给定两个阵列A []和B []由N个整数的，则任务是更新数组A []通过向一个单一的元素B [j]与更新A [1]到A [分配每个数组元素A [i]于i] + B [j]或A [i] * B [j] ，从而使数组A []的乘积最大化。
注意：两个数组中的每个数组元素只能与另一个数组中的单个元素配对。
例子：
Input: A[] = {1, 1, 6}, B[] = {1, 2, 3}
Output: 108
Explanation:
- Update A[0] = A[0] + B[0], A[] modifies to {2, 1, 6}
- Update A[1] = A[1] + B[1], A[] modifies to {2, 3, 6}
- Update A[0] = A[0] * B[2], A[] modifies to {6, 3, 6}
Therefore, the product of the array A[] is 6 * 3 * 6 = 108.
Input: A[] = {1, 1, 10}, B[] ={1, 1, 1}
Output: 60
Explanation:
- Update A[0] = A[0] + B[0], A[] modifies to {2, 1, 10}
- Update A[1] = A[1] + B[1], A[] modifies to {2, 2, 10}
- Update A[0] = A[0] * B[2], A[] modifies to {3, 2, 10}
方法：可以通过使用优先级队列(最小堆)来解决上述问题。请按照以下步骤解决问题：
- 对数组B []排序。
- 将数组A []的所有元素插入优先级队列，以便每次都获得最少的元素。
- 使用变量j遍历给定的数组B []，并从优先级队列中弹出一个元素，作为minE + B [j]或minE * B [j]的最大值，并将该最大值推入优先级队列。
- 完成上述步骤后，优先级队列中元素的乘积就是所需的结果。
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 largest
// product of array A[]
int largeProduct(vector A,
vector B, int N)
{
// Base Case
if (N == 0)
return 0;
// Store all the elements of
// the array A[]
priority_queue,
greater> pq;
for(int i = 0; i < N; i++)
pq.push(A[i]);
// Sort the Array B[]
sort(B.begin(), B.end());
// Traverse the array B[]
for(int i = 0; i < N; i++)
{
// Pop minimum element
int minn = pq.top();
pq.pop();
// Check which operation is
// producing maximum element
int maximized_element = max(minn * B[i],
minn + B[i]);
// Insert resultant element
// into the priority queue
pq.push(maximized_element);
}
// Evaluate the product
// of the elements of A[]
int max_product = 1;
while (pq.size() > 0)
{
max_product *= pq.top();
pq.pop();
}
// Return the maximum product
return max_product;
}
// Driver Code
int main()
{
// Given arrays
vector A = { 1, 1, 10 };
vector B = { 1, 1, 1 };
int N = 3;
// Function Call
cout << largeProduct(A, B, N);
}
// This code is contributed by mohit kumar 29 Java
// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG {
// Function to find the largest
// product of array A[]
public static int largeProduct(
int A[], int B[], int N)
{
// Base Case
if (N == 0)
return 0;
// Store all the elements of
// the array A[]
PriorityQueue pq
= new PriorityQueue<>();
for (int i = 0; i < N; i++)
pq.add(A[i]);
// Sort the Array B[]
Arrays.sort(B);
// Traverse the array B[]
for (int i = 0; i < N; i++) {
// Pop minimum element
int minn = pq.poll();
// Check which operation is
// producing maximum element
int maximized_element
= Math.max(minn * B[i],
minn + B[i]);
// Insert resultant element
// into the priority queue
pq.add(maximized_element);
}
// Evaluate the product
// of the elements of A[]
int max_product = 1;
while (pq.size() > 0) {
max_product *= pq.poll();
}
// Return the maximum product
return max_product;
}
// Driver Code
public static void main(String[] args)
{
// Given arrays
int A[] = { 1, 1, 10 };
int B[] = { 1, 1, 1 };
int N = 3;
// Function Call
System.out.println(
largeProduct(A, B, N));
}
} Python3
# Python program for the above approach
# Function to find the largest
# product of array A[]
def largeProduct(A, B, N):
# Base Case
if(N == 0):
return 0
# Store all the elements of
# the array A[]
pq = []
for i in range(N):
pq.append(A[i])
# Sort the Array B[]
B.sort()
pq.sort(reverse = True)
# Traverse the array B[]
for i in range(N):
# Pop minimum element
minn = pq.pop()
# Check which operation is
# producing maximum element
maximized_element = max(minn * B[i], minn + B[i])
# Insert resultant element
# into the priority queue
pq.append(maximized_element)
pq.sort(reverse = True)
# Evaluate the product
# of the elements of A[]
max_product = 1
while(len(pq) > 0):
max_product *= pq.pop();
# Return the maximum product
return max_product
# Driver Code
# Given arrays
A = [1, 1, 10]
B = [1, 1, 1]
N = 3
# Function Call
print(largeProduct(A, B, N))
# This code is contributed by avanitrachhadiya2155C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to find the largest
// product of array A[]
public static int largeProduct(int[] A, int[] B, int N)
{
// Base Case
if(N == 0)
{
return 0;
}
// Store all the elements of
// the array A[]
List pq = new List();
for(int i = 0; i < N; i++)
{
pq.Add(A[i]);
}
// Sort the Array B[]
Array.Sort(B);
pq.Sort();
// Traverse the array B[]
for(int i = 0; i < N; i++)
{
int min = pq[0];
// Pop minimum element
pq.RemoveAt(0);
// Check which operation is
// producing maximum element
int maximized_element = Math.Max(min* B[i], min + B[i]);
// Insert resultant element
// into the priority queue
pq.Add(maximized_element);
pq.Sort();
}
// Evaluate the product
// of the elements of A[]
int max_product = 1;
while(pq.Count > 0)
{
max_product *= pq[0];
pq.RemoveAt(0);
}
// Return the maximum product
return max_product;
}
// Driver Code
static public void Main ()
{
// Given arrays
int[] A = { 1, 1, 10 };
int[] B = { 1, 1, 1 };
int N = 3;
// Function Call
Console.WriteLine(largeProduct(A, B, N));
}
}
// This code is contributed by rag2127 输出：
60时间复杂度： O(N log N)
辅助空间： O(N)