ПРОБЛЕМЪТ НА МЪРЗЕЛИВИЯ ЗАВЕДЕНИЯ

Като се има предвид цяло число, обозначаващо броя на разфасовки, които могат да бъдат направени на палачинка, да се намерят максималния брой парчета, които могат да бъдат оформени чрез правене на N разрязвания.
Примери:

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

Препоръчителна практика Проблемът на мързеливия заведения Опитайте!

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).

Можем да оптимизираме по -горе формулата.

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

Следователно с тази оптимизация можем да отговорим на всички заявки в O (1).
По -долу е прилагането на горната идея:

C++

// A C++ program to find the solution to // The Lazy Caterer's Problem #include    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# 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.  ?>

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) + '  
');  document.write(findPieces(2) + '  
');  document.write(findPieces(3) + '  
');  document.write(findPieces(50));   </script>

Резултат:

2 4 7 1276

Референции: oeis.org