对数组中的对进行计数，以使至少一个元素为质数

给定一个由不同元素组成的数组arr [] ，任务是计算其中至少一个元素为质数的不同对的总数。
例子：

Input: arr[] = {1, 3, 10, 7, 8}
Output: 7
Pairs with at least one prime are (1, 3), (1, 7), 
(3, 1), (3, 7), (3, 8), (10, 7), (7, 8).

Input:arr[]={4, 6, 8, 2, 9};
Output: 4

方法：首先使用Sieve预先计算所有素数，直到数组的Maximum元素 。遍历每个可能的对，并检查对中是否有任何素数。如果是，则增加计数。
下面是上述方法的实现：

C++

// C++ implementation of the above approach
#include 
using namespace std;
 
// Function to find primes
void sieve(int maxm, int prime[])
{
    prime[0] = prime[1] = 1;
 
    for (int i = 2; i * i <= maxm; i++)
        if (!prime[i])
            for (int j = 2 * i; j <= maxm; j += i)
                prime[j] = 1;
}
 
// Function to count the pair
int countPair(int a[], int n)
{
    // Find the maximum element of the array
    int maxm = *max_element(a, a + n);
    int prime[maxm + 1];
    memset(prime, 0, sizeof(prime));
 
    // Find primes upto maximum
    sieve(maxm, prime);
 
    // Count pairs with at least prime
    int count = 0;
    for (int i = 0; i < n; i++)
        for (int j = i + 1; j < n; j++)
            if (prime[a[i]] == 0 || prime[a[j]] == 0)
                count++;
 
    return count;
}
 
// Driver code
int main()
{
    int arr[] = { 2, 3, 5, 4, 7 };
    int n = 5;
 
    cout << countPair(arr, n);
 
    return 0;
}

Java

// Java implementation of the above approach
 
class GFG
{
    // Function to find primes
    static void sieve(int maxm, int prime[])
    {
        prime[0] = prime[1] = 1;
     
        for (int i = 2; i * i <= maxm; i++)
            if (prime[i] == 0)
                for (int j = 2 * i; j <= maxm; j += i)
                    prime[j] = 1;
    }
     
    // Function to count the pair
    static int countPair(int a[], int n)
    {
        // Find the maximum element of the array
        int maxm = a[0];
         
        for(int i = 1; i < n; i++)
            if(a[i] > maxm)
            maxm = a[i];
         
        int [] prime = new int[maxm + 1];
         
        for(int i = 0; i < maxm + 1; i++)
            prime[i] = 0;
     
        // Find primes upto maximum
        sieve(maxm, prime);
     
        // Count pairs with at least prime
        int count = 0;
        for (int i = 0; i < n; i++)
            for (int j = i + 1; j < n; j++)
                if (prime[a[i]] == 0 || prime[a[j]] == 0)
                    count++;
     
        return count;
    }
     
    // Driver code
    public static void main(String []args)
    {
        int arr[] = { 2, 3, 5, 4, 7 };
        int n = arr.length;
        System.out.println(countPair(arr, n));
    }
}
 
// This code is contributed by ihritik

Python3

# Python 3 implementation of the above approach
from math import sqrt
 
# Function to count the pair
def countPair(a, n):
     
    # Find the maximum element of the array
    maxm = a[0]
    for i in range(len(a)):
        if(a[i] > maxm):
            maxm = a[i]
    prime = [0 for i in range(maxm + 1)]
 
    # Find primes upto maximum
    prime[0] = prime[1] = 1;
 
    for i in range(2, int(sqrt(maxm)) + 1, 1):
        if (prime[i] == 0):
            for j in range(2 * i, maxm + 1, i):
                prime[j] = 1
 
    # Count pairs with at least prime
    count = 0
    for i in range(n):
        for j in range(i + 1, n, 1):
            if (prime[a[i]] == 0 or
                prime[a[j]] == 0):
                count += 1
 
    return count
 
# Driver code
if __name__ == '__main__':
    arr = [2, 3, 5, 4, 7]
    n = 5
 
    print(countPair(arr, n))
 
# This code is contributed by
# Sanjit_Prasad

C#

// C# implementation of the above approach
using System;
 
class GFG
{
    // Function to find primes
    static void sieve(int maxm, int []prime)
    {
        prime[0] = prime[1] = 1;
     
        for (int i = 2; i * i <= maxm; i++)
            if (prime[i] == 0)
                for (int j = 2 * i; j <= maxm; j += i)
                    prime[j] = 1;
    }
     
    // Function to count the pair
    static int countPair(int []a, int n)
    {
        // Find the maximum element of the array
        int maxm = a[0];
         
        for(int i = 1; i < n; i++)
            if(a[i] > maxm)
            maxm = a[i];
         
        int [] prime = new int[maxm + 1];
         
        for(int i = 0; i < maxm + 1; i++)
            prime[i] = 0;
     
        // Find primes upto maximum
        sieve(maxm, prime);
     
        // Count pairs with at least prime
        int count = 0;
        for (int i = 0; i < n; i++)
            for (int j = i + 1; j < n; j++)
                if (prime[a[i]] == 0 || prime[a[j]] == 0)
                    count++;
     
        return count;
    }
     
    // Driver code
    public static void Main()
    {
        int []arr = { 2, 3, 5, 4, 7 };
        int n = arr.Length;
        Console.WriteLine(countPair(arr, n));
    }
}
 
// This code is contributed by ihritik

Javascript

 
// Javascript implementation of the above approach
     
    // Function to find primes
    function sieve(maxm,prime)
    {
        prime[0] = prime[1] = 1;
       
        for (let i = 2; i * i <= maxm; i++)
            if (prime[i] == 0)
                for (let j = 2 * i; j <= maxm; j += i)
                    prime[j] = 1;
    }
     
    // Function to count the pair
    function countPair(a,n)
    {
        // Find the maximum element of the array
        let maxm = a[0];
           
        for(let i = 1; i < n; i++)
            if(a[i] > maxm)
            maxm = a[i];
           
        let prime = new Array(maxm + 1);
           
        for(let i = 0; i < maxm + 1; i++)
            prime[i] = 0;
       
        // Find primes upto maximum
        sieve(maxm, prime);
       
        // Count pairs with at least prime
        let count = 0;
        for (let i = 0; i < n; i++)
            for (let j = i + 1; j < n; j++)
                if (prime[a[i]] == 0 || prime[a[j]] == 0)
                    count++;
       
        return count;
    }
     
    // Driver code
    let arr=[2, 3, 5, 4, 7];
    let n = arr.length;
    document.write(countPair(arr, n));
     
    // This code is contributed by unknown2108
     
     

输出：

10

高效方法：
方法：首先使用Sieve预先计算所有素数，直到数组的Maximum元素 。保持素数和非素数的计数。然后计算具有单个质数的对，即nonPrimes *质数，并计算具有两个质数的对(Primes *(Primes – 1))/ 2 。返回两个计数之和。
下面是上述方法的实现：

C++

// C++ implementation of the above approach
#include 
#define ll long long int
using namespace std;
 
// Function to find primes
void sieve(int maxm, int prime[])
{
    prime[0] = prime[1] = 1;
 
    for (int i = 2; i * i <= maxm; i++)
        if (!prime[i])
            for (int j = 2 * i; j <= maxm; j += i)
                prime[j] = 1;
}
 
ll countPair(int a[], int n)
{
    // Find the maximum element of the array
    int maxm = *max_element(a, a + n);
    int prime[maxm + 1];
    memset(prime, 0, sizeof(prime));
 
    // Find primes upto maximum
    sieve(maxm, prime);
 
    // Count number of primes
    int countPrimes = 0;
    for (int i = 0; i < n; i++)
        if (prime[a[i]] == 0)
            countPrimes++;
 
    int nonPrimes = n - countPrimes;
    ll pairswith1Prime = nonPrimes * countPrimes;
    ll pairsWith2Primes = (countPrimes * (countPrimes - 1)) / 2;
 
    return pairswith1Prime + pairsWith2Primes;
}
 
// Driver code
int main()
{
    int arr[] = { 2, 3, 5, 4, 7 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << countPair(arr, n);
 
    return 0;
}

Java

// Java implementation of the above approach
 
class GFG
{
    // Function to find primes
    static void sieve(int maxm, int []prime)
    {
        prime[0] = prime[1] = 1;
 
        for (int i = 2; i * i <= maxm; i++)
            if (prime[i]==0)
                for (int j = 2 * i; j <= maxm; j += i)
                    prime[j] = 1;
    }
 
    static long countPair(int []a, int n)
    {
        // Find the maximum element of the array
        int maxm = a[0];
         
        int i;
        for( i = 1; i < n ; i++)
            if(a[i] > maxm)
                maxm = a[i];
         
        int [] prime = new int[maxm + 1];
         
        for( i = 0; i < maxm + 1 ;i++)
            prime[i] = 0;
             
        // Find primes upto maximum
        sieve(maxm, prime);
     
        // Count number of primes
        int countPrimes = 0;
        for ( i = 0; i < n; i++)
            if (prime[a[i]] == 0)
                countPrimes++;
     
        int nonPrimes = n - countPrimes;
        long pairswith1Prime = nonPrimes *
                                countPrimes;
        long pairsWith2Primes = (countPrimes *
                            (countPrimes - 1)) / 2;
     
        return pairswith1Prime + pairsWith2Primes;
    }
     
    // Driver code
    public static void main(String []args)
    {
        int [] arr = { 2, 3, 5, 4, 7 };
        int n = arr.length;
     
        System.out.println(countPair(arr, n));
    }
}
 
// This code is contributed by ihritik

Python3

# Python3 implementation of the above approach
 
# Function to find primes
def sieve(maxm, prime):
 
    prime[0] = prime[1] = 1;
    i = 2;
 
    while (i * i <= maxm):
        if (prime[i] == 0):
            for j in range(2 * i, maxm + 1, i):
                prime[j] = 1;
        i += 1;
 
def countPair(a, n):
 
    # Find the maximum element
    # of the array
    maxm = max(a);
    prime = [0] * (maxm + 1);
 
    # Find primes upto maximum
    sieve(maxm, prime);
 
    # Count number of primes
    countPrimes = 0;
    for i in range(n):
        if (prime[a[i]] == 0):
            countPrimes += 1;
 
    nonPrimes = n - countPrimes;
    pairswith1Prime = nonPrimes * countPrimes;
    pairsWith2Primes = (countPrimes *
                       (countPrimes - 1)) // 2;
 
    return pairswith1Prime + pairsWith2Primes;
 
# Driver code
arr = [ 2, 3, 5, 4, 7 ];
n = len(arr);
 
print(countPair(arr, n));
 
# This code is contributed by mits

C＃

// C# implementation of the above approach
using System;
 
class GFG
{
    // Function to find primes
    static void sieve(int maxm, int []prime)
    {
        prime[0] = prime[1] = 1;
 
        for (int i = 2; i * i <= maxm; i++)
            if (prime[i] == 0)
                for (int j = 2 * i; j <= maxm; j += i)
                    prime[j] = 1;
    }
 
    static long countPair(int []a, int n)
    {
        // Find the maximum element of the array
        int maxm = a[0];
         
        int i;
        for( i = 1; i < n ;i++)
            if(a[i] > maxm)
                maxm = a[i];
         
        int [] prime = new int[maxm + 1];
         
        for( i = 0; i < maxm + 1 ;i++)
            prime[i] = 0;
             
        // Find primes upto maximum
        sieve(maxm, prime);
     
        // Count number of primes
        int countPrimes = 0;
        for ( i = 0; i < n; i++)
            if (prime[a[i]] == 0)
                countPrimes++;
     
        int nonPrimes = n - countPrimes;
        long pairswith1Prime = nonPrimes *
                                countPrimes;
        long pairsWith2Primes = (countPrimes *
                            (countPrimes - 1)) / 2;
     
        return pairswith1Prime + pairsWith2Primes;
    }
     
    // Driver code
    public static void Main()
    {
        int [] arr = { 2, 3, 5, 4, 7 };
        int n = arr.Length;
        Console.WriteLine(countPair(arr, n));
    }
}
 
// This code is contributed by ihritik

输出：

10