总和为Prime的数组中的对数

给定一个大小为N个元素的数组arr ，任务是计算数组中总和为素数的元素对的数量。

例子：

Input: arr = {1, 2, 3, 4, 5} 
Output: 5 
Explanation: Pairs with sum as a prime number are: {1, 2}, {1, 4}, {2, 3}, {2, 5} and {3, 4}

Input: arr = {10, 20, 30, 40} 
Output: 0 
Explanation: No pair whose sum is a prime number exists. 
 

天真的方法：
计算数组中每对元素的和，并检查该和是否为质数。

下面的代码是上述方法的实现：

C++

// C++ code to count of pairs
// of elements in an array
// whose sum is prime
#include 
using namespace std;
 
// Function to check whether a
// number is prime or not
bool isPrime(int num)
{
    if (num == 0 || num == 1) {
        return false;
    }
    for (int i = 2; i * i <= num; i++) {
        if (num % i == 0) {
            return false;
        }
    }
    return true;
}
 
// Function to count total number of pairs
// of elements whose sum is prime
int numPairsWithPrimeSum(int* arr, int n)
{
    int count = 0;
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            int sum = arr[i] + arr[j];
            if (isPrime(sum)) {
                count++;
            }
        }
    }
    return count;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << numPairsWithPrimeSum(arr, n);
    return 0;
}

Java

// Java code to find number of pairs of
// elements in an array whose sum is prime
import java.io.*;
import java.util.*;
 
class GFG {
 
    // Function to check whether a number
    // is prime or not
    public static boolean isPrime(int num)
    {
        if (num == 0 || num == 1) {
            return false;
        }
        for (int i = 2; i * i <= num; i++) {
            if (num % i == 0) {
                return false;
            }
        }
        return true;
    }
 
    // Function to count total number of pairs
    // of elements whose sum is prime
    public static int numPairsWithPrimeSum(
        int[] arr, int n)
    {
        int count = 0;
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                int sum = arr[i] + arr[j];
                if (isPrime(sum)) {
                    count++;
                }
            }
        }
        return count;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int[] arr = { 1, 2, 3, 4, 5 };
        int n = arr.length;
        System.out.println(
            numPairsWithPrimeSum(arr, n));
    }
}

Python3

# Python3 code to find number of pairs of
# elements in an array whose sum is prime
import math
 
# Function to check whether a
# number is prime or not
def isPrime(num):
     
    sq = int(math.ceil(math.sqrt(num)))
 
    if num == 0 or num == 1:
        return False
 
    for i in range(2, sq + 1):
        if num % i == 0:
            return False
 
    return True
 
# Function to count total number of pairs
# of elements whose sum is prime
def numPairsWithPrimeSum(arr, n):
     
    count = 0
     
    for i in range(n):
        for j in range(i + 1, n):
            sum = arr[i] + arr[j]
             
            if isPrime(sum):
                count += 1
                 
    return count
 
# Driver Code
arr = [ 1, 2, 3, 4, 5 ]
n = len(arr)
 
print(numPairsWithPrimeSum(arr, n))
 
# This code is contributed by grand_master

C#

// C# code to find number of pairs of
// elements in an array whose sum is prime
using System;
class GFG{
 
// Function to check whether a number
// is prime or not
public static bool isPrime(int num)
{
    if (num == 0 || num == 1)
    {
        return false;
    }
    for (int i = 2; i * i <= num; i++)
    {
        if (num % i == 0)
        {
            return false;
        }
    }
    return true;
}
 
// Function to count total number of pairs
// of elements whose sum is prime
public static int numPairsWithPrimeSum(int[] arr,
                                       int n)
{
    int count = 0;
    for (int i = 0; i < n; i++)
    {
        for (int j = i + 1; j < n; j++)
        {
            int sum = arr[i] + arr[j];
            if (isPrime(sum))
            {
                count++;
            }
        }
    }
    return count;
}
 
// Driver code
public static void Main()
{
    int[] arr = { 1, 2, 3, 4, 5 };
    int n = arr.Length;
    Console.Write(numPairsWithPrimeSum(arr, n));
}
}
 
// This code is contributed by Nidhi_Biet

输出：

5

时间复杂度：

$O(N^2 * \sqrt{sum})$

高效方法：
使用Eratosthenes筛子预先计算并存储素数。现在，对于每对元素，检查它们的和是否为素数。

下面的代码是上述方法的实现：

C++

// C++ code to find number of pairs
// of elements in an array whose
// sum is prime
#include 
using namespace std;
 
// Function for Sieve Of Eratosthenes
bool* sieveOfEratosthenes(int N)
{
    bool* isPrime = new bool[N + 1];
    for (int i = 0; i < N + 1; i++) {
        isPrime[i] = true;
    }
    isPrime[0] = false;
    isPrime[1] = false;
    for (int i = 2; i * i <= N; i++) {
        if (isPrime[i] == true) {
            int j = 2;
            while (i * j <= N) {
                isPrime[i * j] = false;
                j++;
            }
        }
    }
    return isPrime;
}
 
// Function to count total number of pairs
// of elements whose sum is prime
int numPairsWithPrimeSum(int* arr, int n)
{
    int N = 2 * 1000000;
    bool* isPrime = sieveOfEratosthenes(N);
    int count = 0;
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            int sum = arr[i] + arr[j];
            if (isPrime[sum]) {
                count++;
            }
        }
    }
    return count;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << numPairsWithPrimeSum(arr, n);
    return 0;
}

Java

// Java code to find number of pairs of
// elements in an array whose sum is prime
import java.io.*;
import java.util.*;
 
class GFG {
    // Function for Sieve Of Eratosthenes
    public static boolean[] sieveOfEratosthenes(int N)
    {
        boolean[] isPrime = new boolean[N + 1];
        for (int i = 0; i < N + 1; i++) {
            isPrime[i] = true;
        }
        isPrime[0] = false;
        isPrime[1] = false;
        for (int i = 2; i * i <= N; i++) {
            if (isPrime[i] == true) {
                int j = 2;
                while (i * j <= N) {
                    isPrime[i * j] = false;
                    j++;
                }
            }
        }
        return isPrime;
    }
 
    // Function to count total number of pairs
    // of elements whose sum is prime
    public static int numPairsWithPrimeSum(
        int[] arr, int n)
    {
        int N = 2 * 1000000;
        boolean[] isPrime = sieveOfEratosthenes(N);
        int count = 0;
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                int sum = arr[i] + arr[j];
                if (isPrime[sum]) {
                    count++;
                }
            }
        }
        return count;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int[] arr = { 1, 2, 3, 4, 5 };
        int n = arr.length;
        System.out.println(
            numPairsWithPrimeSum(arr, n));
    }
}

Python3

# Python3 code to find number of pairs of
# elements in an array whose sum is prime
 
# Function for Sieve Of Eratosthenes
def sieveOfEratosthenes(N):
 
    isPrime = [True for i in range(N + 1)]
         
    isPrime[0] = False
    isPrime[1] = False
     
    i = 2
     
    while((i * i) <= N):
        if (isPrime[i]):
            j = 2
            while (i * j <= N):
                isPrime[i * j] = False
                j += 1
                 
        i += 1
         
    return isPrime
     
# Function to count total number of pairs
# of elements whose sum is prime
def numPairsWithPrimeSum(arr, n):
     
    N = 2 * 1000000
    isPrime = sieveOfEratosthenes(N)
    count = 0
     
    for i in range(n):
        for j in range(i + 1, n):
            sum = arr[i] + arr[j]
             
            if (isPrime[sum]):
                count += 1
                 
    return count
 
# Driver code   
if __name__=="__main__":
     
    arr = [ 1, 2, 3, 4, 5 ]
    n = len(arr)
     
    print(numPairsWithPrimeSum(arr, n))
 
# This code is contributed by rutvik_56

C＃

// C# code to find number of pairs of
// elements in an array whose sum is prime
using System;
 
class GFG{
     
// Function for Sieve Of Eratosthenes
public static bool[] sieveOfEratosthenes(int N)
{
    bool[] isPrime = new bool[N + 1];
    for (int i = 0; i < N + 1; i++)
    {
        isPrime[i] = true;
    }
    isPrime[0] = false;
    isPrime[1] = false;
    for (int i = 2; i * i <= N; i++)
    {
        if (isPrime[i] == true)
        {
            int j = 2;
            while (i * j <= N)
            {
                isPrime[i * j] = false;
                j++;
            }
        }
    }
    return isPrime;
}
 
// Function to count total number of pairs
// of elements whose sum is prime
public static int numPairsWithPrimeSum(int[] arr,
                                       int n)
{
    int N = 2 * 1000000;
    bool[] isPrime = sieveOfEratosthenes(N);
    int count = 0;
    for (int i = 0; i < n; i++)
    {
        for (int j = i + 1; j < n; j++)
        {
            int sum = arr[i] + arr[j];
            if (isPrime[sum])
            {
                count++;
            }
        }
    }
    return count;
}
 
// Driver code
public static void Main(String[] args)
{
    int[] arr = { 1, 2, 3, 4, 5 };
    int n = arr.Length;
    Console.WriteLine(numPairsWithPrimeSum(arr, n));
}
}
 
// This code is contributed by 29AjayKumar

输出：

5

时间复杂度： O(N ^ 2)