计算将A转换为B所需的最小移动

给定两个整数A和B ，通过多次执行以下操作之一，将A转换为B ：

• A = A + K
• A = A – K ，其中K属于[1，10]

任务是找到使用上述操作将A转换为B所需的最少操作数。

例子：

Input: A = 13, B = 42
Output: 3
Explanation:
The following sequence of moves can be performed: 13 → 23 → 32 → 42(add 10, add 9, add 10).

Input: A = 18, B = 4
Output: 2
Explanation:
The following sequence of moves can be performed: 18 → 10 → 4 (subtract 8, subtract 6).

方法：这个想法是通过将A和B的绝对差除以[1…10]范围内的所有数字，并将其加到结果变量中来简单地计算所需的移动次数。请按照以下步骤解决问题：

• 初始化变量required_moves以存储所需的最少移动次数。
• 求出A和B的绝对差。
• 迭代范围[ 1，10 ]并执行以下操作：
  • 将数字除以i，然后将其添加到结果变量中。
  • 用i计算绝对差的模。
• 最后，打印required_moves的值。

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 number
// of moves to obtained B from A
void convertBfromA(int a, int b)
{
    // Stores the minimum
    // number of moves
    int moves = 0;
 
    // Absolute difference
    int x = abs(a - b);
 
    // K is in range [0, 10]
    for (int i = 10; i > 0; i--) {
        moves += x / i;
        x = x % i;
    }
 
    // Print the required moves
    cout << moves << " ";
}
 
// Driver Code
int main()
{
    int A = 188, B = 4;
 
    convertBfromA(A, B);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
 
class GFG{
 
// Function to find minimum number
// of moves to obtained B from A
static void convertBfromA(int a, int b)
{
     
    // Stores the minimum
    // number of moves
    int moves = 0;
 
    // Absolute difference
    int x = Math.abs(a - b);
 
    // K is in range [0, 10]
    for(int i = 10; i > 0; i--)
    {
        moves += x / i;
        x = x % i;
    }
 
    // Print the required moves
    System.out.print(moves + " ");
}
 
// Driver Code
public static void main (String[] args)
{
    int A = 188, B = 4;
 
    convertBfromA(A, B);
}
}
 
// This code is contributed by code_hunt

Python3

# Python3 program for the above approach
 
# Function to find minimum number
# of moves to obtained B from A
def convertBfromA(a, b):
     
    # Stores the minimum
    # number of moves
    moves = 0
 
    # Absolute difference
    x = abs(a - b)
 
    # K is in range [0, 10]
    for i in range(10, 0, -1):
        moves += x // i
        x = x % i
     
    # Print the required moves
    print(moves, end = " ")
 
# Driver Code
A = 188
B = 4
 
convertBfromA(A, B)
 
# This code is contributed by code_hunt

C#

// C# program for the above approach 
using System;
 
class GFG{
 
// Function to find minimum number
// of moves to obtained B from A
static void convertBfromA(int a, int b)
{
     
    // Stores the minimum
    // number of moves
    int moves = 0;
 
    // Absolute difference
    int x = Math.Abs(a - b);
 
    // K is in range [0, 10]
    for(int i = 10; i > 0; i--)
    {
        moves += x / i;
        x = x % i;
    }
 
    // Print the required moves
    Console.Write(moves + " ");
}
 
// Driver Code
public static void Main ()
{
    int A = 188, B = 4;
 
    convertBfromA(A, B);
}
}
 
// This code is contributed by code_hunt

输出：

19

时间复杂度： O(K)，其中K在[0，10]范围内
辅助空间： O(1)