使用给定的操作最大化 Array 的除法结果

给定一个由N 个整数组成的数组arr[] ，任务是通过重复以下两个步骤来找到数组中可能剩余的最大值：

• 删除任意两个数组元素。
• 将它们除法的商插入数组中。

注意：我们可以改变元素的顺序。

例子：

Input: arr[] = {100, 1000, 10, 2} 
Output: 200 
Explanation: 

• Remove 100 and 10 from the array. Insert 10 (= 100/10) back into the array.
• Remove 10 and 2 from the array. Insert 5 into the array.
• Remove 1000 and 5 from the array. Insert 200 into the array

Hence, the maximum result is 200.

Input: arr[] = {2, 100, 1} 
Output: 50 

方法：
为了解决上述问题，我们可以观察到，我们将获得最大的结果，当元件以递减顺序和分割操作进行排序发生在序列ARR [0] /(ARR [1] / ARR [2] / ARR [3]…..arr[n-1])的反向排序数组。因此，相应地对数组进行排序并计算适当的结果。

下面的代码是上述方法的实现：

C++

// C++ implementation to maximize the
// result of division of the given
// array elements
#include 
using namespace std;
 
// Function to find the max result
float maxDivision(int arr[], int n)
{
 
    // Sort the array in descending order
    sort(arr, arr + n, greater());
 
    float mxdiv = arr[1];
 
    // loop to divide in this order
    // arr[0] / ( arr[1] / arr[2] / ....
    // arr[n-2] / arr[n-1])
    for (int i = 2; i < n; ++i)
        mxdiv = mxdiv / arr[i];
 
    // return the final result
    return arr[0] / mxdiv;
}
 
// Driver code
int main()
{
 
    int arr[] = { 100, 1000, 10, 2 };
 
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << maxDivision(arr, n);
 
    return 0;
}

Java

// Java implementation to maximize the
// result of division of the given
// array elements
import java.util.*;
 
class GFG{
 
// Function to find the max result
static float maxDivision(Integer arr[], int n)
{
 
    // Sort the array in descending order
    Arrays.sort(arr, Collections.reverseOrder());
 
    float mxdiv = arr[1];
 
    // Loop to divide in this order
    // arr[0] / ( arr[1] / arr[2] / ....
    // arr[n-2] / arr[n-1])
    for(int i = 2; i < n; ++i)
       mxdiv = mxdiv / arr[i];
 
    // Return the final result
    return arr[0] / mxdiv;
}
 
// Driver code
public static void main(String[] args)
{
 
    Integer arr[] = { 100, 1000, 10, 2 };
    int n = arr.length;
 
    System.out.print((int)maxDivision(arr, n));
}
}
 
// This code is contributed by amal kumar choubey

Python3

# Python3 implementation to maximize
# the result of division of the
# given array elements
 
# Function to find the max result
def maxDivision(arr, n):
     
    # Sort the array in descending order
    arr.sort(reverse = True)
    mxdiv = arr[1]
     
    # Loop to divide in this order
    # arr[0] / ( arr[1] / arr[2] / ....
    # arr[n-2] / arr[n-1])
    for i in range(2, n):
        mxdiv = mxdiv / arr[i]
         
    # Return the final result
    return arr[0] / mxdiv
 
# Driver code
arr = [ 100, 1000, 10, 2 ]
n = len(arr)
 
print(maxDivision(arr, n))
 
# This code is contributed by ishayadav181

C#

// C# implementation to maximize the
// result of division of the given
// array elements
using System;
 
class GFG{
 
// Function to find the max result
static float maxDivision(int []arr, int n)
{
 
    // Sort the array in descending order
    Array.Sort(arr);
    Array.Reverse(arr);
 
    float mxdiv = arr[1];
 
    // Loop to divide in this order
    // arr[0] / ( arr[1] / arr[2] / ....
    // arr[n-2] / arr[n-1])
    for(int i = 2; i < n; ++i)
       mxdiv = mxdiv / arr[i];
 
    // Return the readonly result
    return arr[0] / mxdiv;
}
 
// Driver code
public static void Main(String[] args)
{
    int []arr = { 100, 1000, 10, 2 };
    int n = arr.Length;
 
    Console.Write((int)maxDivision(arr, n));
}
}
 
// This code is contributed by amal kumar choubey

Javascript

 
// Javascript implementation to maximize
// the result of division of the given
// array elements
 
// Function to find the max result
function maxDivision(arr, n)
{
     
    // Sort the array in descending order
    arr.sort((a, b) => b - a);
   
    let mxdiv = arr[1];
   
    // Loop to divide in this order
    // arr[0] / ( arr[1] / arr[2] / ....
    // arr[n-2] / arr[n-1])
    for(let i = 2; i < n; ++i)
        mxdiv = mxdiv / arr[i];
   
    // Return the final result
    return arr[0] / mxdiv;
}
  
// Driver Code
let arr = [ 100, 1000, 10, 2 ];
let n = arr.length;
 
document.write(maxDivision(arr, n));
 
// This code is contributed by susmitakundugoaldanga
 

输出：

200

时间复杂度： O(N logN)

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