n块的最小周长

给定n个大小为1 x 1的块，我们需要找到由这些块构成的网格的最小周长。

例子 ：

输入：n = 4输出：8带有4个块的最小可能周长是8。请参见以下说明。输入：n = 11输出：14上面示例的正方形网格为

让我们以一个例子来看一个模式。假设我们有4个区块，以下是不同的可能性

+--+--+--+--+
  |  |  |  |  |  Perimeter = 10
  +--+--+--+--+

  +--+--+--+
  |  |  |  |     Perimeter = 10
  +--+--+--+
        |  |
        +--+

  +--+--+--+
  |  |  |  |     Perimeter = 10
  +--+--+--+
     |  |
     +--+


  +--+--+
  |  |  |        Perimeter = 8
  +--+--+
  |  |  |
  +--+--+

如果我们使用笔和纸做一些示例，我们会注意到，当所形成的形状最接近正方形时，周长会最小。原因是，我们希望块的最大边面向形状内部，以使形状的周长最小。

如果块数是一个完美的正方形，则周长将简单地为4 * sqrt(n)。
但是，如果“块数”不是理想的平方根，则我们将计算最接近平方根的行数和列数。在将块排列成矩形后，我们仍然剩下块，然后我们只需将2加到周长，因为只剩下2个额外的边。
下面给出上述想法的实现。

C++

// CPP program to find minimum 
// perimeter using n blocks.
#include 
using namespace std;
  
int minPerimeter(int n)
{
    int l = sqrt(n);
    int sq = l * l;
  
    // if n is a perfect square
    if (sq == n) 
        return l * 4;
    else
    {
        // Number of rows 
        long long int row = n / l; 
  
        // perimeter of the 
        // rectangular grid 
        long long int perimeter 
                      = 2 * (l + row); 
  
        // if there are blocks left 
        if (n % l != 0) 
            perimeter += 2;
        return perimeter;
    }
}
  
// Driver code
int main()
{
    int n = 10;
    cout << minPerimeter(n);
    return 0;
}

Java

// JAVA Code to find minimum 
// perimeter using n blocks
import java.util.*;
  
class GFG 
{
    public static long minPerimeter(int n)
    {
        int l = (int) Math.sqrt(n);
        int sq = l * l;
      
        // if n is a perfect square
        if (sq == n) 
            return l * 4;
        else
        {
            // Number of rows 
            long row = n / l; 
      
            // perimeter of the 
            // rectangular grid 
            long perimeter 
                  = 2 * (l + row); 
      
            // if there are blocks left 
            if (n % l != 0) 
                perimeter += 2;
            return perimeter;
        }
    }
      
    // Driver code
    public static void main(String[] args) 
    {
        int n = 10;
        System.out.println(minPerimeter(n));
    }
}
  
// This code is contributed by Arnav Kr. Mandal

Python3

# Python3 program to find minimum 
# perimeter using n blocks.
import math
  
def minPerimeter(n):
    l = math.sqrt(n)
    sq = l * l
   
    # if n is a perfect square
    if (sq == n): 
        return l * 4
    else :
        # Number of rows 
        row = n / l
   
        # perimeter of the 
        # rectangular grid 
        perimeter = 2 * (l + row)
                        
        # if there are blocks left 
        if (n % l != 0): 
            perimeter += 2
        return perimeter
  
# Driver code
n = 10
print(int(minPerimeter(n)))
  
# This code is contributed by 
# Prasad Kshirsagar

C#

// C# Code to find minimum 
// perimeter using n blocks
using System;
  
class GFG 
{
    public static long minPerimeter(int n)
    {
        int l = (int) Math.Sqrt(n);
        int sq = l * l;
      
        // if n is a perfect square
        if (sq == n) 
            return l * 4;
        else
        {
            // Number of rows 
            long row = n / l; 
          
            // perimeter of the 
            // rectangular grid 
            long perimeter
                  = 2 * (l + row); 
      
            // if there are blocks left 
            if (n % l != 0) 
                perimeter += 2;
            return perimeter;
        }
    }
      
    // Driver code
    public static void Main() 
    {
        int n = 10;
        Console.Write(minPerimeter(n));
    }
}
  
// This code is contributed by nitin mittal

输出 ：

14

参考 ：
http://mathforum.org/library/drmath/view/61595.html
intermath.coe.uga.edu/tweb/gcsu-geo-spr06/aheath/aheath_rectperimeter.doc