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

pizza-cut


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++

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// 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;
}

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Java














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// 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 



















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Python3














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# 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 



















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C#














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// 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. 



















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PHP














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<?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.  


?> 



















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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 contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.


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