满足方程1 / X + 1 / Y = 1 / N的有序对(X，Y)的数量

给定正整数N ，任务是查找有序对(X，Y)的数量，其中X和Y均为正整数，这样它们满足方程1 / X + 1 / Y = 1 / N。

例子：

Input: N = 5 
Output: 3 
Explanation: Only 3 pairs {(30,6), (10,10), (6,30)} satisfy the given equation.
 

Input: N = 360 
Output: 105 
 

方法：
请按照以下步骤解决问题：

• 使用给定的方程式求解X。

  1/X + 1/Y = 1/N 
  => 1/X = 1/N – 1/Y 
  => 1/X = (Y – N) / NY 
  => X = NY / (Y – N) 
  => X = N + N² / (Y – N)

• 因此，可以看出，要有一个正整数X ， N ²除以(Y – N)时的余数必须为0 。
• 可以看出， Y的最小值可以为N + 1 (因此分母Y – N> 0) ， Y的最大值可以为N ² + N，从而N ² /(Y – N)保持为正整数≥1 。
• 然后遍历最大和Y的最小可能值，并且用于Y的各值，对于这个N ^2％(Y – N)== 0，递增计数。
• 最后，将count作为有序对的数量返回。

下面是上述方法的实现：

C++

// C++ Program for the above approach
#include 
using namespace std;
  
// Function to find number of ordered
// positive integer pairs (x,y) such
// that  they satisfy the equation
void solve(int n)
{
    // Initialize answer variable
    int ans = 0;
  
// Iterate over all possible values of y
    for (int y = n + 1; y <= n * n + n; y++) {
  
        // For valid x and y,
        // (n*n)%(y - n) has to be 0
        if ((n * n) % (y - n) == 0) {
  
            // Increment count of ordered pairs
            ans += 1;
        }
    }
  
    // Print the answer
    cout << ans;
}
  
// Driver Code
int main()
{
    int n = 5;
    // Function call
    solve(n);
    return 0;
}

Java

// Java program for the above approach
class GFG{
  
// Function to find number of ordered
// positive integer pairs (x,y) such
// that they satisfy the equation
static void solve(int n)
{
      
    // Initialize answer variable
    int ans = 0;
  
    // Iterate over all possible values of y
    for(int y = n + 1; y <= n * n + n; y++) 
    {
          
        // For valid x and y,
        // (n*n)%(y - n) has to be 0
        if ((n * n) % (y - n) == 0)
        {
              
            // Increment count of ordered pairs
            ans += 1;
        }
    }
  
    // Print the answer
    System.out.print(ans);
}
  
// Driver Code
public static void main(String[] args)
{
    int n = 5;
      
    // Function call
    solve(n);
}
}
  
// This code is contributed by Amit Katiyar

Python3

# Python3 program for the above approach 
  
# Function to find number of ordered
# positive integer pairs (x,y) such
# that they satisfy the equation
def solve(n):
  
    # Initialize answer variable
    ans = 0
  
    # Iterate over all possible values of y
    y = n + 1
    while(y <= n * n + n):
  
        # For valid x and y,
        # (n*n)%(y - n) has to be 0
        if ((n * n) % (y - n) == 0):
              
            # Increment count of ordered pairs
            ans += 1
  
        y += 1
  
    # Print the answer
    print(ans)
  
# Driver Code
n = 5
  
# Function call 
solve(n)
  
# This code is contributed by Shivam Singh

C#

// C# program for the above approach
using System;
  
class GFG{
  
// Function to find number of ordered
// positive integer pairs (x,y) such
// that they satisfy the equation
static void solve(int n)
{
      
    // Initialize answer variable
    int ans = 0;
  
    // Iterate over all possible values of y
    for(int y = n + 1; y <= n * n + n; y++) 
    {
          
        // For valid x and y,
        // (n*n)%(y - n) has to be 0
        if ((n * n) % (y - n) == 0)
        {
              
            // Increment count of ordered pairs
            ans += 1;
        }
    }
  
    // Print the answer
    Console.Write(ans);
}
  
// Driver Code
public static void Main(String[] args)
{
    int n = 5;
      
    // Function call
    solve(n);
}
}
  
// This code is contributed by Amit Katiyar

输出：

3

时间复杂度： O(logN)
辅助空间： O(1)