El problema del catering perezoso

Dado un número entero n, que denota el número de cortes que se pueden hacer en un panqueque, encuentre el número máximo de piezas que se pueden formar al hacer n cortes. 
Ejemplos: 
 

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). 

Podemos optimizar la fórmula anterior. 
 

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

Por lo tanto, con esta optimización, podemos responder todas las consultas en O(1).
A continuación se muestra la implementación de la idea anterior:
 

// 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 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

# 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# 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
// 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.
?>

<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>

Producción :  

2
4
7
1276

Referencias: oeis.org
Este artículo es una contribución de Ashutosh Kumar . Si te gusta GeeksforGeeks y te gustaría contribuir, también puedes escribir un artículo usando write.geeksforgeeks.org o enviar tu artículo por correo a review-team@geeksforgeeks.org. Vea su artículo que aparece en la página principal de GeeksforGeeks y ayude a otros Geeks.
Escriba comentarios si encuentra algo incorrecto o si desea compartir más información sobre el tema tratado anteriormente.