Q的可能值，使得对于R的任何值，其乘积等于其乘积的X倍

给定一个整数X ，任务是找到该对(Q，R)的Q的可能值，以使它们的乘积等于X乘以它们的总和，其中Q≤R且X <10 ⁷ 。打印这些Q值的总数以及值。

例子：

Input: X = 3
Output: 
2
4, 6
Explanation: 
On taking Q = 4 and R = 12, 
LHS = 12 x 4 = 48 
RHS = 3(12 + 4) = 3 x 16 = 48 = LHS
Similarly, the equation also holds for value Q = 6 and R = 6. 
LHS = 6 x 6 = 36 
RHS = 3(6 + 6) = 3 x 12 = 36 = LHS

Input:  X = 16
Output: 
5
17, 18, 20, 24, 32 
Explanation: 
If Q = 17 and R = 272, 
LHS = 17 x 272 = 4624 
RHS = 16(17 + 272) = 16(289) = 4624 = LHS. 
Similarly, there exists a value R for all other values of Q given in the output. 

方法：想法是理解问题以形成一个方程，即(Q x R)= X(Q + R)。

• 这个想法是从1迭代到X并检查是否每个if ((((X + i)* X)％i)== 0 。
• 初始化结果向量，并从1迭代X的所有值。
• 检查以上条件是否成立。如果确实存在，则将值X + i推入向量中。
• 让我们打破等式，以便更清楚地理解它，

 
 

The given expression is (Q x R) = X(Q + R) 
On simplifying this we get,  

 

=>  QR – QX = RX
or,  QR – RX = QX
or,  R = QX / (Q – X)

 

• 因此，观察到(X + i)是Q的可能值，而(X + i)* X是R的可能值。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to find all possible values of Q
void values_of_Q(int X)
{
    // Vector initialization
    // to store all numbers
    // satisfying the given condition
    vector val_Q;
 
    // Iterate for all the values of X
    for (int i = 1; i <= X; i++) {
 
        // Check if condition satisfied
        // then push the number
        if ((((X + i) * X)) % i == 0) {
 
            // Possible value of Q
            val_Q.push_back(X + i);
        }
    }
 
    cout << val_Q.size() << endl;
 
    // Print all the numbers
    for (int i = 0; i < val_Q.size(); i++) {
        cout << val_Q[i] << " ";
    }
}
 
// Driver code
int main()
{
    int X = 3;
 
    values_of_Q(X);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG{
   
// Function to find all possible values of Q
static void values_of_Q(int X)
{
    // Vector initialization
    // to store all numbers
    // satisfying the given condition
    ArrayList val_Q = new ArrayList();
  
    // Iterate for all the values of X
    for (int i = 1; i <= X; i++)
    {
  
        // Check if condition satisfied
        // then push the number
        if ((((X + i) * X)) % i == 0)
        {
  
            // Possible value of Q
            val_Q.add(X + i);
        }
    }
  
    System.out.println(val_Q.size());
  
    // Print all the numbers
    for (int i = 0; i < val_Q.size(); i++)
    {
        System.out.print(val_Q.get(i)+" ");
    }
}
  
// Driver code
public static void main(String[] args)
{
    int X = 3;
  
    values_of_Q(X);
}
}
 
// This code is contributed by Ritik Bansal

Python3

# Python3 program for the above approach
 
# Function to find all possible values of Q
def values_of_Q(X):
 
    # Vector initialization
    # to store all numbers
    # satisfying the given condition
    val_Q = []
 
    # Iterate for all the values of X
    for i in range(1, X + 1):
 
        # Check if condition satisfied
        # then push the number
        if ((((X + i) * X)) % i == 0):
 
            # Possible value of Q
            val_Q.append(X + i)
 
    print(len(val_Q))
 
    # Print all the numbers
    for i in range(len(val_Q)):
        print(val_Q[i], end = " ")
 
# Driver Code       
X = 3
 
values_of_Q(X)
 
# This code is contributed by divyeshrabadiya07

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find all possible
// values of Q
static void values_of_Q(int X)
{
     
    // List initialization
    // to store all numbers
    // satisfying the given condition
    List val_Q = new List();
 
    // Iterate for all the values of X
    for(int i = 1; i <= X; i++)
    {
         
        // Check if condition satisfied
        // then push the number
        if ((((X + i) * X)) % i == 0)
        {
             
            // Possible value of Q
            val_Q.Add(X + i);
        }
    }
     
    Console.WriteLine(val_Q.Count);
 
    // Print all the numbers
    for(int i = 0; i < val_Q.Count; i++)
    {
        Console.Write(val_Q[i] + " ");
    }
}
 
// Driver code
public static void Main(String[] args)
{
    int X = 3;
 
    values_of_Q(X);
}
}
 
// This code is contributed by Amit Katiyar

Javascript

    // Javascript program for the above approach
     
    // Function to find all possible values of Q
    function values_of_Q(X)
    {
     
        // Vector initialization
        // to store all numbers
        // satisfying the given condition
        let val_Q = [];
 
        // Iterate for all the values of X
        for (let i = 1; i <= X; i++)
        {
 
            // Check if condition satisfied
            // then push the number
            if ((((X + i) * X)) % i == 0)
            {
 
                // Possible value of Q
                val_Q.push(X + i);
            }
        }
 
        document.write(val_Q.length + "</br>");
 
        // Print all the numbers
        for (let i = 0; i < val_Q.length; i++) {
            document.write(val_Q[i] + " ");
        }
    }
     
    let X = 3;
    values_of_Q(X);
     
    // This code is contributed by divyesh072019.

输出：

2
4 6

时间复杂度： O(N)

辅助空间： O(X)