披萨切问题(或按行圈划)

给定切割数量，找到可能的最大数量。
例子：

Input  : 2
Output : 4

Input  : 3
Output : 7

这个问题不过是懒惰的餐饮者问题，具有以下公式。
最大件数= 1 + n *(n + 1)/ 2
请参阅此作为证明。

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)
{
   return 1 + n*(n+1)/2;   
}
 
// Driver code
int main()
{
   cout << findMaximumPieces(3);
   return 0;
}

Java

// Java program to find maximum no of
// pieces by given number of cuts
class GFG {
     
    // Function for finding maximum pieces
    // with n cuts.
    static int findMaximumPieces(int n)
    {
        return 1 + n * (n + 1) / 2;
    }
     
    // Driver Program to test above function
    public static void main(String arg[])
    {
         
        System.out.print(findMaximumPieces(3));
    }
}
 
// This code is contributed by Anant Agarwal.

Python3

# Python3 program to find maximum
# no. of pieces by given
# number of cuts
 
# Function for finding maximum
# pieces with n cuts.
def findMaximumPieces(n):
    return int(1 + n * (n + 1) / 2)
 
# Driver code
print(findMaximumPieces(3))
 
# This code is contributed 29AjayKumar

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.
    static int findMaximumPieces(int n)
    {
        return 1 + n * (n + 1) / 2;
    }
      
    // Driver Program to test above function
    public static void Main()
    {
          
        Console.Write(findMaximumPieces(3));
    }
}
  
// This code is contributed by nitin mittal.

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)
    {
        return 1 + n * (n + 1) / 2;
    }
 
// Driver Code
 
    document.write(findMaximumPieces(3));
 

输出：

7