通过重复添加偶数除数将M转换为N的最低成本

给定两个整数M和N ，任务是通过重复添加M的当前值的偶数除数( M除外)来找到将M转换为N的最小成本。

The cost to add an even divisor of the current value of M, say d, is equal to M / d. 
 

如果无法将M转换为N ，则打印“ -1” 。

例子：

Input: M = 6, N = 24
Output: 10
Explanation: 
Step 1: M = 6 + 2 = 8, Cost = (6 / 2) = 3 
Step 2: M = 8 + 4 = 12, Cost = 3 + (8 / 2) = 5 
Step 3: M = 12 + 6 = 18, Cost = 5 + (12/ 6) = 7 
Step 4: M = 18 + 6 = 24, Cost = 7 + (18 / 6) = 10 
Therefore, the minimum cost to convert M to N is equal to 10.

Input: M = 9, N = 17
Output: -1
Explanation:
Since there are no even divisors of 9, therefore, conversion is not possible.

天真的方法：最简单的方法是遍历给定数M的所有可能偶数除数，然后递归计算将M更改为N的最小成本。形成的递归关系由下式给出：

min_cost = Math.min(min_cost, m / i + minSteps(m + i, n))

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
int inf = 1000000008;
 
// Function to find the value of
// minimum steps to convert m to n
int minSteps(int m, int n)
{
     
    // Base Case
    if (n == m)
        return 0;
 
    // If n exceeds m
    if (m > n)
        return inf;
 
    int min_cost = inf;
 
    // Iterate through all possible
    // even divisors of m
    for(int i = 2; i < m; i += 2)
    {
         
        // If m is divisible by i,
        // then find the minimum cost
        if (m % i == 0)
        {
             
            // Add the cost to convert
            // m to m+i and recursively
            // call next state
            min_cost = min(min_cost,
                           m / i +
                           minSteps(m + i, n));
        }
    }
 
    // Return min_cost
    return min_cost;
}
 
// Driver code
int main()
{
    int M = 6;
    int N = 24;
 
    // Function call
    int minimum_cost = minSteps(M, N);
 
    // If conversion is
    // not possible
    if (minimum_cost == inf)
        minimum_cost = -1;
 
    // Print the cost
    cout << minimum_cost;
     
    return 0;
}
 
// This code is contributed by akhilsaini

Java

// Java program for the above approach
 
import java.util.*;
 
public class GFG {
 
    static int inf = 1000000008;
 
    // Function to find the value of
    // minimum steps to convert m to n
    public static int
    minSteps(int m, int n)
    {
        // Base Case
        if (n == m)
            return 0;
 
        // If n exceeds m
        if (m > n)
            return inf;
 
        int min_cost = inf;
 
        // Iterate through all possible
        // even divisors of m
        for (int i = 2; i < m; i += 2) {
 
            // If m is divisible by i,
            // then find the minimum cost
            if (m % i == 0) {
 
                // Add the cost to convert
                // m to m+i and recursively
                // call next state
                min_cost
                    = Math.min(
                        min_cost,
                        m / i
                            + minSteps(m + i, n));
            }
        }
 
        // Return min_cost
        return min_cost;
    }
 
    // Driver Code
    public static void
    main(String args[])
    {
        int M = 6;
        int N = 24;
 
        // Function Call
        int minimum_cost
            = minSteps(M, N);
 
        // If conversion is
        // not possible
        minimum_cost
            = minimum_cost
                      == inf
                  ? -1
                  : minimum_cost;
 
        // Print the cost
        System.out.println(minimum_cost);
    }
}

Python3

# Python3 program for the above approach
inf = 1000000008
 
# Function to find the value of
# minimum steps to convert m to n
def minSteps(m, n):
     
    # Base Case
    if (n == m):
        return 0
 
    # If n exceeds m
    if (m > n):
        return inf
 
    min_cost = inf
 
    # Iterate through all possible
    # even divisors of m
    for i in range(2, m, 2):
 
        # If m is divisible by i,
        # then find the minimum cost
        if (m % i == 0):
 
            # Add the cost to convert
            # m to m+i and recursively
            # call next state
            min_cost = min(min_cost, m / i +
                            minSteps(m + i, n))
 
    # Return min_cost
    return min_cost
 
# Driver Code
if __name__ == '__main__':
     
    M = 6
    N = 24
 
    # Function call
    minimum_cost = minSteps(M, N)
 
    # If conversion is
    # not possible
    if minimum_cost == inf:
        minimum_cost = -1
 
    # Print the cost
    print(minimum_cost)
     
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
class GFG{
 
  static int inf = 1000000008;
 
  // Function to find the value of
  // minimum steps to convert m to n
  public static int minSteps(int m,
                             int n)
  {
    // Base Case
    if (n == m)
      return 0;
 
    // If n exceeds m
    if (m > n)
      return inf;
 
    int min_cost = inf;
 
    // Iterate through all possible
    // even divisors of m
    for (int i = 2; i < m; i += 2)
    {
      // If m is divisible by i,
      // then find the minimum cost
      if (m % i == 0)
      {
        // Add the cost to convert
        // m to m+i and recursively
        // call next state
        min_cost = Math.Min(min_cost, m / i +
                            minSteps(m + i, n));
      }
    }
 
    // Return min_cost
    return min_cost;
  }
 
  // Driver Code
  public static void Main(String []args)
  {
    int M = 6;
    int N = 24;
 
    // Function Call
    int minimum_cost = minSteps(M, N);
 
    // If conversion is
    // not possible
    minimum_cost = minimum_cost == inf ? -1 :
                   minimum_cost;
 
    // Print the cost
    Console.WriteLine(minimum_cost);
  }
}
 
// This code is contributed by Rajput-Ji

输出：

10

时间复杂度： O(2 ^N )
辅助空间： O(1)

高效的方法：可以通过对上述实现使用动态编程和备注来优化上述方法。不必一次又一次地计算状态，而是将其存储在数组dp []中，并在需要时使用它。

dp(m) = min(dp(m), (m/i) + dp(m+i)) for all even divisors of m less than m

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
int inf = 1000000008;
 
// Utility function to calculate the
// minimum cost
int minStepsUtil(int m, int n, int dp[])
{
     
    // Positive base case
    if (n == m)
        return 0;
 
    // Negative base case
    if (m > n)
        return inf;
 
    // If current state is already
    // computed then return the
    // current state value
    if (dp[m] != inf)
        return dp[m];
 
    int min_cost = inf;
 
    // Iterate through all possible
    // even divisors
    for(int i = 2; i < m; i += 2)
    {
        if (m % i == 0)
        {
            min_cost = min(min_cost,
                           m / i +
                           minStepsUtil(m + i,
                                        n, dp));
        }
    }
 
    // Store the precomputed answer
    return dp[m] = min_cost;
}
 
void minSteps(int M, int N)
{
     
    // Initialise the dp array
    // with infinity
    int dp[N + 5];
    for(int i = 0; i < N + 5; i++)
    {
        dp[i] = inf;
    }
 
    // Function call
    int minimum_cost = minStepsUtil(M, N, dp);
 
    if (minimum_cost == inf)
        minimum_cost = -1;
 
    // Print the minimum cost
    cout << minimum_cost;
}
 
// Driver code
int main()
{
    int M = 6;
    int N = 24;
 
    // Function call
    minSteps(M, N);
}
 
// This code is contributed by akhilsaini

Java

// Java program for the above approach
 
import java.util.*;
 
public class GFG {
 
    static int inf = 1000000008;
 
    // Utility function to calculate the
    // minimum cost
    public static int
    minStepsUtil(int m, int n, int dp[])
    {
        // Positive base case
        if (n == m)
            return 0;
 
        // Negative base case
        if (m > n)
            return inf;
 
        // If current state is already
        // computed then return the
        // current state value
        if (dp[m] != inf)
            return dp[m];
 
        int min_cost = inf;
 
        // Iterate through all possible
        // even divisors
        for (int i = 2; i < m; i += 2) {
            if (m % i == 0) {
                min_cost = Math.min(
                    min_cost,
                    m / i
                        + minStepsUtil(m + i, n, dp));
            }
        }
 
        // Store the precomputed answer
        return dp[m] = min_cost;
    }
 
    public static void
    minSteps(int M, int N)
    {
 
        // Initialise the dp array
        // with infinity
        int dp[] = new int[N + 5];
 
        Arrays.fill(dp, inf);
 
        // Function Call
        int minimum_cost
            = minStepsUtil(M, N, dp);
 
        minimum_cost
            = minimum_cost
                      == inf
                  ? -1
                  : minimum_cost;
 
        // Print the minimum cost
        System.out.println(minimum_cost);
    }
 
    // Driver code
    public static void main(String args[])
    {
        int M = 6;
        int N = 24;
 
        // Function Call
        minSteps(M, N);
    }
}

C＃

// C# program for the above approach
using System;
class GFG{
 
static int inf = 1000000008;
 
// Utility function to calculate the
// minimum cost
public static int minStepsUtil(int m,
                               int n, int []dp)
{
    // Positive base case
    if (n == m)
        return 0;
 
    // Negative base case
    if (m > n)
        return inf;
 
    // If current state is already
    // computed then return the
    // current state value
    if (dp[m] != inf)
        return dp[m];
 
    int min_cost = inf;
 
    // Iterate through all possible
    // even divisors
    for (int i = 2; i < m; i += 2)
    {
        if (m % i == 0)
        {
            min_cost = Math.Min(min_cost, m / i +
                                minStepsUtil(m + i,
                                             n, dp));
        }
    }
 
    // Store the precomputed answer
    return dp[m] = min_cost;
    }
 
    public static void minSteps(int M,
                                int N)
    {
        // Initialise the dp array
        // with infinity
        int []dp = new int[N + 5];
        for(int i = 0; i < dp.GetLength(0);
                i++)
            dp[i] = inf;
 
        // Function Call
        int minimum_cost = minStepsUtil(M, N, dp);
 
        minimum_cost = minimum_cost ==
                       inf ? -1 : minimum_cost;
 
        // Print the minimum cost
        Console.WriteLine(minimum_cost);
    }
 
    // Driver code
    public static void Main(String []args)
    {
        int M = 6;
        int N = 24;
 
        // Function Call
        minSteps(M, N);
    }
}
 
// This code is contributed by 29AjayKumar

Python3

# Python program for the above approach
 
 
inf = 1000000008;
 
 
# Utility function to calculate the
# minimum cost
def minStepsUtil(m, n, dp):
    # Positive base case
    if (n == m):
        return 0;
 
    # Negative base case
    if (m > n):
        return inf;
 
    # If current state is already
    # computed then return the
    # current state value
    if (dp[m] != inf):
        return dp[m];
 
    min_cost = inf;
 
    # Iterate through all possible
    # even divisors
    for i in range(2,m,2):
        if (m % i == 0):
            min_cost = min(min_cost, m // i + minStepsUtil(m + i, n, dp));
 
    # Store the precomputed answer
    dp[m] = min_cost
    return dp[m];
 
 
def minSteps(M, N):
    # Initialise the dp array
    # with infinity
    dp = [inf]*(N + 5);
 
    # Function Call
    minimum_cost = minStepsUtil(M, N, dp);
 
    minimum_cost = -1 if minimum_cost == inf else minimum_cost;
 
    # Prthe minimum cost
    print(minimum_cost);
 
 
# Driver code
if __name__ == '__main__':
    M = 6;
    N = 24;
 
    # Function Call
    minSteps(M, N);
 
# This code contributed by shikhasingrajput

输出：

10

时间复杂度： O(Nlog(M))
辅助空间： O(N)