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
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