需要对两个给定数字之一执行的最小递增/递减数，以使其非互质

给定两个正整数A和B ，任务是找到需要在A或B上执行的最小递增/递减数，以使这两个数字互质。

例子：

Input: A = 12, B = 3
Output: 0
Explanation:
As 12 & 3 are already non-coprimes, so the minimum count of increment/decrement operation required is 0.

Input: A = 7, B = 17
Output: 2

方法：可以根据以下观察解决给定的问题：

• 如果A和B的最大公约数大于1 ，则不会执行递增或递减，因为数字已经是非互质的。
• 现在，检查A和B在两个方向上的差异1 。因此，只需一步即可将任何数转换为偶数。
• 如果上述两种情况均不适用，则需要进行2 次递增/递减操作以使数字A和B成为最接近的偶数，从而使数字变为非共素数。

根据以上观察，按照以下步骤解决问题：

• 如果A和B的 GCD 不等于1 ，则打印0 ，因为不需要任何操作。
• 否则，如果 { {A + 1, B} , {A – 1, B} , {A, B + 1} , {A, B – 1} } 对之一的 GCD 不等于1 ，则打印1因为只需要一项操作。
• 否则，打印2 。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to find the minimum number of
// increments/decrements operations required
// to make both the numbers non-coprime
int makeCoprime(int a, int b)
{
    // If a & b are already non-coprimes
    if (__gcd(a, b) != 1)
        return 0;
 
    // If a & b can become non-coprimes
    // by only incrementing/decrementing
    // a number only once
    if (__gcd(a - 1, b) != 1
        or __gcd(a + 1, b) != 1
        or __gcd(b - 1, a) != 1
        or __gcd(b + 1, a) != 1)
        return 1;
 
    // Otherwise
    return 2;
}
 
// Driver Code
int main()
{
    int A = 7, B = 17;
    cout << makeCoprime(A, B);
 
    return 0;
}

Java

// Java program for the above approach
public class GFG
{
 
  // function to calculate gcd
  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 the minimum number of
  // increments/decrements operations required
  // to make both the numbers non-coprime
  static int makeCoprime(int a, int b)
  {
 
    // If a & b are already non-coprimes
    if (__gcd(a, b) != 1)
      return 0;
 
    // If a & b can become non-coprimes
    // by only incrementing/decrementing
    // a number only once
    if (__gcd(a - 1, b) != 1 || __gcd(a + 1, b) != 1
        || __gcd(b - 1, a) != 1
        || __gcd(b + 1, a) != 1)
      return 1;
 
    // Otherwise
    return 2;
  }
 
  // Driver code
  public static void main(String args[])
  {
    int A = 7, B = 17;
    System.out.println(makeCoprime(A, B));
  }
}
 
// This code is contributed by SoumikMondal

Python3

# Python3 program for the above approach
from math import gcd
 
# Function to find the minimum number of
# increments/decrements operations required
# to make both the numbers non-coprime
def makeCoprime(a, b):
     
    # If a & b are already non-coprimes
    if (gcd(a, b) != 1):
        return 0
 
    # If a & b can become non-coprimes
    # by only incrementing/decrementing
    # a number only once
    if (gcd(a - 1, b) != 1 or
        gcd(a + 1, b) != 1 or
        gcd(b - 1, a) != 1 or
        gcd(b + 1, a) != 1):
        return 1
 
    # Otherwise
    return 2
 
# Driver Code
if __name__ == '__main__':
     
    A = 7
    B = 17
     
    print(makeCoprime(A, B))
 
# This code is contributed by SURENDRA_GANGWAR

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to calculate gcd
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 the minimum number of
// increments/decrements operations required
// to make both the numbers non-coprime
static int makeCoprime(int a, int b)
{
     
    // If a & b are already non-coprimes
    if (__gcd(a, b) != 1)
        return 0;
     
    // If a & b can become non-coprimes
    // by only incrementing/decrementing
    // a number only once
    if (__gcd(a - 1, b) != 1 ||
        __gcd(a + 1, b) != 1 ||
        __gcd(b - 1, a) != 1 ||
        __gcd(b + 1, a) != 1)
        return 1;
     
    // Otherwise
    return 2;
}
 
// Driver Code
public static void Main(String[] args)
{
    int A = 7, B = 17;
     
    Console.Write(makeCoprime(A, B));
}
}
 
// This code is contributed by sanjoy_62

Javascript

 
        // JavaScript program for the above approach
 
        function __gcd(a, b) {
            if (b == 0)
                return a;
            return __gcd(b, a % b);
 
        }
 
        // Function to find the minimum number of
        // increments/decrements operations required
        // to make both the numbers non-coprime
        function makeCoprime(a, b) {
            // If a & b are already non-coprimes
            if (__gcd(a, b) != 1)
                return 0;
 
            // If a & b can become non-coprimes
            // by only incrementing/decrementing
            // a number only once
            if (__gcd(a - 1, b) != 1
                || __gcd(a + 1, b) != 1
                || __gcd(b - 1, a) != 1
                || __gcd(b + 1, a) != 1)
                return 1;
 
            // Otherwise
            return 2;
        }
 
        // Driver Code
 
        let A = 7, B = 17;
        document.write(makeCoprime(A, B));
 
    // This code is contributed by Potta Lokesh
  

输出：

2

时间复杂度： O(log(A, B))
辅助空间： O(1)