查找小于或等于N的三个整数，以使它们的LCM最大

给定一个数字N(> = 3)。任务是找到三个整数(<= N)，以使这三个整数的LCM最大。
例子：

Input: N = 3
Output: 1 2 3

Input: N = 5
Output: 3 4 5

方法：由于任务是使LCM最大化，因此，如果所有三个数字都没有任何公因子，则LCM将是这三个数字的乘积，并且该乘积将是最大的。

• If n is odd then the answer will be n, n-1, n-2.
• If n is even, 
  1. If gcd of n and n-3 is 1 then answer will be n, n-1, n-3.
  2. Otherwise, n-1, n-2, n-3 will be required answer.

下面是上述方法的实现：

C++

// CPP Program to find three integers
// less than N whose LCM is maximum
#include 
using namespace std;
 
// function to find three integers
// less than N whose LCM is maximum
void MaxLCM(int n)
{
    // if n is odd
    if (n % 2 != 0)
        cout << n << " " << (n - 1) << " " << (n - 2);
 
    // if n is even and n, n-3 gcd is 1
    else if (__gcd(n, (n - 3)) == 1)
        cout << n << " " << (n - 1) << " " << (n - 3);
 
    else
        cout << (n - 1) << " " << (n - 2) << " " << (n - 3);
}
 
// Driver code
int main()
{
    int n = 12;
 
    // function call
    MaxLCM(n);
 
    return 0;
}

Java

// Java Program to find three integers
// less than N whose LCM is maximum
 
import java.io.*;
 
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 find three integers
// less than N whose LCM is maximum
static void MaxLCM(int n)
{
    // if n is odd
    if (n % 2 != 0)
        System.out.print(n + " " + (n - 1) + " " + (n - 2));
 
    // if n is even and n, n-3 gcd is 1
    else if (__gcd(n, (n - 3)) == 1)
        System.out.print( n + " " +(n - 1)+ " " + (n - 3));
 
    else
        System.out.print((n - 1) + " " + (n - 2) + " " + (n - 3));
}
 
// Driver code
public static void main (String[] args) {
    int n = 12;
 
    // function call
    MaxLCM(n);
    }
}
// This code is contributed by anuj_67..

Python3

# Python 3 Program to find three integers
# less than N whose LCM is maximum
from math import gcd
 
# function to find three integers
# less than N whose LCM is maximum
def MaxLCM(n) :
 
    # if n is odd
    if (n % 2 != 0) :
        print(n, (n - 1), (n - 2))
 
    # if n is even and n, n-3 gcd is 1
    elif (gcd(n, (n - 3)) == 1) :
        print(n, (n - 1), (n - 3))
 
    else :
        print((n - 1), (n - 2), (n - 3))
 
# Driver Code
if __name__ == "__main__" :
     
    n = 12
 
    # function call
    MaxLCM(n)
 
# This code is contributed by Ryuga

C#

// C# Program to find three integers
// less than N whose LCM is maximum
 
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 find three integers
// less than N whose LCM is maximum
static void MaxLCM(int n)
{
    // if n is odd
    if (n % 2 != 0)
        Console.Write(n + " " + (n - 1) + " " + (n - 2));
 
    // if n is even and n, n-3 gcd is 1
    else if (__gcd(n, (n - 3)) == 1)
        Console.Write( n + " " +(n - 1)+ " " + (n - 3));
 
    else
        Console.Write((n - 1) + " " + (n - 2) + " " + (n - 3));
}
 
// Driver code
public static void Main () {
    int n = 12;
 
    // function call
    MaxLCM(n);
    }
}
// This code is contributed by anuj_67..

PHP

 $b)
        return __gcd($a - $b, $b);
    return __gcd($a, $b - $a);
} 
 
// function to find three integers
// less than N whose LCM is maximum
function MaxLCM($n)
{
    // if n is odd
    if ($n % 2 != 0)
        echo $n , " " , ($n - 1) ,
                  " " , ($n - 2);
 
    // if n is even and n, n-3 gcd is 1
    else if (__gcd($n, ($n - 3)) == 1)
        echo $n , " " , ($n - 1),
                  " " , ($n - 3);
  
    else
        echo ($n - 1) , " " , ($n - 2),
                        " " , ($n - 3);
}
 
// Driver code
$n = 12;
 
// function call
MaxLCM($n);
 
// This code is contributed by Sachin
?>

Javascript

 
// JavaScript Program to find three integers
// less than N whose LCM is maximum
 
    // 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 find three integers
    // less than N whose LCM is maximum
    function MaxLCM(n) {
        // if n is odd
        if (n % 2 != 0)
            document.write(n + " " + (n - 1) + " " + (n - 2));
 
        // if n is even and n, n-3 gcd is 1
        else if (__gcd(n, (n - 3)) == 1)
            document.write(n + " " + (n - 1) + " " + (n - 3));
 
        else
            document.write((n - 1) + " " + (n - 2) + " " + (n - 3));
    }
 
    // Driver code
     
        var n = 12;
 
        // function call
        MaxLCM(n);
         
// This code contributed by Rajput-Ji
 

输出：

11 10 9