前N个自然数的所有对(i，j)中的最大LCM

给定正整数N> 1 ，任务是在所有对(i，j)中找到最大LCM，以使i 。
例子：

Input: N = 3 
Output: 6 
LCM(1, 2) = 2 
LCM(1, 3) = 3 
LCM(2, 3) = 6
Input: N = 4 
Output: 12 
 

方法：由于两个连续元素的LCM等于它们的倍数，因此很明显，最大LCM将为(N，N – 1)对，即(N *(N – 1)) 。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the maximum LCM
// among all the pairs(i, j) of
// first n natural numbers
int maxLCM(int n)
{
    return (n * (n - 1));
}
 
// Driver code
int main()
{
    int n = 3;
 
    cout << maxLCM(n);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
     
// Function to return the maximum LCM
// among all the pairs(i, j) of
// first n natural numbers
static int maxLCM(int n)
{
    return (n * (n - 1));
}
 
// Driver code
public static void main(String[] args)
{
    int n = 3;
 
    System.out.println(maxLCM(n));
}
}
 
// This code is contributed by Code_Mech

Python3

# Python3 implementation of the approach
 
# Function to return the maximum LCM
# among all the pairs(i, j) of
# first n natural numbers
def maxLCM(n) :
 
    return (n * (n - 1));
 
# Driver code
if __name__ == "__main__" :
 
    n = 3;
 
    print(maxLCM(n));
 
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
     
class GFG
{
     
// Function to return the maximum LCM
// among all the pairs(i, j) of
// first n natural numbers
static int maxLCM(int n)
{
    return (n * (n - 1));
}
 
// Driver code
public static void Main(String[] args)
{
    int n = 3;
 
    Console.WriteLine(maxLCM(n));
}
}
 
// This code is contributed by Rajput-Ji

Javascript

 
// Javascript implementation of the approach
 
// Function to return the maximum LCM
// among all the pairs(i, j) of
// first n natural numbers
function maxLCM(n)
{
    return (n * (n - 1));
}
 
// Driver code
var n = 3;
document.write(maxLCM(n));
 

输出：

6

时间复杂度： O(1)