📜  从1到N的互素对数的数量等于N

📅  最后修改于: 2021-06-26 16:47:33             🧑  作者: Mango

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


PHP


Javascript


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

Number of coprime pairs : 3