通过重复除法和乘法将X转换为Y的最小步骤

给定两个整数X和Y ，任务是找到在每个步骤中使用任何操作将整数X转换为Y的最小步骤数：

• 用数字除以任何自然数
• 将数字乘以任何自然数

例子：

Input: X = 8, Y = 12 
Output: 2 
Explanation: 
First divide 8 by 2: 8/2 = 4 
Then multiply by 3: 4*3 = 12

Input: X = 4, Y = 8 
Output: 1 
Explanation: 
To convert 4 to 8 multiply 4 by 2: 4 * 2 = 8 

方法：解决上述问题：

• 确保X包含X和Y中较小的值。现在，如果X大于Y，则我们知道将较小的数字更改为较大的数字总是比较容易。因此，我们只交换X和Y的值，然后执行下面提到的步骤。
• 如果两个整数都相同，则答案将为零，因为不进行任何转换。
  例如，

If X = 4, Y = 4

Here 4 = 4
Therefore, answer = 0
(as they both are already same)

• 但是，如果X小于Y，则：
  • 我们必须检查Y％X是否给出0。
  • 如果是，则Y可以表示为X *(Y / X) ，我们一步就可以得到所需的输出。
    例如，

If X = 4, Y = 12

Here 12 % 4 = 0
Therefore, answer = 1
(4 * 3 = 12)

• 否则，答案将为2，因为它需要两个步骤，一个步骤用于除法(X = X / X)，另一个步骤用于相乘(X = X * Y)。
  例如，

If X = 8, Y = 13

Here 13 % 8 != 0
Therefore, 
1. X = X/X = 8/8 = 1
2. X = X*Y = 1*13 = 13

Hence, answer = 2

下面是上述方法的实现：

C++

// C++ implementation to find minimum
// steps to convert X to Y by repeated
// division and multiplication
 
#include 
using namespace std;
 
int solve(int X, int Y)
{
    // Check if X is greater than Y
    // then swap the elements
    if (X > Y) {
        int temp = X;
        X = Y;
        Y = temp;
    }
 
    // Check if X equals Y
    if (X == Y)
        cout << 0 << endl;
 
    else if (Y % X == 0)
        cout << 1 << endl;
    else
        cout << 2 << endl;
}
 
// Driver code
int main()
{
    int X = 8, Y = 13;
    solve(X, Y);
 
    return 0;
}

Java

// Java implementation to find minimum
// steps to convert X to Y by repeated
// division and multiplication
class GFG{
 
static int solve(int X, int Y)
{
    // Check if X is greater than Y
    // then swap the elements
    if (X > Y)
    {
        int temp = X;
        X = Y;
        Y = temp;
    }
 
    // Check if X equals Y
    if (X == Y)
        System.out.println(0 );
 
    else if (Y % X == 0)
        System.out.println( 1 );
    else
        System.out.println(2 );
        return 0;
}
 
// Driver code
public static void main(String args[])
{
    int X = 8, Y = 13;
    solve(X, Y);
}
}
 
// This code is contributed by shivanisinghss2110

Python3

# Python3 implementation to find minimum
# steps to convert X to Y by repeated
# division and multiplication
def solve(X, Y):
     
    # Check if X is greater than Y
    # then swap the elements
    if (X > Y):
        temp = X
        X = Y
        Y = temp
   
    # Check if X equals Y
    if (X == Y):
        print(0)
   
    elif (Y % X == 0):
        print(1)
    else:
        print(2)
  
# Driver code
X = 8
Y = 13
 
solve(X, Y)
 
# This code is contributed by code_hunt

C#

// C# implementation to find minimum
// steps to convert X to Y by repeated
// division and multiplication
using System;
 
class GFG{
 
static int solve(int X, int Y)
{
     
    // Check if X is greater than Y
    // then swap the elements
    if (X > Y)
    {
        int temp = X;
        X = Y;
        Y = temp;
    }
 
    // Check if X equals Y
    if (X == Y)
        Console.WriteLine(0);
 
    else if (Y % X == 0)
        Console.WriteLine(1);
    else
        Console.WriteLine(2);
    return 0;
}
 
// Driver code
public static void Main(String[] args)
{
    int X = 8, Y = 13;
    solve(X, Y);
}
}
 
// This code is contributed by amal kumar choubey

Javascript

 
    // Javascript implementation to find minimum
    // steps to convert X to Y by repeated
    // division and multiplication
     
    function solve(X, Y)
    {
        // Check if X is greater than Y
        // then swap the elements
        if (X > Y) {
            let temp = X;
            X = Y;
            Y = temp;
        }
 
        // Check if X equals Y
        if (X == Y)
            document.write(0);
 
        else if (Y % X == 0)
            document.write(1);
        else
            document.write(2);
    }
     
    let X = 8, Y = 13;
    solve(X, Y);
 
 

输出：

2