将n转换为m所需的最小操作数|套装2

给定两个整数n和m以及a和b ，在一次运算中， n可以乘以a或b 。任务是用最少的给定操作数将n转换为m 。如果无法通过给定的操作将n转换为m，则打印-1。

例子：

Input: n = 120, m = 51840, a = 2, b = 3
Output: 7
120 * 2 * 2 * 2 * 2 * 3 * 3 * 3 = 51840

Input: n = 10, m = 50, a = 5, b = 7
Output: 1
10 * 5 = 50 

在上一篇文章中，我们讨论了使用除法的方法。

在本文中，我们将使用一种方法来使用递归查找最少数量的操作。递归将包含两个状态，数字乘以a或b并计算步数。这两个步骤中最少的一个就是答案。

下面是上述方法的实现：

C++

// C++ implementation of the above approach
  
#include 
using namespace std;
#define MAXN 10000000
  
// Function to find the minimum number of steps
int minimumSteps(int n, int m, int a, int b)
{
    // If n exceeds M
    if (n > m)
        return MAXN;
  
    // If N reaches the target
    if (n == m)
        return 0;
  
    // The minimum of both the states
    // will be the answer
    return min(1 + minimumSteps(n * a, m, a, b),
               1 + minimumSteps(n * b, m, a, b));
}
  
// Driver code
int main()
{
    int n = 120, m = 51840;
    int a = 2, b = 3;
    cout << minimumSteps(n, m, a, b);
    return 0;
}

Java

// Java implementation of the above approach
class GFG
{
    static int MAXN = 10000000;
      
    // Function to find the minimum number of steps
    static int minimumSteps(int n, int m, int a, int b)
    {
        // If n exceeds M
        if (n > m)
            return MAXN;
      
        // If N reaches the target
        if (n == m)
            return 0;
      
        // The minimum of both the states
        // will be the answer
        return Math.min(1 + minimumSteps(n * a, m, a, b),
                1 + minimumSteps(n * b, m, a, b));
    }
      
    // Driver code
    public static void main (String[] args)
    {
        int n = 120, m = 51840;
        int a = 2, b = 3;
        System.out.println(minimumSteps(n, m, a, b));
    }
}
  
// This code is contributed by ihritik

Python3

# Python 3 implementation of the 
# above approach
MAXN = 10000000
  
# Function to find the minimum 
# number of steps
def minimumSteps(n, m, a, b):
      
    # If n exceeds M
    if (n > m):
        return MAXN
  
    # If N reaches the target
    if (n == m):
        return 0
  
    # The minimum of both the states
    # will be the answer
    return min(1 + minimumSteps(n * a, m, a, b), 
               1 + minimumSteps(n * b, m, a, b))
  
# Driver code
if __name__ == '__main__':
    n = 120
    m = 51840
    a = 2
    b = 3
    print(minimumSteps(n, m, a, b))
  
# This code is contributed by
# Surendra_Gangwar

C#

// C# implementation of the above approach
using System;
  
class GFG
{
    static int MAXN = 10000000;
      
    // Function to find the minimum number of steps
    static int minimumSteps(int n, int m, int a, int b)
    {
        // If n exceeds M
        if (n > m)
            return MAXN;
      
        // If N reaches the target
        if (n == m)
            return 0;
      
        // The minimum of both the states
        // will be the answer
        return Math.Min(1 + minimumSteps(n * a, m, a, b),
                1 + minimumSteps(n * b, m, a, b));
    }
      
    // Driver code
    public static void Main ()
    {
        int n = 120, m = 51840;
        int a = 2, b = 3;
        Console.WriteLine(minimumSteps(n, m, a, b));
    }
}
  
// This code is contributed by ihritik

PHP

 $m)
        return $MAXN;
  
    // If N reaches the target
    if ($n == $m)
        return 0;
  
    // The minimum of both the states
    // will be the answer
    return min(1 + minimumSteps($n * $a, $m, $a, $b),
               1 + minimumSteps($n * $b, $m, $a, $b));
}
  
// Driver code
$n = 120; $m = 51840;
$a = 2; $b = 3;
echo minimumSteps($n, $m, $a, $b);
  
// This code is contributed by Akanksha Rai
?>

输出：

7