计算将甲转换为乙所需的最小移动次数

原文:https://www . geesforgeks . org/count-最小移动-需要转换为 a-b/

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 所需的最小操作数。

示例:

输入: A = 13，B = 42 输出: 3 说明: 可以进行以下顺序的招式:13 → 23 → 32 → 42(加 10，加 9，加 10)。

输入: A = 18，B = 4 输出: 2 说明: 可以进行如下的招式顺序:18 → 10 → 4(减 8，减 6)。

方法:思路是将 A 和 B 的绝对差除以【1…10】范围内的所有数字，并将其加到合成变量中，简单计算出所需的移动次数。按照以下步骤解决问题:

  • 初始化一个变量 required_moves 来存储所需的最小移动次数。
  • 求 A 和 B 的绝对差。
  • 迭代范围【1，10】并执行以下操作:
    • 将该数除以 i ，并将其加到结果变量中。
    • 通过 i 计算绝对差的模。
  • 最后，打印所需 _ 移动的值。

下面是上述方法的实现:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
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 语言(一种计算机语言，尤用于创建网站)

// 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

Python 3

# 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

java 描述语言

<script>

// Javascript program to implement
// the above approach

// Function to find minimum number
// of moves to obtained B from A
function convertBfromA(a, b)
{

    // Stores the minimum
    // number of moves
    let moves = 0;

    // Absolute difference
    let x = Math.abs(a - b);

    // K is in range [0, 10]
    for(let i = 10; i > 0; i--)
    {
        moves += Math.floor(x / i);
        x = x % i;
    }

    // Print the required moves
    document.write(moves + " ");
}

// Driver Code

    let A = 188, B = 4;

    convertBfromA(A, B);

</script>

Output: 

19

时间复杂度: O(K)，其中 K 在范围【0，10】 辅助空间: O(1)

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