使一个数组的所有元素成为另一个数组的倍数所需的最小前缀增量
原文:https://www . geeksforgeeks . org/最小前缀增量-要求将数组中的所有元素设为另一个数组的倍数/
给定两个大小为 N 的数组 A[] 和 B[] ，任务是通过将前缀子数组增加 1 来找到使 A[i] 成为 B[i] 的倍数所需的最小操作数。
示例:
输入: A[ ] = {3，2，9}，B[ ] = {5，7，4}，N = 3 输出: 7 解释: 递增{A[0]}两次将 A[ ]修改为{ 5 ，2，9} 递增子阵列{A[0]，A[1]两次将 A[]修改为{ 7，4， 9
输入: A[ ] = {3，4，5，2，5，5，9}，B[ ] = {1，1，9，6，3，8，7}，N = 7 T3】输出: 22
方法:使用贪婪技术可以解决问题。为了使运算次数最少，思路是寻找 A[i] 的最近最小的大于或等于元素，它是 B[i] 的倍数:
- 遍历阵T3【A】从结束到这个开始。
- 求的最近倍数与 B[] 的对应元素之间的最小差 K 。
- 由于 K 等于在 i 第T10】元素处运算的次数，因此所有元素的值从 0 第T16】索引到 (i-1) 第 索引将增加 K.****
- 现在维护一个变量进位，它将存储累积增量，因此如果 i th 元素递增 K 次，那么将 K 加到进位上。
- 的值携带变量将被用来到找到(I-1)的新值 A[]的第 元素。
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 count of operations
// required to make A[i] multiple of B[i] by
// incrementing prefix subarray
int MinimumMoves(int A[], int B[], int N)
{
// Stores minimum count of operations
// required to make A[i] multiple of B[i]
// by incrementing prefix subarray
int totalOperations = 0;
// Stores the carry
int carry = 0;
// Stores minimum difference of
// correspoinding element in
// prefix subarray
int K = 0;
// Traverse the array
for (int i = N - 1; i >= 0; i--) {
// Stores the closest greater or equal number
// to A[i] which is a multiple of B[i]
int nearestMultiple = ceil((double)(A[i] + carry)
/ (double)(B[i]))
* B[i];
// Stores minimum difference
K = nearestMultiple - (A[i] + carry);
// Update totalOperations
totalOperations += K;
// Update carry
carry += K;
}
return totalOperations;
}
// Driver Code
int main()
{
// Input arrays A[] and B[]
int A[] = { 3, 4, 5, 2, 5, 5, 9 };
int B[] = { 1, 1, 9, 6, 3, 8, 7 };
// Length of arrays
int N = sizeof(A) / sizeof(A[0]);
cout << MinimumMoves(A, B, N) << endl;
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.util.*;
class GFG
{
// Function to find minimum count of operations
// required to make A[i] multiple of B[i] by
// incrementing prefix subarray
static int MinimumMoves(int A[], int B[], int N)
{
// Stores minimum count of operations
// required to make A[i] multiple of B[i]
// by incrementing prefix subarray
int totalOperations = 0;
// Stores the carry
int carry = 0;
// Stores minimum difference of
// correspoinding element in
// prefix subarray
int K = 0;
// Traverse the array
for (int i = N - 1; i >= 0; i--)
{
// Stores the closest greater or equal number
// to A[i] which is a multiple of B[i]
int nearestMultiple = (int) (Math.ceil((double)(A[i] + carry)
/ (double)(B[i]))
* B[i]);
// Stores minimum difference
K = nearestMultiple - (A[i] + carry);
// Update totalOperations
totalOperations += K;
// Update carry
carry += K;
}
return totalOperations;
}
// Driver Code
public static void main(String[] args)
{
// Input arrays A[] and B[]
int A[] = { 3, 4, 5, 2, 5, 5, 9 };
int B[] = { 1, 1, 9, 6, 3, 8, 7 };
// Length of arrays
int N = A.length;
System.out.print(MinimumMoves(A, B, N) +"\n");
}
}
// This code is contributed by 29AjayKumar
Python 3
# Python3 program for the above approach
from math import ceil,floor
# Function to find minimum count of operations
# required to make A[i] multiple of B[i] by
# incrementing prefix subarray
def MinimumMoves(A, B, N):
# Stores minimum count of operations
# required to make A[i] multiple of B[i]
# by incrementing prefix subarray
totalOperations = 0
# Stores the carry
carry = 0
# Stores minimum difference of
# correspoinding element in
# prefix subarray
K = 0
# Traverse the array
for i in range(N - 1, -1, -1):
# Stores the closest greater or equal number
# to A[i] which is a multiple of B[i]
nearestMultiple = ceil((A[i] + carry)/ B[i])* B[i]
# Stores minimum difference
K = nearestMultiple - (A[i] + carry)
# Update totalOperations
totalOperations += K
# Update carry
carry += K
return totalOperations
# Driver Code
if __name__ == '__main__':
# Input arrays A[] and B[]
A = [3, 4, 5, 2, 5, 5, 9]
B = [1, 1, 9, 6, 3, 8, 7]
# Length of arrays
N = len(A)
print (MinimumMoves(A, B, N))
# This code is contributed by mohit kumar 29.
C
// C# program for the above approach
using System;
class GFG{
// Function to find minimum count of operations
// required to make A[i] multiple of B[i] by
// incrementing prefix subarray
static int MinimumMoves(int[] A, int[] B, int N)
{
// Stores minimum count of operations
// required to make A[i] multiple of B[i]
// by incrementing prefix subarray
int totalOperations = 0;
// Stores the carry
int carry = 0;
// Stores minimum difference of
// correspoinding element in
// prefix subarray
int K = 0;
// Traverse the array
for (int i = N - 1; i >= 0; i--)
{
// Stores the closest greater or equal number
// to A[i] which is a multiple of B[i]
int nearestMultiple = (int) (Math.Ceiling((double)(A[i] + carry)
/ (double)(B[i]))
* B[i]);
// Stores minimum difference
K = nearestMultiple - (A[i] + carry);
// Update totalOperations
totalOperations += K;
// Update carry
carry += K;
}
return totalOperations;
}
// Driver Code
public static void Main(string[] args)
{
// Input arrays A[] and B[]
int[] A = { 3, 4, 5, 2, 5, 5, 9 };
int[] B = { 1, 1, 9, 6, 3, 8, 7 };
// Length of arrays
int N = A.Length;
Console.Write(MinimumMoves(A, B, N) +"\n");
}
}
// This code is contributed by sanjoy_62.
java 描述语言
<script>
// javascript program of the above approach
// Function to find minimum count of operations
// required to make A[i] multiple of B[i] by
// incrementing prefix subarray
function MinimumMoves(A, B, N)
{
// Stores minimum count of operations
// required to make A[i] multiple of B[i]
// by incrementing prefix subarray
let totalOperations = 0;
// Stores the carry
let carry = 0;
// Stores minimum difference of
// correspoinding element in
// prefix subarray
let K = 0;
// Traverse the array
for (let i = N - 1; i >= 0; i--)
{
// Stores the closest greater or equal number
// to A[i] which is a multiple of B[i]
let nearestMultiple = (Math.ceil((A[i] + carry)
/ (B[i]))
* B[i]);
// Stores minimum difference
K = nearestMultiple - (A[i] + carry);
// Update totalOperations
totalOperations += K;
// Update carry
carry += K;
}
return totalOperations;
}
// Driver Code
// Input arrays A[] and B[]
let A = [ 3, 4, 5, 2, 5, 5, 9 ];
let B = [ 1, 1, 9, 6, 3, 8, 7 ];
// Length of arrays
let N = A.length;
document.write(MinimumMoves(A, B, N) + "<br/>");
</script>
Output:
22
时间复杂度:O(N) T5辅助空间:** O(1)
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