通过将相同的值添加到两个数字中，可能的最大 GCD

给定两个数字A和B ，任务是找到通过将数字X添加到A和B可以获得的最大公约数（GCD） 。

例子

Input: A = 1, B = 5
Output: 4
Explanation: Adding X = 15, the obtained numbers are A = 16 and B = 20. Therefore, GCD of A, B is 4.

Input: A = 2, B = 23
Output: 21

方法：这个问题可以使用欧几里得 GCD 算法的特性以非常优化的方式解决。请按照以下步骤解决问题：

• 如果 a > b： GCD(a, b)= GCD(a – b, b) 。因此， GCD(a, b) = GCD(a – b, b )。
• 将x添加到A, B，gcd(a + x, b + x) = gcd(a – b, b + x)。可以看出（a-b）保持不变。
• 可以肯定地说，这些数字的 GCD 永远不会超过(a – b) 。因为(b + x)可以通过添加一个可能的x值成为(a – b)的倍数。
• 因此，可以得出结论，GCD 仍然存在（a-b）。

下面是上述方法的实现。

C++

// C++ implementation of above approach.
#include 
using namespace std;
 
// Function to calculate maximum
// gcd of two numbers possible by
// adding same value to both a and b
void maxGcd(int a, int b)
{
    cout << abs(a - b);
}
 
// Driver Code
int main()
{
    // Given Input
    int a = 2231;
    int b = 343;
 
    maxGcd(a, b);
 
    return 0;
}

Java

// Java implementation of above approach.
import java.io.*;
 
class GFG
{
 
  // Function to calculate maximum
  // gcd of two numbers possible by
  // adding same value to both a and b
  static void maxGcd(int a, int b)
  {
    System.out.println(Math.abs(a - b));
  }
 
  // Driver Code
  public static void main (String[] args)
  {
 
    // Given Input
    int a = 2231;
    int b = 343;
 
    maxGcd(a, b);
 
  }
}
 
// This code is contributed by Potta Lokesh

Python3

# Python3 program for the above approach
 
# Function to calculate maximum
# gcd of two numbers possible by
# adding same value to both a and b
def maxGcd(a, b):
     
    print(abs(a - b))
 
# Driver code
 
# Given Input
a = 2231
b = 343
 
maxGcd(a, b)
 
# This code is contributed by Parth Manchanda

C#

// C# program for the above approach
using System;
 
class GFG{
 
 // Function to calculate maximum
  // gcd of two numbers possible by
  // adding same value to both a and b
  static void maxGcd(int a, int b)
  {
    Console.Write(Math.Abs(a - b));
  }
 
// Driver Code
static public void Main ()
{
     
     // Given Input
    int a = 2231;
    int b = 343;
 
    maxGcd(a, b);
}
}
 
// This code is contributed by code_hunt.

Javascript

       // JavaScript Program for the above approach
 
       // Function to calculate maximum
       // gcd of two numbers possible by
       // adding same value to both a and b
       function maxGcd(a, b) {
           document.write(Math.abs(a - b));
       }
 
       // Driver Code
 
       // Given Input
       let a = 2231;
       let b = 343;
 
       maxGcd(a, b);
 
   // This code is contributed by Potta Lokesh
   

输出：

1888

时间复杂度： O(1)
辅助空间： O(1)