找到两个不同的数字，使它们的LCM处于给定范围内

给定两个数字L和R ，任务是找到两个不同的最小正整数X和Y ，以使它们的LCM在[L，R]范围内。如果不存在任何X和Y值，则打印“ -1” 。

例子：

Input: L = 3, R = 8 
Output: x = 3, y=6
Explanation:
LCM of 3 and 6 is 6 which is in range 3, 8

Input: L = 88, R = 90
Output: -1
Explanation:
Minimum possible x and y are 88 and 176 respectively, but 176 is greater than 90.    

方法：想法是以X和Y的LCM在给定范围[L，R]中的方式选择X和Y的值。步骤如下：

1. 对于X的最小值，请选择L作为最小值，因为这是给定范围内的最小值。
2. 现在为Y的值选择2 * L，因为这是LCM为L的Y的最小值。
3. 现在，如果X和Y的上述两个值在[L，R]范围内，则这是必需的整数对，其中X和Y的最小可能值。
4. 否则，请打印“ -1” ，因为不存在其他任何对。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find two distinct numbers
// X and Y s.t. their LCM lies between
// L and R  and X, Y are minimum possible
void answer(int L, int R)
{
 
    // Check if 2*L lies in range L, R
    if (2 * L <= R)
 
        // Print the answer
        cout << L << ", "
             << 2 * L << "\n";
    else
        cout << -1;
}
 
// Driver Code
int main()
{
    // Given value of ranges
    int L = 3, R = 8;
 
    // Function call
    answer(L, R);
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
 
class GFG{
 
// Function to find two distinct numbers
// X and Y s.t. their LCM lies between
// L and R and X, Y are minimum possible
static void answer(int L, int R)
{
 
    // Check if 2*L lies in range L, R
    if (2 * L <= R)
 
        // Print the answer
        System.out.println(L + ", " + (2 * L));
     
    else
        System.out.println("-1");
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given value of ranges
    int L = 3, R = 8;
 
    // Function call
    answer(L, R);
}
}
 
// This code is contributed by sanjoy_62

Python3

# Python3 program for the above approach
 
# Function to find two distinct numbers
# X and Y s.t. their LCM lies between
# L and R and X, Y are minimum possible
def answer(L, R):
 
    # Check if 2*L lies in range L, R
    if (2 * L <= R):
 
        # Print the answer
        print(L, ",", 2 * L)
 
    else:
        print(-1)
 
# Driver Code
 
# Given value of ranges
L = 3
R = 8
 
# Function call
answer(L, R)
 
# This code is contributed by sanjoy_62

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to find two distinct numbers
// X and Y s.t. their LCM lies between
// L and R and X, Y are minimum possible
static void answer(int L, int R)
{
 
    // Check if 2*L lies in range L, R
    if (2 * L <= R)
     
        // Print the answer
        Console.WriteLine(L + ", " + (2 * L));
     
    else
        Console.WriteLine("-1");
}
 
// Driver Code
public static void Main()
{
     
    // Given value of ranges
    int L = 3, R = 8;
 
    // Function call
    answer(L, R);
}
}
 
// This code is contributed by sanjoy_62

Javascript

// JavaScript program for the above approac
 
// Function to find two distinct numbers
// X and Y s.t. their LCM lies between
// L and R and X, Y are minimum possible
function answer(L, R)
{
 
    // Check if 2*L lies in range L, R
    if (2 * L <= R)
 
        // Print the answer
        document.write(L + ", "
            + 2 * L + "<br>");
    else
        document.write(-1);
}
 
// Driver Code
 
    // Given value of ranges
    let L = 3, R = 8;
 
    // Function call
    answer(L, R);
 
// This code is contributed by Manoj.

输出：

3, 6

时间复杂度： O(1)
辅助空间： O(1)