使两个数相等所需的最少操作

给定两个整数A和B。任务是找到使A和B相等所需的最小操作数。在每个操作中，可以执行以下任一步骤：

• 用其初始值递增A或B。
• 将A和B都增加其初始值

例子：

Input: A = 4, B = 10
Output: 4
Explanation:
Initially A = 4, B = 10
Operation 1: Increment A only: A = A + 4 = 8
Operation 2: Increment A only: A = A + 4 = 12
Operation 3: Increment A only: A = A + 4 = 16
Operation 4: Increment A and B: A = A + 4 = 20 and B = B + 10 = 20
They are equal now.

Input: A = 7, B = 23
Output: 22
Explanation:
Initially A = 7, B = 23
Operation 1 – 7: Increment A and B: A = 56 and B = 161
Operation 8 – 22: Increment A: A = 161 and B = 161
They are equal now.

方法：可以使用GCD解决此问题。

1. 如果A大于B，则交换A和B。
2. 现在减少B，使A和B的gcd变为1。
3. 因此，达到相等值所需的最小操作为(B – 1)。

例如：如果A = 4，B = 10：

• 第1步：比较4和10，因为我们始终需要B作为更大的值。在这里，B已经大于A。因此，现在不需要交换。
• 步骤2： GCD(4，10)=2。因此，我们将B减小为B / 2。现在，A = 4，B = 5。
  GCD(4，5)= 1，这是目标。
• 步骤3： (B – 1的当前值)将是所需的计数。此处，当前B =5。因此(5 – 1 = 4)，即总共需要进行4次操作。

下面是上述方法的实现。

C++

// C++ program to find minimum
// operations required to
// make two numbers equal
  
#include 
using namespace std;
  
// Function to return the
// minimum operations required
long long int minOperations(
    long long int A,
    long long int B)
{
  
    // Keeping B always greater
    if (A > B)
        swap(A, B);
  
    // Reduce B such that
    // gcd(A, B) becomes 1.
    B = B / __gcd(A, B);
  
    return B - 1;
}
  
// Driver code
int main()
{
    long long int A = 7, B = 15;
  
    cout << minOperations(A, B)
         << endl;
  
    return 0;
}

Java

// Java program to find minimum
// operations required to
// make two numbers equal
class GFG{
   
// Function to return the
// minimum operations required
static int  minOperations(
    int  A,
    int  B)
{
   
    // Keeping B always greater
    if (A > B) {
        A = A+B;
        B = A-B;
        A = A-B;
    }
   
    // Reduce B such that
    // gcd(A, B) becomes 1.
    B = B / __gcd(A, B);
   
    return B - 1;
}
static int __gcd(int a, int b)  
{  
    return b == 0? a:__gcd(b, a % b);     
} 
  
// Driver code
public static void main(String[] args)
{
    int  A = 7, B = 15;
   
    System.out.print(minOperations(A, B)
         +"\n");
   
}
}
  
// This code contributed by sapnasingh4991

Python3

# Python program to find minimum 
# operations required to 
# make two numbers equal 
import math
  
# Function to return the 
# minimum operations required 
def minOperations(A, B):
  
    # Keeping B always greater 
    if (A > B):
        swap(A, B)
  
    # Reduce B such that 
    # gcd(A, B) becomes 1. 
    B = B // math.gcd(A, B); 
  
    return B - 1
  
# Driver code 
A = 7
B = 15
  
print(minOperations(A, B))
  
# This code is contributed by Sanjit_Prasad

C#

// C# program to find minimum
// operations required to
// make two numbers equal
using System;
  
class GFG{
    
// Function to return the
// minimum operations required
static int  minOperations(
    int  A,
    int  B)
{
    
    // Keeping B always greater
    if (A > B) {
        A = A+B;
        B = A-B;
        A = A-B;
    }
    
    // Reduce B such that
    // gcd(A, B) becomes 1.
    B = B / __gcd(A, B);
    
    return B - 1;
}
static int __gcd(int a, int b)  
{  
    return b == 0? a:__gcd(b, a % b);     
} 
   
// Driver code
public static void Main(String[] args)
{
    int  A = 7, B = 15;
    
    Console.Write(minOperations(A, B)
         +"\n");
}
}
  
// This code is contributed by sapnasingh4991

输出：

14

时间复杂度： O(log(max(A，B))