打印总和等于 N 和 LCM 至多 N/2 的三元组

给定一个正整数N ，任务是找到一个三元组{X, Y, Z} ，使得X 、 Y和Z 的最小公倍数小于或等于N/2并且X 、 Y 、并且Z等于N 。如果存在多个答案，则打印其中任何一个。

例子：

Input: N = 8
Output: 4 2 2
Explanation: 
One possible triplet is {4, 2, 2}. The sum of all the integers is equal to (4+2+2 = 8) and LCM is equal 4 which is equal to N/2( =4).

Input: N = 5
Output: 1 2 2
Explanation: 
One possible triplet is {1, 2, 2}. The sum of all the integers is equal to (1+2+2 = 5) and LCM is equal 2 which is equal to N/2( =2).

处理方法：可以根据以下事实解决问题：

Suppose, N = X+Y+Z, then:

1. If N is odd then either any one of X, Y or Z is odd or all three are odd numbers.
2. If N is even, then all numbers must be even.
3. Therefore, the idea is to include minimum possible value in the answer according to the above facts which will decrease the LCM of X, Y and Z.

请按照以下步骤解决问题：

• 将3 个变量x、y和z初始化为0以存储三元组的值。
• 如果N%2不等于0，即N是奇数，则执行以下步骤：
  • 如果N是奇数，那么x、y或z中至少有一个应该是奇数，并且在最坏的情况下， x、y、z的 LCM 是N/2 。
  • 将x和y的值设置为N/2 ，将z的值设置为1。
• 否则，如果N%4不等于0，则z的值可以是2 ， x和y的值可以是N/2-1。否则， x的值可以是N/2 ， y和z的值可以是N/4。
• 执行上述步骤后，打印x、y和z 的值。

下面是上述方法的实现。

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find a triplet  with the
// sum equal to N and LCM less than or
// equal to N
void validTriplet(int N)
{
    // Stores the triplets
    int x, y, z;
 
    // If N is odd
    if ((N % 2) != 0) {
        x = N / 2;
        y = N / 2;
        z = 1;
    }
    // If N is even
    else {
 
        // If N is not a multiple of 4
        if ((N % 4) != 0) {
            x = (N / 2) - 1;
            y = (N / 2) - 1;
            z = 2;
        }
 
        // If N is a multiple of 4
        else {
            x = N / 2;
            y = N / 4;
            z = N / 4;
        }
    }
 
    // Finally, print the answer
    cout << x << " " << y << " " << z << '\n';
}
 
// Driver Code
int main()
{
    // Given Input
    int N = 5;
 
    // Function Call
    validTriplet(N);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
 
class GFG
{
   
    // Function to find a triplet  with the
    // sum equal to N and LCM less than or
    // equal to N
    static void validTriplet(int N)
    {
       
        // Stores the triplets
        int x, y, z;
 
        // If N is odd
        if ((N % 2) != 0) {
            x = N / 2;
            y = N / 2;
            z = 1;
        }
       
        // If N is even
        else {
 
            // If N is not a multiple of 4
            if ((N % 4) != 0) {
                x = (N / 2) - 1;
                y = (N / 2) - 1;
                z = 2;
            }
 
            // If N is a multiple of 4
            else {
                x = N / 2;
                y = N / 4;
                z = N / 4;
            }
        }
 
        // Finally, print the answer
        System.out.print(x + " " + y + " " + z);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
       
        // Given Input
        int N = 5;
 
        // Function Call
        validTriplet(N);
    }
}
 
// This code is contributed by Potta Lokesh

Python3

# Function to find a triplet  with the
# sum equal to N and LCM less than or
# equal to N
def validTriplet(N):
   
     # If N is odd
    if (N % 2) != 0:
        x = N//2
        y = N//2
        z = 1
         
        # if N is even
    else:
       
         # If N is not a multiple of 4
        if (N % 4) != 0:
            x = N//2 - 1
            y = N//2 - 1
            z = 2
             
            # if N is multiple of 4
        else:
            x = N//2
            y = N//4
            z = N//4
             
     # Print the result
    print(str(x) + " " + str(y) + " " + str(z))
 
# Driver code
N = 5
validTriplet(N)
 
# This code is contributed by Parth Manchanda

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find a triplet  with the
// sum equal to N and LCM less than or
// equal to N
static void validTriplet(int N)
{
   
    // Stores the triplets
    int x, y, z;
 
    // If N is odd
    if ((N % 2) != 0) {
        x = N / 2;
        y = N / 2;
        z = 1;
    }
    // If N is even
    else {
 
        // If N is not a multiple of 4
        if ((N % 4) != 0) {
            x = (N / 2) - 1;
            y = (N / 2) - 1;
            z = 2;
        }
 
        // If N is a multiple of 4
        else {
            x = N / 2;
            y = N / 4;
            z = N / 4;
        }
    }
 
    // Finally, print the answer
    Console.Write(x + " " + y + " " + z );
}
 
// Driver Code
public static void Main()
{
    // Given Input
    int N = 5;
 
    // Function Call
    validTriplet(N);
     
}
}
 
// This code is contributed by ipg2016107.

Javascript

// Java program for the above approach
// Function to find a triplet  with the
// sum equal to N and LCM less than or
// equal to N
function validTriplet(N)
    {
       
        // Stores the triplets
        var x, y, z;
 
        // If N is odd
        if ((N % 2) == 0) {
            x = N / 2;
            y = N / 2;
            z = 1;
        }
       
        // If N is even
        else {
 
            // If N is not a multiple of 4
            if ((N % 4) != 0) {
                x = (N / 2) - 1;
                y = (N / 2) - 1;
                z = 1;
            }
 
            // If N is a multiple of 4
            else {
                x = N / 2;
                y = N / 4;
                z = N / 4;
            }
        }
 
        // Finally, print the answer
        document.write(x + " " + y + " " + z);
    }
 
    // Driver Code
    // Given Input
       var N = 4;
 
       // Function Call
       validTriplet(N);
 
// This code is contributed by shivanisinghss2110

输出

2 2 1

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