计算数组中的孪生素数对

给定一个由N 个自然数组成的数组arr[] 。任务是计算arr[]中所有可能的孪生素数对。
孪生素数是那些素数并且两个素数之间相差二 (2) 的数。换句话说，孪生素数是素数间隙为2的素数。

例子：

Input: arr[] = { 2, 3, 5, 7} 
Output: 2 
Explanation: 
The 2 pairs are (3, 5) and (5, 7).

Input: arr[] = { 2, 4, 6, 11} 
Output: 0 
Explanation: 
There are no such pairs forming Twin Primes. 

天真的方法：
这个想法是在给定的数组arr[] 中找到所有可能的对，并检查对中的两个元素是否都是素数并且它们相差2 ，然后当前对形成孪生素数。

下面是上述方法的实现：

C++

// C++ program to count Twin
// Prime pairs in array
#include 
using namespace std;
 
// A utility function to check if
// the number n is prime or not
bool isPrime(int n)
{
    // Base Cases
    if (n <= 1)
        return false;
    if (n <= 3)
        return true;
 
    // Check to skip middle five
    // numbers in below loop
    if (n % 2 == 0 || n % 3 == 0)
        return false;
 
    for (int i = 5; i * i <= n; i += 6) {
 
        // If n is divisible by i and i+2
        // then it is not prime
        if (n % i == 0 || n % (i + 2) == 0) {
            return false;
        }
    }
 
    return true;
}
 
// A utility function that check
// if n1 and n2 are Twin Primes
// or not
bool twinPrime(int n1, int n2)
{
    return (isPrime(n1)
            && isPrime(n2)
            && abs(n1 - n2) == 2);
}
 
// Function to find Twin Prime
// pairs from the given array
int countTwinPairs(int arr[], int n)
{
    int count = 0;
 
    // Iterate through all pairs
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
 
            // Increment count if
            // twin prime pair
            if (twinPrime(arr[i], arr[j])) {
                count++;
            }
        }
    }
 
    return count;
}
 
// Driver's code
int main()
{
    int arr[] = { 2, 3, 5, 11 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    // Function call to find
    // Twin Primes pair
    cout << countTwinPairs(arr, n);
    return 0;
}

Java

// Java program to count Twin
// Prime pairs in array
import java.util.*;
 
class GFG{
  
// A utility function to check if
// the number n is prime or not
static boolean isPrime(int n)
{
    // Base Cases
    if (n <= 1)
        return false;
    if (n <= 3)
        return true;
  
    // Check to skip middle five
    // numbers in below loop
    if (n % 2 == 0 || n % 3 == 0)
        return false;
  
    for (int i = 5; i * i <= n; i += 6) {
  
        // If n is divisible by i and i+2
        // then it is not prime
        if (n % i == 0 || n % (i + 2) == 0) {
            return false;
        }
    }
  
    return true;
}
  
// A utility function that check
// if n1 and n2 are Twin Primes
// or not
static boolean twinPrime(int n1, int n2)
{
    return (isPrime(n1)
            && isPrime(n2)
            && Math.abs(n1 - n2) == 2);
}
  
// Function to find Twin Prime
// pairs from the given array
static int countTwinPairs(int arr[], int n)
{
    int count = 0;
  
    // Iterate through all pairs
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
  
            // Increment count if
            // twin prime pair
            if (twinPrime(arr[i], arr[j])) {
                count++;
            }
        }
    }
  
    return count;
}
  
// Driver's code
public static void main(String[] args)
{
    int arr[] = { 2, 3, 5, 11 };
    int n = arr.length;
  
    // Function call to find
    // Twin Primes pair
    System.out.print(countTwinPairs(arr, n));
}
}
 
// This code is contributed by Rajput-Ji

Python 3

# Python 3 program to count Twin
# Prime pairs in array
from math import sqrt
 
# A utility function to check if
# the number n is prime or not
def isPrime(n):
     
    # Base Cases
    if (n <= 1):
        return False
    if (n <= 3):
        return True
 
    # Check to skip middle five
    # numbers in below loop
    if (n % 2 == 0 or n % 3 == 0):
        return False
 
    for i in range(5,int(sqrt(n))+1,6):
         
        # If n is divisible by i and i+2
        # then it is not prime
        if (n % i == 0 or n % (i + 2) == 0):
            return False
 
    return True
 
# A utility function that check
# if n1 and n2 are Twin Primes
# or not
def twinPrime(n1, n2):
    return (isPrime(n1) and isPrime(n2) and abs(n1 - n2) == 2)
 
# Function to find Twin Prime
# pairs from the given array
def countTwinPairs(arr, n):
    count = 0
 
    # Iterate through all pairs
    for i in range(n):
        for j in range(i + 1,n):
             
            # Increment count if
            # twin prime pair
            if (twinPrime(arr[i], arr[j])):
                count += 1
 
    return count
 
# Driver's code
if __name__ == '__main__':
    arr = [2, 3, 5, 11]
    n = len(arr)
 
    # Function call to find
    # Twin Primes pair
    print(countTwinPairs(arr, n))
 
# This code is contributed by Surendra_Gangwar

C#

// C# program to count Twin
// Prime pairs in array
using System;
 
class GFG{
   
// A utility function to check if
// the number n is prime or not
static bool isPrime(int n)
{
    // Base Cases
    if (n <= 1)
        return false;
    if (n <= 3)
        return true;
   
    // Check to skip middle five
    // numbers in below loop
    if (n % 2 == 0 || n % 3 == 0)
        return false;
   
    for (int i = 5; i * i <= n; i += 6) {
   
        // If n is divisible by i and i+2
        // then it is not prime
        if (n % i == 0 || n % (i + 2) == 0) {
            return false;
        }
    }
   
    return true;
}
   
// A utility function that check
// if n1 and n2 are Twin Primes
// or not
static bool twinPrime(int n1, int n2)
{
    return (isPrime(n1)
            && isPrime(n2)
            && Math.Abs(n1 - n2) == 2);
}
   
// Function to find Twin Prime
// pairs from the given array
static int countTwinPairs(int []arr, int n)
{
    int count = 0;
   
    // Iterate through all pairs
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
   
            // Increment count if
            // twin prime pair
            if (twinPrime(arr[i], arr[j])) {
                count++;
            }
        }
    }
   
    return count;
}
   
// Driver's code
public static void Main(String[] args)
{
    int []arr = { 2, 3, 5, 11 };
    int n = arr.Length;
   
    // Function call to find
    // Twin Primes pair
    Console.Write(countTwinPairs(arr, n));
}
}
 
// This code is contributed by 29AjayKumar

Javascript

// Javascipt implementation of the
// above approach
 
// A utility function to check if
// the number n is prime or not
function isPrime(n)
{
    // Base Cases
    if (n <= 1)
        return false;
    if (n <= 3)
        return true;
   
    // Check to skip middle five
    // numbers in below loop
    if (n % 2 == 0 || n % 3 == 0)
        return false;
   
    for (var i = 5; i * i <= n; i += 6) {
   
        // If n is divisible by i and i+2
        // then it is not prime
        if (n % i == 0 || n % (i + 2) == 0) {
            return false;
        }
    }
   
    return true;
}
   
// A utility function that check
// if n1 and n2 are Twin Primes
// or not
function twinPrime(n1, n2)
{
    return (isPrime(n1)
            && isPrime(n2)
            && Math.abs(n1 - n2) == 2);
}
   
// Function to find Twin Prime
// pairs from the given array
function countTwinPairs(arr, n)
{
    var count = 0;
   
    // Iterate through all pairs
    for (var i = 0; i < n; i++) {
        for (var j = i + 1; j < n; j++) {
   
            // Increment count if
            // twin prime pair
            if (twinPrime(arr[i], arr[j])) {
                count++;
            }
        }
    }
   
    return count;
}
 
// Driver code
var arr=[ 2, 3, 5, 11 ];
var n = arr.length;
 
document.write(countTwinPairs(arr, n));
 
// This code is contributed by Shivanisingh

输出：

1

时间复杂度： O(sqrt(M)*N ² )，其中 N 是给定数组中的元素数，M 是数组中的最大元素。

有效的方法：

1. 使用 Eratosthenes 筛法预先计算所有质数直到给定数组arr[]中的最大数。
2. 存储给定数组arr[]的所有元素的所有频率。
3. 对给定的数组arr[] 进行排序。
4. 对于数组中的每个元素，检查该元素是否为素数。
5. 如果元素是素数，则检查 (element+2) 是否是素数并且是否存在于给定的数组arr[] 中。
6. 如果 (element+2) 存在，则 (element+2) 的频率将给出当前元素的对数。
7. 对数组中的所有元素重复上述步骤。