给定两个由N个整数组成的数组A []和B [] ，任务是将增加或减少数组元素的总成本减1 ，以使得对于每个第i个元素， A [i]都是B [的倍数] i] ，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，例如cost ，以存储所需的最低成本。
- 同时遍历数组A []和B []并执行以下步骤：
- 情况1：找出更新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：找出更新B [i]使其成为A [i]的倍数的成本，这是(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 souravmahato348Python3
# 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 shubhamsingh10C#
// 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输出：
4时间复杂度： O(N)
辅助空间： O(1)