找到给定总和为 N 的三元组 (X, Y, Z)，两个数字的 GCD 是第三个数字

给定一个正整数N ，任务是找到三个不同的正整数 (X, Y, Z) 的三元组，使得 X + Y + Z = N 和 X = GCD (Y, Z)。

例子：

Input: N = 12
Output: 2 4 6
Explanation: The triplet (2, 4, 6) is set of distinct integers such that 2 + 4 + 6 = 12 and 2 = GCD(4, 6).

Input: N = 5675
Output:1 2835 2839

朴素方法：基本思想是用总和 N 迭代 (X, Y, Z) 的所有可能的三元组，并且对于每个三元组，检查 GCD(Y, Z) = X。

时间复杂度： O(N ² )
辅助空间： O(1)

有效方法：上述方法可以通过观察进一步优化，对于任何给定的 N，存在以下三种情况：

• 情况 1：如果N是偶数，则有效的三元组是 ( 1 , N/2 , N/2 -1 )。
• 情况 2：如果N为奇数且 ( N/2 ) 为偶数，则有效的三元组为 ( 1 , N/2 + 1 , N/2 -1 )。
• 情况 3：如果N是奇数并且 ( N/2 ) 也是奇数，则有效的三元组是 ( 1 , N/2 – 2 , N/2 + 2 )。

因此，对于任何给定的N ，识别案例并打印其各自的三元组。
下面是该方法的实现：

C++

// C++ program of the above approach
#include 
using namespace std;
  
// Function to find a triplet (X, Y, Z)
// of distinct integers with their sum
// as N and GCD(Y, Z) = X
int printTriplet(int N)
{
    // Case 1 where N is even
    if (N % 2 == 0) {
        cout << 1 << " " << (N / 2) 
<< " " << (N / 2) - 1;
    }
    else {
  
        // Case 2 where N is Odd
        // and N/2 is even
        if ((N / 2) % 2 == 0) {
            cout << 1 << " "
 << (N / 2) - 1 << " "
                 << (N / 2) + 1;
        }
  
        // Case 3 where N is Odd
        // and N/2 is also odd
        else {
            cout << 1 << " "
 << (N / 2) - 2 << " "
                 << (N / 2) + 2;
        }
    }
}
  
// Driver Code
int main()
{
    int N = 5875;
    printTriplet(N);
  
    return 0;
}

Java

// Java program of the above approach
import java.util.*;
class GFG 
{
  
  // Function to find a triplet (X, Y, Z)
  // of distinct integers with their sum
  // as N and GCD(Y, Z) = X
  static void printTriplet(int N) 
  {
  
    // Case 1 where N is even
    if (N % 2 == 0) {
      System.out.print(1 + " " + (N / 2) + 
                       " " + ((N / 2) - 1));
    } else {
  
      // Case 2 where N is Odd
      // and N/2 is even
      if ((N / 2) % 2 == 0) {
        System.out.print(1 + " " + ((N / 2) - 1) +
                         " " + ((N / 2) + 1));
      }
  
      // Case 3 where N is Odd
      // and N/2 is also odd
      else {
        System.out.print(1 + " " + ((N / 2) - 2) +
                         " " + ((N / 2) + 2));
      }
    }
  }
  
  // Driver Code
  public static void main(String[] args) {
    int N = 5875;
    printTriplet(N);
  
  }
}
  
// This code is contributed by 29AjayKumar

Python3

# python3 program of the above approach
  
# Function to find a triplet (X, Y, Z)
# of distinct integers with their sum
# as N and GCD(Y, Z) = X
def printTriplet(N):
  
    # Case 1 where N is even
    if (N % 2 == 0):
        print(f"{1} {(N / 2)} {(N / 2) - 1}")
  
    else:
  
        # Case 2 where N is Odd
        # and N/2 is even
        if ((N // 2) % 2 == 0):
            print(f"{1} {(N // 2) - 1} {(N // 2) + 1}")
  
        # Case 3 where N is Odd
        # and N/2 is also odd
        else:
            print(f"{1} {(N // 2) - 2} {(N // 2) + 2}")
  
# Driver Code
if __name__ == "__main__":
  
    N = 5875
    printTriplet(N)
  
# This code is contributed by rakeshsahni

C#

// C# program of the above approach
using System;
  
class GFG{
  
// Function to find a triplet (X, Y, Z)
// of distinct integers with their sum
// as N and GCD(Y, Z) = X
static void printTriplet(int N)
{
      
    // Case 1 where N is even
    if (N % 2 == 0)
    {
        Console.Write(1 + " " + (N / 2) + " " + 
                               ((N / 2) - 1));
    }
    else
    {
          
        // Case 2 where N is Odd
        // and N/2 is even
        if ((N / 2) % 2 == 0) 
        {
            Console.Write(1 + " " + ((N / 2) - 1) + " " + 
                                    ((N / 2) + 1));
        }
  
        // Case 3 where N is Odd
        // and N/2 is also odd
        else
        {
            Console.Write(1 + " " + ((N / 2) - 2) + " " + 
                                    ((N / 2) + 2));
        }
    }
}
  
// Driver Code
public static void Main()
{
    int N = 5875;
      
    printTriplet(N);
}
}
  
// This code is contributed by ukasp

Javascript

       // JavaScript code for the above approach
 
       // Function to find a triplet (X, Y, Z)
       // of distinct integers with their sum
       // as N and GCD(Y, Z) = X
       function printTriplet(N)
       {
         
           // Case 1 where N is even
           if (N % 2 == 0) {
               document.write(1 + " " + Math.floor(N / 2)
                   + " " + (Math.floor(N / 2) - 1));
           }
           else {
 
               // Case 2 where N is Odd
               // and N/2 is even
               if ((N / 2) % 2 == 0) {
                   document.write(1 + " "
                       + (Math.floor(N / 2) - 1) + " "
                       + (Math.floor(N / 2) + 1));
               }
 
               // Case 3 where N is Odd
               // and N/2 is also odd
               else {
                   document.write(1 + " "
                       + (Math.floor(N / 2) - 2) + " "
                       + (Math.floor(N / 2) + 2));
               }
           }
       }
 
       // Driver Code
       let N = 5875;
       printTriplet(N);
 
 // This code is contributed by Potta Lokesh
   

输出

1 2935 2939

时间复杂度： O(1)
辅助空间： O(1)