The Lazy Caterer’s Problem

Given an integer n, denoting the number of cuts that can be made on a pancake, find the maximum number of pieces that can be formed by making n cuts. 
Examples :

Input :  n = 1
Output : 2
With 1 cut we can divide the pancake in 2 pieces

Input :  2
Output : 4
With 2 cuts we can divide the pancake in 4 pieces

Input : 3
Output : 7
We can divide the pancake in 7 parts with 3 cuts

Input : 50
Output : 1276

Let f(n) denote the maximum number of pieces
that can be obtained by making n cuts.
Trivially,
f(0) = 1                                 
As there'd be only 1 piece without any cut.

Similarly,
f(1) = 2

Proceeding in similar fashion we can deduce 
the recursive nature of the function.
The function can be represented recursively as :
f(n) = n + f(n-1)

Hence a simple solution based on the above 
formula can run in O(n). 

We can optimize above formula. 

We now know ,
f(n) = n + f(n-1) 

Expanding f(n-1) and so on we have ,
f(n) = n + n-1 + n-2 + ...... + 1 + f(0)

which gives,
f(n) = (n*(n+1))/2 + 1

Hence with this optimization, we can answer all the queries in O(1).
Below is the implementation of above idea :

C++

// A C++ program to find the solution to
// The Lazy Caterer's Problem
#include <iostream>
using namespace std;

// This function receives an integer n
// and returns the maximum number of
// pieces that can be made form pancake
// using n cuts
int findPieces(int n)
{
    // Use the formula
    return (n * ( n + 1)) / 2 + 1;
}

// Driver Code
int main()
{
    cout << findPieces(1) << endl;
    cout << findPieces(2) << endl;
    cout << findPieces(3) << endl;
    cout << findPieces(50) << endl;
    return 0;
}

Java

// Java program to find the solution to
// The Lazy Caterer's Problem
import java.io.*;

class GFG
{
    // This function returns the maximum
    // number of pieces that can be made
    //  form pancake using n cuts
    static int findPieces(int n)
    {
        // Use the formula
        return (n * (n + 1)) / 2 + 1;
    }

    // Driver program to test above function
    public static void main (String[] args)
    {
        System.out.println(findPieces(1));
        System.out.println(findPieces(2));
        System.out.println(findPieces(3));
        System.out.println(findPieces(50));
    }
}

// This code is contributed by Pramod Kumar

Python3

# A Python 3 program to
# find the solution to
# The Lazy Caterer's Problem

# This function receives an
# integer n and returns the
# maximum number of pieces
# that can be made form
# pancake using n cuts
def findPieces( n ):

    # Use the formula
    return (n * ( n + 1)) // 2 + 1

# Driver Code
print(findPieces(1))
print(findPieces(2))
print(findPieces(3))
print(findPieces(50))

# This code is contributed
# by ihritik

C#

// C# program to find the solution
// to The Lazy Caterer's Problem
using System;

class GFG
{
    // This function returns the maximum
    // number of pieces that can be made
    // form pancake using n cuts
    static int findPieces(int n)
    {
        // Use the formula
        return (n * (n + 1)) / 2 + 1;
    }

    // Driver code
    public static void Main ()
    {
        Console.WriteLine(findPieces(1));
        Console.WriteLine(findPieces(2));
        Console.WriteLine(findPieces(3));
        Console.Write(findPieces(50));
    }
}

// This code is contributed by Nitin Mittal.

PHP

<?php
// A php program to find
// the solution to The
// Lazy Caterer's Problem

// This function receives
// an integer n and returns
// the maximum number of
// pieces that can be made
// form pancake using n cuts
function findPieces($n)
{
    // Use the formula
    return ($n * ( $n + 1)) / 2 + 1;
}

// Driver Code
echo findPieces(1) , "\n" ;
echo findPieces(2) , "\n" ;
echo findPieces(3) , "\n" ;
echo findPieces(50) ,"\n";

// This code is contributed
// by nitin mittal.
?>

Javascript

<script>

// Javascript program to find the solution to
// The Lazy Caterer's Problem

    // This function returns the maximum
    // number of pieces that can be made
    //  form pancake using n cuts
    function findPieces(n)
    {
        // Use the formula
        return (n * (n + 1)) / 2 + 1;
    }


// Driver Code

        document.write(findPieces(1) + "<br/>");
        document.write(findPieces(2) + "<br/>");
        document.write(findPieces(3) + "<br/>");
        document.write(findPieces(50));

</script>

Output :

2
4
7
1276

References : oeis.org
This article is contributed by Ashutosh Kumar .If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.