最小化增量或减量的成本，以便相同的索引元素成为彼此的倍数

给定两个由N 个整数组成的数组A[]和B[] ，任务是将数组元素递增或递减1的总成本最小化，使得对于每个^第i^个元素， A[i]是B[的倍数i]或反之亦然。

例子：

Input: A[] = {3, 6, 3}, B[] = {4, 8, 13}
Output: 4
Explanation:
Incrementing A[0] by 1 (3 + 1 = 4) makes it multiple of B[0](= 4).
Incrementing A[1] by 2 (6 + 2 = 8) makes it a multiple of B[1](= 8).
Decrementing B[2] by 1 (13 – 1 = 12) makes it a multiple of A[2](= 3).
Therefore, the total cost = 1 + 2 + 1 = 4.

Input: A[] = {13, 2, 31, 7}, B[] = {6, 8, 11, 3}
Output: 4

方法：可以贪婪地解决给定的问题。请按照以下步骤解决问题：

• 初始化一个变量，比如cost ，以存储所需的最小成本。
• 同时遍历数组A[]和B[]并执行以下步骤：
  • 案例 1：找到更新A[i]使其成为B[i]的倍数的成本，这是(B[i] % A[i])和(A[i] – B[i] )的最小值% A[i]) 。
  • 情况 2：找到更新B[i]使其成为A[i]的倍数的成本，即(A[i] % B[i])和(B[i] – A[i] )的最小值% B[i]) 。
  • 将上述两个成本中的最小值添加到每个数组元素的可变成本中。
• 完成上述步骤后，打印成本值作为结果。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the minimum cost to
// make A[i] multiple of B[i] or
// vice-versa for every array element
int MinimumCost(int A[], int B[],
                int N)
{
    // Stores the minimum cost
    int totalCost = 0;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
        // Case 1: Update A[i]
        int mod_A = B[i] % A[i];
        int totalCost_A = min(mod_A,
                              A[i] - mod_A);
 
        // Case 2: Update B[i]
        int mod_B = A[i] % B[i];
        int totalCost_B = min(mod_B,
                              B[i] - mod_B);
 
        // Add the minimum of
        // the above two cases
        totalCost += min(totalCost_A,
                         totalCost_B);
    }
 
    // Return the resultant cost
    return totalCost;
}
 
// Driver Code
int main()
{
    int A[] = { 3, 6, 3 };
    int B[] = { 4, 8, 13 };
    int N = sizeof(A) / sizeof(A[0]);
 
    cout << MinimumCost(A, B, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
 
class GFG{
 
// Function to find the minimum cost to
// make A[i] multiple of B[i] or
// vice-versa for every array element
static int MinimumCost(int A[], int B[], int N)
{
     
    // Stores the minimum cost
    int totalCost = 0;
 
    // Traverse the array
    for(int i = 0; i < N; i++)
    {
         
        // Case 1: Update A[i]
        int mod_A = B[i] % A[i];
        int totalCost_A = Math.min(mod_A,
                            A[i] - mod_A);
 
        // Case 2: Update B[i]
        int mod_B = A[i] % B[i];
        int totalCost_B = Math.min(mod_B,
                            B[i] - mod_B);
 
        // Add the minimum of
        // the above two cases
        totalCost += Math.min(totalCost_A,
                              totalCost_B);
    }
 
    // Return the resultant cost
    return totalCost;
}
 
// Driver Code
public static void main(String[] args)
{
    int A[] = { 3, 6, 3 };
    int B[] = { 4, 8, 13 };
    int N = A.length;
 
    System.out.print(MinimumCost(A, B, N));
}
}
 
// This code is contributed by souravmahato348

Python3

# Python program for the above approach
 
# Function to find the minimum cost to
# make A[i] multiple of B[i] or
# vice-versa for every array element
def MinimumCost(A, B, N):
     
    # Stores the minimum cost
    totalCost = 0
     
    # Traverse the array
    for i in range(N):
         
        # Case 1: Update A[i]
        mod_A = B[i] % A[i]
        totalCost_A = min(mod_A, A[i] - mod_A)
         
        # Case 2: Update B[i]
        mod_B = A[i] % B[i]
        totalCost_B = min(mod_B, B[i] - mod_B)
         
        # Add the minimum of
        # the above two cases
        totalCost += min(totalCost_A, totalCost_B)
         
    # Return the resultant cost
    return totalCost
 
# Driver Code
A = [3, 6, 3]
B =  [4, 8, 13]
N = len(A)
 
print(MinimumCost(A, B, N))
 
# This code is contributed by shubhamsingh10

C#

// C# program for the above approach
using System;
 
class GFG {
 
    // Function to find the minimum cost to
    // make A[i] multiple of B[i] or
    // vice-versa for every array element
    static int MinimumCost(int[] A, int[] B, int N)
    {
 
        // Stores the minimum cost
        int totalCost = 0;
 
        // Traverse the array
        for (int i = 0; i < N; i++) {
 
            // Case 1: Update A[i]
            int mod_A = B[i] % A[i];
            int totalCost_A = Math.Min(mod_A, A[i] - mod_A);
 
            // Case 2: Update B[i]
            int mod_B = A[i] % B[i];
            int totalCost_B = Math.Min(mod_B, B[i] - mod_B);
 
            // Add the minimum of
            // the above two cases
            totalCost += Math.Min(totalCost_A, totalCost_B);
        }
 
        // Return the resultant cost
        return totalCost;
    }
 
    // Driver Code
    public static void Main()
    {
        int[] A = { 3, 6, 3 };
        int[] B = { 4, 8, 13 };
        int N = A.Length;
 
        Console.Write(MinimumCost(A, B, N));
    }
}
 
// This code is contributed by rishavmahato348

Javascript

// javascript program for the above approach
 
    // Function to find the minimum cost to
    // make A[i] multiple of B[i] or
    // vice-versa for every array element
    function MinimumCost(A , B , N) {
 
        // Stores the minimum cost
        var totalCost = 0;
 
        // Traverse the array
        for (i = 0; i < N; i++) {
 
            // Case 1: Update A[i]
            var mod_A = B[i] % A[i];
            var totalCost_A = Math.min(mod_A, A[i] - mod_A);
 
            // Case 2: Update B[i]
            var mod_B = A[i] % B[i];
            var totalCost_B = Math.min(mod_B, B[i] - mod_B);
 
            // Add the minimum of
            // the above two cases
            totalCost += Math.min(totalCost_A, totalCost_B);
        }
 
        // Return the resultant cost
        return totalCost;
    }
 
    // Driver Code
     
        var A = [ 3, 6, 3 ];
        var B = [ 4, 8, 13 ];
        var N = A.length;
 
        document.write(MinimumCost(A, B, N));
 
// This code contributed by aashish1995

输出：

4

时间复杂度： O(N)
辅助空间： O(1)

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