N 次切割的最大件数

给定一块正方形和可用的切割总数 n，找出用 n 次切割可以获得的相同尺寸的矩形或正方形的最大数量。允许的切割是水平切割和垂直切割。
注意：不允许堆叠和折叠。
例子：

Input : n = 1
Output : 2
Explanation : 

Input : n = 2
Output : 4
Explanation : 

Input : n = 3
Output : 6
Explanation : 

给定的是 n，它是允许的切割次数。由于需要在 n 次切割后最大化片数，所以水平切割的数量将等于垂直切割的数量。这可以使用微分来证明。所以水平切割的数量将是 n/2。和垂直切割将是 nn/2。
所以件数=（水平切割+1）*（垂直切割+1）。
程序：

C++

// C++ program to find maximum no of pieces
// by given number of cuts
#include 
using namespace std;
 
// Function for finding maximum pieces
// with n cuts.
int findMaximumPieces(int n)
{
    // to maximize number of pieces
    // x is the horizontal cuts
    int x = n / 2;
 
    // Now (x) is the horizontal cuts
    // and (n-x) is vertical cuts, then
    // maximum number of pieces = (x+1)*(n-x+1)
    return ((x + 1) * (n - x + 1));
}
 
// Driver code
int main()
{
 
    // Taking the maximum number of cuts allowed as 3
    int n = 3;
 
    // Finding and printing the max number of pieces
    cout << "Max number of pieces for n = " << n
         << " is " << findMaximumPieces(3);
 
    return 0;
}

Java

// Java program to find maximum
// no of pieces by given number
// of cuts
import java.util.*;
 
class GFG
{
// Function for finding maximum
// pieces with n cuts.
public static int findMaximumPieces(int n)
{
    // to maximize number of pieces
    // x is the horizontal cuts
    int x = n / 2;
 
    // Now (x) is the horizontal cuts
    // and (n-x) is vertical cuts, then
    // maximum number of pieces = (x+1)*(n-x+1)
    return ((x + 1) * (n - x + 1));
}
 
// Driver code
public static void main (String[] args)
{
    // Taking the maximum number
    // of cuts allowed as 3
    int n = 3;
     
    // Finding and printing the
    // max number of pieces
    System.out.print("Max number of pieces for n = " +
                   n + " is " + findMaximumPieces(3));
         
}
}
 
// This code is contributed by Kirti_Mangal

Python 3

# Python 3 program to find maximum no of pieces
# by given number of cuts
  
# Function for finding maximum pieces
# with n cuts.
def findMaximumPieces(n):
 
    # to maximize number of pieces
    # x is the horizontal cuts
    x = n // 2
  
    # Now (x) is the horizontal cuts
    # and (n-x) is vertical cuts, then
    # maximum number of pieces = (x+1)*(n-x+1)
    return ((x + 1) * (n - x + 1))
  
# Driver code
if __name__ == "__main__":
  
    #Taking the maximum number of cuts allowed as 3
    n = 3
  
    # Finding and printing the max number of pieces
    print("Max number of pieces for n = " +str( n)
         +" is " + str(findMaximumPieces(3)))
 
# This code is contributed by ChitraNayal

C#

// C# program to find maximum
// no of pieces by given number
// of cuts
using System;
 
class GFG
{
 
// Function for finding maximum
// pieces with n cuts.
public static int findMaximumPieces(int n)
{
    // to maximize number of pieces
    // x is the horizontal cuts
    int x = n / 2;
 
    // Now (x) is the horizontal 
    // cuts and (n-x) is vertical
    // cuts, then maximum number
    // of pieces = (x+1)*(n-x+1)
    return ((x + 1) * (n - x + 1));
}
 
// Driver code
static public void Main ()
{
    // Taking the maximum number
    // of cuts allowed as 3
    int n = 3;
     
    // Finding and printing the
    // max number of pieces
    Console.Write("Max number of pieces for n = " +
                n + " is " + findMaximumPieces(3));
}
}
 
// This code is contributed by Mahadev

Javascript

 
// Javascript program to find maximum no of pieces
// by given number of cuts
 
// Function for finding maximum pieces
// with n cuts.
function findMaximumPieces(n)
{
 
    // to maximize number of pieces
    // x is the horizontal cuts
    var x = parseInt(n / 2);
 
    // Now (x) is the horizontal cuts
    // and (n-x) is vertical cuts, then
    // maximum number of pieces = (x+1)*(n-x+1)
    return ((x + 1) * (n - x + 1));
}
 
// Driver code
// Taking the maximum number of cuts allowed as 3
var n = 3;
 
// Finding and printing the max number of pieces
document.write("Max number of pieces for n = " + n
    + " is " + findMaximumPieces(3));
 
// This code is contributed by noob2000.

输出：

Max number of pieces for n = 3 is 6