计算所有与N互质数为N的所有对数除数

给定整数N ，任务是计算N的所有对数，使得每对的和与N互质。
例子：

Input: N = 24 
Output: 2 
Explaination: 
There are 2 pairs (1, 24) and (2, 3) whose sum is coprime with 24
Input: 105 
Output: 4 
Explaination: 
There are 4 pairs (1, 105), (3, 35), (5, 21) and (7, 15) whose sum is coprime with 105 
 

方法：
为了解决上述问题，我们可以通过找到所有√N复杂度的除数，并检查每对数的和是否与N互质，来轻松计算结果。
下面是上述方法的实现：

C++

// C++ program to count all pairs
// of divisors such that their sum
// is coprime with N
#include 
using namespace std;
 
// Function to calculate GCD
int gcd(int a, int b)
{
    if (b == 0)
        return a;
 
    return (gcd(b, a % b));
}
 
// Function to count all valid pairs
int CountPairs(int n)
{
 
    // Initialize count
    int cnt = 0;
 
    for (int i = 1; i * i <= n; i++) {
        if (n % i == 0) {
            int div1 = i;
 
            int div2 = n / i;
 
            int sum = div1 + div2;
 
            // Check if sum of pair
            // and n are coprime
            if (gcd(sum, n) == 1)
 
                cnt += 1;
        }
    }
 
    // Return the result
    return cnt;
}
 
// Driver code
int main()
{
 
    int n = 24;
 
    cout << CountPairs(n) << endl;
 
    return 0;
}

Java

// Java program to count all pairs
// of divisors such that their sum
// is coprime with N
import java.util.*;
 
class GFG{
 
// Function to calculate GCD
public static int gcd(int a, int b)
{
    if (b == 0)
        return a;
 
    return (gcd(b, a % b));
}
 
// Function to count all valid pairs
public static int CountPairs(int n)
{
 
    // Initialize count
    int cnt = 0;
 
    for(int i = 1; i * i <= n; i++)
    {
       if (n % i == 0)
       {
           int div1 = i;
           int div2 = n / i;
           int sum = div1 + div2;
            
           // Check if sum of pair
           // and n are coprime
           if (gcd(sum, n) == 1)
               cnt += 1;
       }
    }
 
    // Return the result
    return cnt;
}
 
// Driver code
public static void main(String[] args)
{
    int n = 24;
 
    System.out.println(CountPairs(n));
}
}
 
// This code is contributed by equbalzeeshan

Python3

# Python3 program to count all pairs
# of divisors such that their sum
# is coprime with N
import math as m
 
 
# Function to count all valid pairs
def CountPairs(n):
     
    # initialize count
    cnt = 0
     
    i = 1
     
    while i * i <= n :
         
        if(n % i == 0):
             
            div1 = i
            div2 = n//i
             
            sum = div1 + div2;
             
            # Check if sum of pair
            # and n are coprime
            if( m.gcd(sum, n) == 1):
                cnt += 1
         
        i += 1
     
    # Return the result
    return cnt
     
 
# Driver code
n = 24
 
print(CountPairs(n))

C#

// C# program to count all pairs of
// divisors such that their sum
// is coprime with N
using System;
 
class GFG {
 
    // Function to find gcd of a and b
    static int gcd(int a, int b)
    {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }
 
    // Function to count all valid pairs
    static int CountPairs(int n)
    {
 
        // Initialize count
        int cnt = 0;
 
        for (int i = 1; i * i <= n; i++) {
            if (n % i == 0) {
                int div1 = i;
 
                int div2 = n / i;
 
                int sum = div1 + div2;
 
                // Check if sum of pair
                // and n are coprime
                if (gcd(sum, n) == 1)
                    cnt += 1;
            }
        }
        // Return the result
        return cnt;
    }
 
    // Driver method
    public static void Main()
    {
        int n = 24;
 
        Console.WriteLine(CountPairs(n));
    }
}

Javascript

 
// JavaScript program to count all pairs
// of divisors such that their sum
// is coprime with N
 
// Function to calculate GCD
function gcd(a, b)
{
    if (b == 0)
        return a;
 
    return (gcd(b, a % b));
}
 
// Function to count all valid pairs
function CountPairs(n)
{
 
    // Initialize count
    let cnt = 0;
 
    for (let i = 1; i * i <= n; i++) {
        if (n % i == 0) {
            let div1 = i;
 
            let div2 = Math.floor(n / i);
 
            let sum = div1 + div2;
 
            // Check if sum of pair
            // and n are coprime
            if (gcd(sum, n) == 1)
 
                cnt += 1;
        }
    }
 
    // Return the result
    return cnt;
}
 
// Driver code
    let n = 24;
 
    document.write(CountPairs(n) + "<br>");
 
// This code is contributed by Surbhi Tyagi.
 

输出：

2

时间复杂度： O(√N* log(N))