查找给定乘积的两个不同质数

给定数字N(大于2)。任务是找到两个不同的质数，它们的乘积将等于给定数。可能有几种组合。仅打印第一对。
如果不可能将N表示为两个不同素数的乘积，请打印“不可能”。

例子：

Input : N = 15
Output : 3, 5
3 and 5 are both primes and,
3*5 = 15

Input : N = 39
Output : 3, 13
3 and 13 are both primes and,
3*13 = 39

这个想法是使用Eratosthenes筛子找到所有小于或等于给定数N的素数。一旦有了一个告诉所有素数的数组，我们就可以遍历该数组以查找具有给定乘积的一对。

下面是上述方法的实现：

C++

// C++ program to find a distinct prime number
// pair whose product is equal to given number
#include 
using namespace std;
 
// Function to generate all prime
// numbers less than n
bool SieveOfEratosthenes(int n, bool isPrime[])
{
    // Initialize all entries of boolean array
    // as true. A value in isPrime[i] will finally
    // be false if i is Not a prime, else true
    // bool isPrime[n+1];
    isPrime[0] = isPrime[1] = false;
    for (int i = 2; i <= n; i++)
        isPrime[i] = true;
 
    for (int p = 2; p * p <= n; p++) {
        // If isPrime[p] is not changed, then it is
        // a prime
        if (isPrime[p] == true) {
            // Update all multiples of p
            for (int i = p * 2; i <= n; i += p)
                isPrime[i] = false;
        }
    }
}
 
// Function to print a prime pair
// with given product
void findPrimePair(int n)
{
    int flag = 0;
 
    // Generating primes using Sieve
    bool isPrime[n + 1];
    SieveOfEratosthenes(n, isPrime);
 
    // Traversing all numbers to find first
    // pair
    for (int i = 2; i < n; i++) {
        int x = n / i;
 
        if (isPrime[i] && isPrime[x] and x != i
            and x * i == n) {
            cout << i << " " << x;
            flag = 1;
            return;
        }
    }
 
    if (!flag)
        cout << "No such pair found";
}
 
// Driven Code
int main()
{
    int n = 39;
 
    findPrimePair(n);
 
    return 0;
}

Java

// Java program to find a distinct prime number
// pair whose product is equal to given number
 
class GFG {
 
    // Function to generate all prime
    // numbers less than n
 
    static void SieveOfEratosthenes(int n,
                                    boolean isPrime[])
    {
        // Initialize all entries of boolean array
        // as true. A value in isPrime[i] will finally
        // be false if i is Not a prime, else true
        // bool isPrime[n+1];
        isPrime[0] = isPrime[1] = false;
        for (int i = 2; i <= n; i++)
            isPrime[i] = true;
 
        for (int p = 2; p * p <= n; p++) {
            // If isPrime[p] is not changed, then it is
            // a prime
            if (isPrime[p] == true) {
                // Update all multiples of p
                for (int i = p * 2; i <= n; i += p)
                    isPrime[i] = false;
            }
        }
    }
 
    // Function to print a prime pair
    // with given product
    static void findPrimePair(int n)
    {
        int flag = 0;
 
        // Generating primes using Sieve
        boolean[] isPrime = new boolean[n + 1];
        SieveOfEratosthenes(n, isPrime);
 
        // Traversing all numbers to find first
        // pair
        for (int i = 2; i < n; i++) {
            int x = n / i;
 
            if (isPrime[i] && isPrime[x] && x != i
                && x * i == n) {
                System.out.println(i + " " + x);
                flag = 1;
                return;
            }
        }
 
        if (flag == 0)
            System.out.println("No such pair found");
    }
 
    // Driven Code
    public static void main(String[] args)
    {
        int n = 39;
 
        findPrimePair(n);
    }
}
 
// This code is contributed by
// ihritik

Python3

# Python3 program to find a distinct
# prime number pair whose product
# is equal to given number
 
# from math lib. import everything
from math import *
 
# Function to generate all prime
# numbers less than n
def SieveOfEratosthenes(n, isPrime) :
 
    # Initialize all entries of boolean
    # array as true. A value in isPrime[i]
    # will finally be false if i is Not a
    # prime, else true bool isPrime[n+1];
    isPrime[0], isPrime[1] = False, False
 
    for i in range(2, n + 1) :
        isPrime[i] = True
 
    for p in range(2, int(sqrt(n)) + 1) :
 
        # If isPrime[p] is not changed,
        # then it is a prime
        if isPrime[p] == True :
 
            for i in range(p * 2, n + 1, p) :
                isPrime[i] = False
 
# Function to print a prime pair
# with given product
def findPrimePair(n) :
 
    flag = 0
     
    # Generating primes using Sieve
    isPrime = [False] * (n + 1)
    SieveOfEratosthenes(n, isPrime)
 
    # Traversing all numbers to
    # find first pair
    for i in range(2, n) :
        x = int(n / i)
 
        if (isPrime[i] & isPrime[x] and
             x != i and x * i == n) :
            print(i, x)
            flag = 1
            break
 
    if not flag :
        print("No such pair found")
     
# Driver code    
if __name__ == "__main__" :
 
    # Function calling
    n = 39;
 
    findPrimePair(n)
 
# This code is contributed by ANKITRAI1

C#

// C# program to find a distinct prime number
// pair whose product is equal to given number
using System;
class GFG
{
     
// Function to generate all
// prime numbers less than n
static void SieveOfEratosthenes(int n,
                                bool[] isPrime)
{
    // Initialize all entries of bool
    // array as true. A value in
    // isPrime[i] will finally be false
    // if i is Not a prime, else true
    // bool isPrime[n+1];
    isPrime[0] = isPrime[1] = false;
    for (int i = 2; i <= n; i++)
        isPrime[i] = true;
 
    for (int p = 2; p * p <= n; p++)
    {
        // If isPrime[p] is not changed,
        // then it is a prime
        if (isPrime[p] == true)
        {
            // Update all multiples of p
            for (int i = p * 2; i <= n; i += p)
                isPrime[i] = false;
        }
    }
}
 
// Function to print a prime
// pair with given product
static void findPrimePair(int n)
{
    int flag = 0;
 
    // Generating primes using Sieve
    bool[] isPrime = new bool[n + 1];
    SieveOfEratosthenes(n, isPrime);
 
    // Traversing all numbers to
    // find first pair
    for (int i = 2; i < n; i++)
    {
        int x = n / i;
 
        if (isPrime[i] && isPrime[x] &&
                x != i && x * i == n)
        {
            Console.Write(i + " " + x);
            flag = 1;
            return;
        }
    }
 
    if (flag == 0)
        Console.Write("No such pair found");
}
 
// Driven Code
public static void Main()
{
    int n = 39;
 
    findPrimePair(n);
}
}
 
// This code is contributed by ChitraNayal

Javascript

 
// Javascript program to find a distinct prime number
// pair whose product is equal to given number
 
// Function to generate all
// prime numbers less than n
function SieveOfEratosthenes(n, isPrime)
{
     
    // Initialize all entries of bool
    // array as true. A value in
    // isPrime[i] will finally be false
    // if i is Not a prime, else true
    //  isPrime[n+1];
    isPrime[0] = isPrime[1] = false;
     
    for(var i = 2; i <= n; i++)
        isPrime[i] = true;
  
    for(var p = 2; p * p <= n; p++)
    {
         
        // If isPrime[p] is not changed,
        // then it is a prime
        if (isPrime[p] == true)
        {
             
            // Update all multiples of p
            for(var i = p * 2; i <= n; i += p)
                isPrime[i] = false;
        }
    }
}
  
// Function to print a prime
// pair with given product
function findPrimePair(n)
{
    var flag = 0;
  
    // Generating primes using Sieve
    var isPrime = []
    SieveOfEratosthenes(n, isPrime);
  
    // Traversing all numbers to
    // find first pair
    for(var i = 2; i < n; i++)
    {
        var x = n / i;
  
        if (isPrime[i] && isPrime[x] &&
                x != i && x * i == n)
        {
            document.write(i + " " + x);
            flag = 1;
            return;
        }
    }
  
    if (flag == 0)
        document.write("No such pair found");
}
  
// Driver code
var n = 39;
 
findPrimePair(n);
 
// This code is contributed by bunnyram19
 

输出：

3 13

时间复杂度： O(N * log log(N))

辅助空间： O(N)

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