满足给定方程的最小正整数 X

给定两个整数N和K ，任务是找到满足等式的最小正整数X ：

(X / K) * (X % K) = N

例子：

Input: N = 6, K = 3 
Output: 11 
Explanation: 
For X = 11, (11 / 3) * (11 % 3) = 3 * 2 = 6 
Therefore, the following equation satisfies.
Input: N = 4, K = 6 
Output: 10 
Explanation: 
For X = 10, (10 / 6) * (10 % 6) = 1 * 4 = 4 
Therefore, the following equation satisfies. 
 

方法：
这个想法是观察到，因为(X / K) * (X % K) = N ，因此， N将被p = X % K整除，它小于K 。因此，对于[1, K)范围内的所有i尝试 p 的所有值，其中：

$x = p \cdot\frac{n - p}{k}$

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
#include 
using namespace std;
 
// Function to find out the smallest
// positive integer for the equation
int findMinSoln(int n, int k)
{
    // Stores the minimum
    int minSoln = INT_MAX;
 
    // Iterate till K
    for (int i = 1; i < k; i++) {
 
        // Check if n is divisible by i
        if (n % i == 0)
            minSoln
                = min(minSoln, (n / i) * k + i);
    }
 
    // Return the answer
    return minSoln;
}
 
// Driver Code
int main()
{
    int n = 4, k = 6;
    cout << findMinSoln(n, k);
}

Java

// Java Program to implement
// the above approach
import java.util.*;
class GFG{
  
// Function to find out the smallest
// positive integer for the equation
static int findMinSoln(int n, int k)
{
    // Stores the minimum
    int minSoln = Integer.MAX_VALUE;
  
    // Iterate till K
    for (int i = 1; i < k; i++)
    {
  
        // Check if n is divisible by i
        if (n % i == 0)
            minSoln = Math.min(minSoln, (n / i) * k + i);
    }
  
    // Return the answer
    return minSoln;
}
  
// Driver Code
public static void main(String[] args)
{
    int n = 4, k = 6;
    System.out.println(findMinSoln(n, k));
}
}
 
// This code is contributed by Ritik Bansal

Python3

# Python3 program to implement
# the above approach
import sys
 
# Function to find out the smallest
# positive integer for the equation
def findMinSoln(n, k):
     
    # Stores the minimum
    minSoln = sys.maxsize;
 
    # Iterate till K
    for i in range(1, k):
 
        # Check if n is divisible by i
        if (n % i == 0):
            minSoln = min(minSoln, (n // i) * k + i);
     
    # Return the answer
    return minSoln;
 
# Driver Code
if __name__ == '__main__':
     
    n = 4;
    k = 6;
     
    print(findMinSoln(n, k));
 
# This code is contributed by amal kumar choubey

C#

// C# program to implement
// the above approach
using System;
 
class GFG{
 
// Function to find out the smallest
// positive integer for the equation
static int findMinSoln(int n, int k)
{
     
    // Stores the minimum
    int minSoln = int.MaxValue;
 
    // Iterate till K
    for (int i = 1; i < k; i++)
    {
 
        // Check if n is divisible by i
        if (n % i == 0)
            minSoln = Math.Min(minSoln,
                              (n / i) * k + i);
    }
 
    // Return the answer
    return minSoln;
}
 
// Driver Code
public static void Main(String[] args)
{
    int n = 4, k = 6;
     
    Console.WriteLine(findMinSoln(n, k));
}
}
 
// This code is contributed by amal kumar choubey

Javascript

 
 
// Javascript Program to implement
// the above approach
 
// Function to find out the smallest
// positive integer for the equation
function findMinSoln(n, k)
{
 
    // Stores the minimum
    var minSoln = 1000000000;
 
    // Iterate till K
    for (var i = 1; i < k; i++) {
 
        // Check if n is divisible by i
        if (n % i == 0)
            minSoln
                = Math.min(minSoln, (n / i) * k + i);
    }
 
    // Return the answer
    return minSoln;
}
 
// Driver Code
var n = 4, k = 6;
document.write( findMinSoln(n, k));
 
// This code is contributed by rutvik_56.

输出：

10

时间复杂度： O(N)
辅助空间： O(1)