max(A，B)的最小可能值，使得LCM(A，B)= C

给定整数C ，任务是找到max(A，B)的最小可能值，以使LCM(A，B)= C。
例子：

Input: C = 6 
Output: 3 
max(1, 6) = 6 
max(2, 3) = 3 
and min(6, 3) = 3
Input: C = 9 
Output: 9 
 

方法：一种解决此问题的方法是使用本文讨论的方法找到给定数量的所有因素，然后找到满足给定条件的对(A，B) ，并取其中最大值中的最小值。对。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the LCM of a and b
int lcm(int a, int b)
{
    return (a / __gcd(a, b) * b);
}
 
// Function to return the minimized value
int getMinValue(int c)
{
    int ans = INT_MAX;
 
    // To find the factors
    for (int i = 1; i <= sqrt(c); i++) {
 
        // To check if i is a factor of c and
        // the minimum possible number
        // satisfying the given conditions
        if (c % i == 0 && lcm(i, c / i) == c) {
 
            // Update the answer
            ans = min(ans, max(i, c / i));
        }
    }
    return ans;
}
 
// Driver code
int main()
{
    int c = 6;
 
    cout << getMinValue(c);
 
    return 0;
}

Java

// Java implementation of the approach
class Solution
{
    // 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 return the LCM of a and b
    static int lcm(int a, int b)
    {
        return (a / __gcd(a, b) * b);
    }
 
    // Function to return the minimized value
    static int getMinValue(int c)
    {
        int ans = Integer.MAX_VALUE;
 
        // To find the factors
        for (int i = 1; i <= Math.sqrt(c); i++)
        {
 
            // To check if i is a factor of c and
            // the minimum possible number
            // satisfying the given conditions
            if (c % i == 0 && lcm(i, c / i) == c)
            {
 
                // Update the answer
                ans = Math.min(ans, Math.max(i, c / i));
            }
        }
        return ans;
    }
 
    // Driver code
    public static void main(String args[])
    {
        int c = 6;
 
        System.out.println(getMinValue(c));
    }
}
 
// This code is contributed by Arnab Kundu

Python3

# Python implementation of the approach
import sys
 
# Recursive function to return gcd of a and b
def __gcd(a, 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 return the LCM of a and b
def lcm(a, b):
    return (a / __gcd(a, b) * b);
 
# Function to return the minimized value
def getMinValue(c):
    ans = sys.maxsize;
 
    # To find the factors
    for i in range(1, int(pow(c, 1/2)) + 1):
 
        # To check if i is a factor of c and
        # the minimum possible number
        # satisfying the given conditions
        if (c % i == 0 and lcm(i, c / i) == c):
 
            # Update the answer
            ans = min(ans, max(i, c / i));
    return int(ans);
 
# Driver code
if __name__ == '__main__':
    c = 6;
 
    print(getMinValue(c));
     
# This code is contributed by 29AjayKumar

C#

// C# implementation of the approach
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 return the LCM of a and b
    static int lcm(int a, int b)
    {
        return (a / __gcd(a, b) * b);
    }
 
    // Function to return the minimized value
    static int getMinValue(int c)
    {
        int ans = int.MaxValue;
 
        // To find the factors
        for (int i = 1; i <= Math.Sqrt(c); i++)
        {
 
            // To check if i is a factor of c and
            // the minimum possible number
            // satisfying the given conditions
            if (c % i == 0 && lcm(i, c / i) == c)
            {
 
                // Update the answer
                ans = Math.Min(ans, Math.Max(i, c / i));
            }
        }
        return ans;
    }
 
    // Driver code
    public static void Main()
    {
        int c = 6;
 
        Console.WriteLine(getMinValue(c));
    }
}
 
// This code is contributed by AnkitRai01

Javascript

 
// javascript implementation of the approach
 
// Recursive function to return gcd of a and b
function __gcd(a , 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 return the LCM of a and b
function lcm(a , b)
{
    return (a / __gcd(a, b) * b);
}
 
// Function to return the minimized value
function getMinValue(c)
{
    var ans = Number.MAX_VALUE;
 
    // To find the factors
    for (i = 1; i <= Math.sqrt(c); i++)
    {
 
        // To check if i is a factor of c and
        // the minimum possible number
        // satisfying the given conditions
        if (c % i == 0 && lcm(i, c / i) == c)
        {
 
            // Update the answer
            ans = Math.min(ans, Math.max(i, c / i));
        }
    }
    return ans;
}
 
// Driver code
 
var c = 6;
 
document.write(getMinValue(c));
 
// This code contributed by shikhasingrajput
 

输出：

3