最小化成本，以使X和Y等于给定的增量

给定两个整数X和Y ，任务是通过多次执行以下操作来使两个整数相等：

• 将X增加M倍。成本= M – 1 。
• 将Y增加N倍。成本= N – 1 。

例子：

Input: X = 2, Y = 4 
Output: 1 
Explanation: 
Increase X by 2 times. Therefore, X = 2 * 2 = 4. Cost = 1. 
Clearly, X = Y. Therefore, total cost = 1.

Input: X = 4, Y = 6 
Output: 3 
Explanation: 
Increase X by 3 times, X = 3 * 4 = 12. Cost = 2. 
increase Y by 2 times, Y = 2 * 6 = 12. Cost = 1. 
Clearly, X = Y. Therefore, total cost = 2 + 1 = 3.

天真的方法：请按照以下步骤解决问题：

1. 对于X的每个值，如果Y小于X ，则增加Y并更新成本。
2. 如果X = Y ，则打印总成本。
3. 如果X ，则增加X并更新成本。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find minimum cost
// to make x and y equal
int minimumCost(int x, int y)
{
    int costx = 0, dup_x = x;
 
    while (true) {
 
        int costy = 0, dup_y = y;
 
        // Check if it possible
        // to make x == y
        while (dup_y != dup_x
               && dup_y < dup_x) {
            dup_y += y;
            costy++;
        }
 
        // Iif both are equal
        if (dup_x == dup_y)
            return costx + costy;
 
        // Otherwise
        else {
 
            // Increment cost of x
            // by 1 and dup_x by x
            dup_x += x;
            costx++;
        }
    }
}
 
// Driver Code
int main()
{
    int x = 5, y = 17;
 
    // Returns the required minimum cost
    cout << minimumCost(x, y) << endl;
}

Java

// Java program for the above approach
import java.util.*;
class GFG{
 
// Function to find minimum cost
// to make x and y equal
static int minimumCost(int x, int y)
{
    int costx = 0, dup_x = x;
    while (true)
    {
        int costy = 0, dup_y = y;
 
        // Check if it possible
        // to make x == y
        while (dup_y != dup_x
               && dup_y < dup_x)
        {
            dup_y += y;
            costy++;
        }
 
        // Iif both are equal
        if (dup_x == dup_y)
            return costx + costy;
 
        // Otherwise
        else {
 
            // Increment cost of x
            // by 1 and dup_x by x
            dup_x += x;
            costx++;
        }
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int x = 5, y = 17;
 
    // Returns the required minimum cost
    System.out.print(minimumCost(x, y) +"\n");
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program for the above approach
 
# Function to find minimum cost
# to make x and y equal
def minimumCost(x, y):
     
    costx, dup_x = 0, x
 
    while (True):
        costy, dup_y = 0, y
         
        # Check if it possible
        # to make x == y
        while (dup_y != dup_x and
               dup_y < dup_x):
            dup_y += y
            costy += 1
 
        # If both are equal
        if (dup_x == dup_y):
            return costx + costy
 
        # Otherwise
        else:
             
            # Increment cost of x
            # by 1 and dup_x by x
            dup_x += x
            costx += 1
 
# Driver Code
if __name__ == '__main__':
     
    x, y = 5, 17
 
    # Returns the required minimum cost
    print(minimumCost(x, y))
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
class GFG
{
 
  // Function to find minimum cost
  // to make x and y equal
  static int minimumCost(int x, int y)
  {
    int costx = 0, dup_x = x;
    while (true)
    {
      int costy = 0, dup_y = y;
 
      // Check if it possible
      // to make x == y
      while (dup_y != dup_x
             && dup_y < dup_x)
      {
        dup_y += y;
        costy++;
      }
 
      // Iif both are equal
      if (dup_x == dup_y)
        return costx + costy;
 
      // Otherwise
      else {
 
        // Increment cost of x
        // by 1 and dup_x by x
        dup_x += x;
        costx++;
      }
    }
  }
 
  // Driver Code
  public static void Main(String[] args)
  {
    int x = 5, y = 17;
 
    // Returns the required minimum cost
    Console.Write(minimumCost(x, y) +"\n");
  }
}
 
// This code is contributed by 29AjayKumar

输出：

20

时间复杂度： O((costx + costy)* Y)
辅助空间： O(1)

高效的方法：这里的想法是使用LCM的概念。制作X和Y的最低成本将等于其LCM。但这还不够。减去X和Y的初始值，以避免在计算答案时将它们相加。
请按照以下步骤解决问题：

1. 查找X和Y的LCM。
2. 从LCM中减去X和Y的值。
3. 现在，将LCM除以X和Y ，然后求和它们的值之和。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find gcd of x and y
int gcd(int x, int y)
{
    if (y == 0)
        return x;
    return gcd(y, x % y);
}
 
// Function to find lcm of x and y
int lcm(int x, int y)
{
    return (x * y) / gcd(x, y);
}
 
// Function to find minimum Cost
int minimumCost(int x, int y)
{
    int lcm_ = lcm(x, y);
 
    // Subtracted intial cost of x
    int costx = (lcm_ - x) / x;
 
    // Subtracted intial cost of y
    int costy = (lcm_ - y) / y;
    return costx + costy;
}
 
// Driver Code
int main()
{
    int x = 5, y = 17;
 
    // Returns the minimum cost required
    cout << minimumCost(x, y) << endl;
}

Java

// Java program for the above approach
import java.util.*;
class GFG
{
 
// Function to find gcd of x and y
static int gcd(int x, int y)
{
    if (y == 0)
        return x;
    return gcd(y, x % y);
}
 
// Function to find lcm of x and y
static int lcm(int x, int y)
{
    return (x * y) / gcd(x, y);
}
 
// Function to find minimum Cost
static int minimumCost(int x, int y)
{
    int lcm_ = lcm(x, y);
 
    // Subtracted intial cost of x
    int costx = (lcm_ - x) / x;
 
    // Subtracted intial cost of y
    int costy = (lcm_ - y) / y;
    return costx + costy;
}
 
// Driver Code
public static void main(String[] args)
{
    int x = 5, y = 17;
 
    // Returns the minimum cost required
    System.out.print(minimumCost(x, y) +"\n");
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program for the above approach
 
# Function to find gcd of x and y
def gcd(x, y):
     
    if (y == 0):
        return x
         
    return gcd(y, x % y)
 
# Function to find lcm of x and y
def lcm(x, y):
 
    return (x * y) // gcd(x, y)
 
# Function to find minimum Cost
def minimumCost(x, y):
     
    lcm_ = lcm(x, y)
 
    # Subtracted intial cost of x
    costx = (lcm_ - x) // x
 
    # Subtracted intial cost of y
    costy = (lcm_ - y) // y
     
    return costx + costy
 
# Driver Code
if __name__ == "__main__":
     
    x = 5
    y = 17
 
    # Returns the minimum cost required
    print(minimumCost(x, y))
 
# This code is contributed by chitranayal

C＃

// C# program for the above approach
using System;
class GFG
{
 
// Function to find gcd of x and y
static int gcd(int x, int y)
{
    if (y == 0)
        return x;
    return gcd(y, x % y);
}
 
// Function to find lcm of x and y
static int lcm(int x, int y)
{
    return (x * y) / gcd(x, y);
}
 
// Function to find minimum Cost
static int minimumCost(int x, int y)
{
    int lcm_ = lcm(x, y);
 
    // Subtracted intial cost of x
    int costx = (lcm_ - x) / x;
 
    // Subtracted intial cost of y
    int costy = (lcm_ - y) / y;
    return costx + costy;
}
 
// Driver Code
public static void Main(String[] args)
{
    int x = 5, y = 17;
 
    // Returns the minimum cost required
    Console.Write(minimumCost(x, y) +"\n");
}
}
 
// This code is contributed by 29AjayKumar

输出：

20

时间复杂度： O(log(min(x，y))
辅助空间： O(1)