从1到N的互素对数的数量等于N

给定数字N。任务是找到从1到N的互质对(a，b)的数量，以使它们的乘积(a * b)等于N。

注意：如果gcd(a，b)= 1，则说对(a，b)是互质的。

例子：

Input: N = 120
Output: No. of co-prime pairs = 3
(3, 40)
(5, 24)
(8, 15) 

Input: N= 250
Output: No. of co-prime pairs = 3
(2, 125) 

方法：假设该对中的元素应互为互质。设一个素数对为(a，b) ，
给定， a * b = N。

所以，
$a, b <= &\sqrt(n)$$

所以这个想法是从1到$&\sqrt(N)$$并检查i和(N / i)是否互质，以及i *(N / i)=N。如果是，则计算这样的对。

下面是上述方法的实现：

C++

// C++ program to count number of Co-prime pairs
// from 1 to N with product equals to N
#include 
using namespace std;
  
// Function to count number of Co-prime pairs
// from 1 to N with product equals to N
void countCoprimePairs(int n)
{
    int count = 0;
  
    cout << "The co- prime pairs are: " << endl;
  
    // find all the co- prime pairs
    // Traverse from 2 to sqrt(N) and check
    // if i, N/i are coprimes
    for (int i = 2; i <= sqrt(n); i++) {
  
        // check if N is divisible by i,
        // so that the other term in pair i.e. N/i
        // is integral
        if (n % i == 0) {
  
            // Check if i and N/i are coprime
            if (__gcd(i, (n / i)) == 1) {
  
                // Display the co- prime pairs
                cout << "(" << i << ", " << (n / i) << ")\n";
                count++;
            }
        }
    }
  
    cout << "\nNumber of coprime pairs : " << count;
}
  
// Driver code
int main()
{
    int N = 120;
  
    countCoprimePairs(N);
  
    return 0;
}

Java

// Java program to count number of Co-prime pairs
// from 1 to N with product equals to N
import java.io.*;
  
public class GFG {
  // Recursive function to return gcd of a and b 
    static int __gcd(int a, int b) 
    { 
        // Everything divides 0  
        if (a == 0) 
          return b; 
        if (b == 0) 
          return a; 
         
        // base case 
        if (a == b) 
            return a; 
         
        // a is greater 
        if (a > b) 
            return __gcd(a-b, b); 
        return __gcd(a, b-a); 
    } 
  
// Function to count number of Co-prime pairs
// from 1 to N with product equals to N
static void countCoprimePairs(int n)
{
    int count = 0;
  
    System.out.println( "The co- prime pairs are: ");
  
    // find all the co- prime pairs
    // Traverse from 2 to sqrt(N) and check
    // if i, N/i are coprimes
    for (int i = 2; i <= Math.sqrt(n); i++) {
  
        // check if N is divisible by i,
        // so that the other term in pair i.e. N/i
        // is integral
        if (n % i == 0) {
  
            // Check if i and N/i are coprime
            if (__gcd(i, (n / i)) == 1) {
  
                // Display the co- prime pairs
                System.out.print( "(" +i + ", " + (n / i) + ")\n");
                count++;
            }
        }
    }
  
    System.out.println("\nNumber of coprime pairs : " + count);
}
  
// Driver code
    public static void main (String[] args) {
            int N = 120;
  
    countCoprimePairs(N);
    }
}
  
// This code is contributed by shs..

Python 3

# Python program to count number 
# of Co-prime pairs from 1 to N 
# with product equals to N 
  
# import everything from math lib
from math import *
  
# Function to count number of 
# Co-prime pairs from 1 to N
# with product equals to N 
def countCoprimePairs(n) :
  
    count = 0
  
    print("The co-prime pairs are: ")
  
    # find all the co- prime pairs 
    # Traverse from 2 to sqrt(N) and 
    # check if i, N//i are coprimes 
    for i in range(2, int(sqrt(n)) + 1) :
  
        # check if N is divisible by i, 
        # so that the other term in pair 
        # i.e. N/i is integral 
        if n % i == 0 :
  
            # Check if i and N/i are coprime 
            if gcd(i, n // i) == 1 :
  
                # Display the co- prime pairs 
                print("(", i,",", (n // i),")")
                count += 1
  
    print("Number of coprime pairs : ", count)
                  
# Driver code     
if __name__ == "__main__" :
  
    N = 120
  
    countCoprimePairs(N)
  
# This code is contributed by ANKITRAI1

C#

// C# program to count number 
// of Co-prime pairs from 1 to N 
// with product equals to N
using System;
  
class GFG
{
// Recursive function to
// return gcd of a and b 
static int __gcd(int a, int b) 
{ 
    // Everything divides 0 
    if (a == 0) 
    return b; 
    if (b == 0) 
    return a; 
      
    // base case 
    if (a == b) 
        return a; 
      
    // a is greater 
    if (a > b) 
        return __gcd(a - b, b); 
    return __gcd(a, b - a); 
} 
  
// Function to count number of 
// Co-prime pairs from 1 to N 
// with product equals to N
static void countCoprimePairs(int n)
{
int count = 0;
  
Console.WriteLine("The co- prime pairs are: ");
  
// find all the co- prime pairs
// Traverse from 2 to sqrt(N) and 
// check if i, N/i are coprimes
for (int i = 2; i <= Math.Sqrt(n); i++) 
{
  
    // check if N is divisible by i,
    // so that the other term in pair 
    // i.e. N/i is integral
    if (n % i == 0) 
    {
  
        // Check if i and N/i are coprime
        if (__gcd(i, (n / i)) == 1) 
        {
  
            // Display the co- prime pairs
            Console.WriteLine( "(" + i + ", " + 
                              (n / i) + ")\n");
            count++;
        }
    }
}
  
Console.WriteLine("\nNumber of coprime" + 
                    " pairs : " + count);
}
  
// Driver code
public static void Main ()
{
    int N = 120;
  
    countCoprimePairs(N);
}
}
  
// This code is contributed by Shashank

输出：

The co- prime pairs are: 
(3, 40)
(5, 24)
(8, 15)

Number of coprime pairs : 3