矩形的长度和宽度使得面积与对角线 ^2 的比率最大

给定一个正整数数组。任务是从给定的数组中选择一对元素，使它们代表矩形的长和宽，并且其面积与对角线²的比率最大。
注意：数组必须包含矩形的所有边。也就是说，您可以从数组中选择至少出现两次的元素，因为矩形的两条边长度相同，两条边宽度相同。
例子：

Input: arr[] = {4, 3, 5, 4, 3, 5, 7} 
Output: 5, 4 
Among all pairs of length and breadth 5, 4 will generate maximum ratio of Area to Diameter².
Input: arr[] = {2, 2, 2, 2, 2, 2} 
Output: 2, 2 
There is only one possible pair of length and breadth 2, 2 which will generate maximum ratio of Area to Diameter².

下面给出的是为了形成矩形而要满足的矩形的一些属性。

• 只有当我们至少有一对相等的整数时才能形成矩形。出现一次的整数不能是任何矩形的一部分。因此，在我们的解决方案中，我们将只考虑出现多次的整数。
• 面积与其直径^{2 之比}为lb/(l ² + b ² )是单调递减函数
  就一个变量而言。这意味着如果我们固定长度，那么当我们减小宽度时，比率会降低，因此固定宽度相同。利用这一点，我们不需要迭代整个数组来找到固定宽度的长度。

算法 ：

1. 对给定的数组进行排序。
2. 创建一个在数组中出现两次的整数数组 (arr_pairs[])。
3. 设置长度 = arr_pairs[0]，宽度 = arr_pairs[0]。
4. 从 i=2 迭代到 sizeof (arr_pairs)
   • 如果（长度/宽度 + 宽度/长度 > arr_pairs[i]/arr_pairs[i-1] + arr_pairs[i-1]/arr_pairs[i]）
     • 更新长度 = arr_pairs[i]，宽度 = arr_pairs[i-1]
5. 打印长度和宽度。

下面是上述方法的实现。

C++

// CPP for finding maximum
// p^2/A ratio of rectangle
#include 
using namespace std;
 
// function to print length and breadth
void findLandB(int arr[], int n)
{
    // sort the input array
    sort(arr, arr + n);
 
    // create array vector of integers occurring in pairs
    vector arr_pairs;
    for (int i = 0; i < n; i++) {
 
        // push the same pairs
        if (arr[i] == arr[i + 1]) {
            arr_pairs.push_back(arr[i]);
            i++;
        }
    }
 
    double length = arr_pairs[0];
    double breadth = arr_pairs[1];
    double size = arr_pairs.size();
 
    // calculate length and breadth as per requirement
    for (int i = 2; i < size; i++) {
 
        // check for given condition
        if ((length / breadth + breadth / length) > (arr_pairs[i] / arr_pairs[i - 1] + arr_pairs[i - 1] / arr_pairs[i])) {
 
            length = arr_pairs[i];
            breadth = arr_pairs[i - 1];
        }
    }
 
    // print the required answer
    cout << length << ", " << breadth << endl;
}
 
// Driver Code
int main()
{
    int arr[] = { 4, 2, 2, 2, 5, 6, 5, 6, 7, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
    findLandB(arr, n);
    return 0;
}

Java

// JAVA for finding maximum
// p^2/A ratio of rectangle
import java.util.*;
 
class GFG
{
 
// function to print length and breadth
static void findLandB(int arr[], int n)
{
    // sort the input array
    Arrays.sort(arr);
 
    // create array vector of integers occurring in pairs
    Vector arr_pairs = new Vector();
    for (int i = 0; i < n - 1; i++)
    {
 
        // push the same pairs
        if (arr[i] == arr[i + 1])
        {
            arr_pairs.add((double) arr[i]);
            i++;
        }
    }
 
    double length = arr_pairs.get(0);
    double breadth = arr_pairs.get(1);
    double size = arr_pairs.size();
 
    // calculate length and breadth as per requirement
    for (int i = 2; i < size; i++)
    {
 
        // check for given condition
        if ((length / breadth + breadth / length) >
            (arr_pairs.get(i) / arr_pairs.get(i - 1) +
            arr_pairs.get(i - 1) / arr_pairs.get(i)))
        {
            length = arr_pairs.get(i);
            breadth = arr_pairs.get(i - 1);
        }
    }
 
    // print the required answer
    System.out.print((int)length + ", " + (int)breadth +"\n");
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 4, 2, 2, 2, 5, 6, 5, 6, 7, 2 };
    int n = arr.length;
    findLandB(arr, n);
}
}
 
// This code is contributed by PrinciRaj1992

Python3

# Python 3 for finding maximum p^2/A
# ratio of rectangle
 
# function to print length and breadth
def findLandB(arr, n):
     
    # sort the input array
    arr.sort(reverse = False)
 
    # create array vector of integers
    # occurring in pairs
    arr_pairs = []
    for i in range(n - 1):
         
        # push the same pairs
        if (arr[i] == arr[i + 1]):
            arr_pairs.append(arr[i])
            i += 1
     
    length = arr_pairs[0]
    breadth = arr_pairs[1]
    size = len(arr_pairs)
 
    # calculate length and breadth as
    # per requirement
    for i in range(1, size - 1):
         
        # check for given condition
        if ((int(length / breadth) +
             int(breadth / length)) >
            (int(arr_pairs[i] / arr_pairs[i - 1]) +
             int(arr_pairs[i - 1] / arr_pairs[i]))):
            length = arr_pairs[i]
            breadth = arr_pairs[i - 1]
 
    # print the required answer
    print(length, ",", breadth)
 
# Driver Code
if __name__== '__main__':
    arr = [4, 2, 2, 2, 5, 6, 5, 6, 7, 2]
    n = len(arr)
    findLandB(arr, n)
     
# This code is contributed by
# Surendra_Gangwar

C#

// C# for finding maximum
// p^2/A ratio of rectangle
using System;
using System.Collections.Generic;
 
class GFG
{
 
// function to print length and breadth
static void findLandB(int []arr, int n)
{
    // sort the input array
    Array.Sort(arr);
 
    // create array vector of integers occurring in pairs
    List arr_pairs = new List();
    for (int i = 0; i < n - 1; i++)
    {
 
        // push the same pairs
        if (arr[i] == arr[i + 1])
        {
            arr_pairs.Add((double) arr[i]);
            i++;
        }
    }
 
    double length = arr_pairs[0];
    double breadth = arr_pairs[1];
    double size = arr_pairs.Count;
 
    // calculate length and breadth as per requirement
    for (int i = 2; i < size; i++)
    {
 
        // check for given condition
        if ((length / breadth + breadth / length) >
            (arr_pairs[i] / arr_pairs[i - 1] +
            arr_pairs[i - 1] / arr_pairs[i]))
        {
            length = arr_pairs[i];
            breadth = arr_pairs[i - 1];
        }
    }
 
    // print the required answer
    Console.Write((int)length + ", " + (int)breadth +"\n");
}
 
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 4, 2, 2, 2, 5, 6, 5, 6, 7, 2 };
    int n = arr.Length;
    findLandB(arr, n);
}
}
 
// This code is contributed by Rajput-Ji

Javascript

 
// Javascript for finding maximum
// p^2/A ratio of rectangle
 
// function to print length and breadth
function findLandB(arr, n)
{
    // sort the input array
    arr.sort();
 
    // create array vector of integers occurring in pairs
    var arr_pairs =[];
    for (var i = 0; i < n; i++) {
 
        // push the same pairs
        if (arr[i] == arr[i + 1]) {
            arr_pairs.push(arr[i]);
            i++;
        }
    }
 
    var length = arr_pairs[0];
    var breadth = arr_pairs[1];
    var size = arr_pairs.length;
 
    // calculate length and breadth as per requirement
    for (var i = 2; i < size; i++) {
 
        // check for given condition
        if ((length / breadth + breadth / length) > (arr_pairs[i] / arr_pairs[i - 1] + arr_pairs[i - 1] / arr_pairs[i])) {
 
            length = arr_pairs[i];
            breadth = arr_pairs[i - 1];
        }
    }
 
    // print the required answer
    document.write( length + ", " + breadth );
}
 
// Driver Code
var arr = [ 4, 2, 2, 2, 5, 6, 5, 6, 7, 2 ];
var n = arr.length;
findLandB(arr, n);
 

输出：

2, 2

时间复杂度： O(N * log N)

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