            cntPairs += 2;
        }
    }
    return cntPairs;
}
 
// Driver Code
 
    let N = 5;
      
    document.write(cntPairsdiffOfPrimeisPrime(N));
      

输出：

2

时间复杂度： O(N * log(log(N)))
辅助空间： O(N)