将给定数字拆分为两个具有 GCD 1 的不同数字

给定一个整数作为N ，任务是将N转换为两个数字a和b ，使得a < b并且GCD(a, b)等于 1。

例子：

Input: N = 12
Output: 5 7

Input: N = 9
Output: 4 5

方法：这个想法是从数字本身开始从循环的后面开始迭代。请按照以下步骤解决问题。

• 如果N小于等于2，则答案不存在。
• 否则，使用变量i遍历范围[N/2, 0)并执行以下任务：
  • 如果i和Ni的 gcd 为1 ，则将其打印为答案并返回。

下面是上述方法的实现：

C++

/// C++ program for the above approach
#include 
using namespace std;
 
// Function to split N into two numbers
void find(int n)
{
 
    // Base Case
    if (n <= 2) {
        cout << "-1";
    }
    else {
 
        // Since a < b
        // start from n/2
        for (int i = n / 2; i > 0; i--) {
 
            // Sum of a and b is n
            int a = i, b = n - i;
 
            // Check the gcd
            if (__gcd(a, b) == 1 && a < b) {
                cout << a <<", " << b;
                return;
            }
        }
    }
}
 
// Driver Code
int main()
{
 
    int N = 9;
 
    find(N);
 
    return 0;
}

Java

/// Java program for the above approach
class GFG {
 
  // Function to split N into two numbers
  public static void find(int n) {
 
    // Base Case
    if (n <= 2) {
      System.out.println("-1");
    } else {
 
      // Since a < b
      // start from n/2
      for (int i = n / 2; i > 0; i--) {
 
        // Sum of a and b is n
        int a = i, b = n - i;
 
        // Check the gcd
        if (__gcd(a, b) == 1 && a < b) {
          System.out.println(a + ", " + b);
          return;
        }
      }
    }
  }
 
  static int __gcd(int a, int b) {
    if (b == 0)
      return a;
    return __gcd(b, a % b);
  }
 
  // Driver Code
  public static void main(String args[]) {
    int N = 9;
    find(N);
  }
}
 
// This code is contributed by saurabh_jaiswal.

Python3

# Python 3 program for the above approach
import math
 
# Function to split N into two numbers
def find(n):
 
    # Base Case
    if (n <= 2):
        print("-1")
 
    else:
 
        # Since a < b
        # start from n/2
        for i in range(n // 2, -1, -1):
 
            # Sum of a and b is n
            a = i
            b = n - i
 
            # Check the gcd
            if (math.gcd(a, b) == 1 and a < b):
                print(a,",", b)
                return
 
# Driver Code
if __name__ == "__main__":
 
    N = 9
    find(N)
 
    #vThis code is contributed by ukasp.

C#

/// C# program for the above approach
using System;
public class GFG {
 
  // Function to split N into two numbers
  public static void find(int n) {
 
    // Base Case
    if (n <= 2) {
      Console.WriteLine("-1");
    } else {
 
      // Since a < b
      // start from n/2
      for (int i = n / 2; i > 0; i--) {
 
        // Sum of a and b is n
        int a = i, b = n - i;
 
        // Check the gcd
        if (__gcd(a, b) == 1 && a < b) {
          Console.WriteLine(a + ", " + b);
          return;
        }
      }
    }
  }
 
  static int __gcd(int a, int b) {
    if (b == 0)
      return a;
    return __gcd(b, a % b);
  }
 
  // Driver Code
  public static void Main(String []args) {
    int N = 9;
    find(N);
  }
}
 
// This code is contributed by 29AjayKumar

Javascript

       // JavaScript code for the above approach
 
       // 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 split N into two numbers
       function find(n) {
 
           // Base Case
           if (n <= 2) {
               document.write("-1");
           }
           else {
 
               // Since a < b
               // start from n/2
               for (let i = Math.floor(n / 2); i > 0; i--) {
 
                   // Sum of a and b is n
                   let a = i, b = n - i;
 
                   // Check the gcd
                   if (__gcd(a, b) == 1 && a < b) {
                       document.write(a + ", " + b);
                       return;
                   }
               }
           }
       }
 
       // Driver Code
       let N = 9;
       find(N);
 
 // This code is contributed by Potta Lokesh
   

输出

4, 5

时间复杂度： O(N log(N))
辅助空间： O(1)