Problema de mosaico – Part 1

Dado un tablero de “2 x n” y fichas de tamaño “2 x 1”, cuente el número de formas de colocar mosaicos en el tablero dado usando las fichas de 2 x 1. Una loseta puede colocarse horizontalmente, es decir, como una loseta de 1 x 2, o verticalmente, es decir, como una loseta de 2 x 1. 

Ejemplos: 

// C++ program to count the
// no. of ways to place 2*1 size
// tiles in 2*n size board.
#include <iostream>
using namespace std;
 
int getNoOfWays(int n)
{
    // Base case
    if (n <= 2)
      return n;
 
    return getNoOfWays(n - 1) + getNoOfWays(n - 2);
}
 
// Driver Function
int main()
{
    cout << getNoOfWays(4) << endl;
    cout << getNoOfWays(3);
    return 0;
}

/* Java program to count the
 no of ways to place 2*1 size
 tiles in 2*n size board. */
import java.io.*;
 
class GFG {
  static int getNoOfWays(int n)
  {
 
    // Base case
    if (n <= 2) {
      return n;
    }
    return getNoOfWays(n - 1) + getNoOfWays(n - 2);
  }
 
  // Driver Function
  public static void main(String[] args)
  {
    System.out.println(getNoOfWays(4));
    System.out.println(getNoOfWays(3));
  }
}
 
// This code is contributed by ashwinaditya21.

# Python3 program to count the
# no. of ways to place 2*1 size
# tiles in 2*n size board.
def getNoOfWays(n):
   
    # Base case
    if n <= 2:
        return n
 
    return getNoOfWays(n - 1) + getNoOfWays(n - 2)
 
# Driver Code
print(getNoOfWays(4))
print(getNoOfWays(3))
 
# This code is contributed by Kevin Joshi

// C# program to implement
// the above approach
using System;
 
class GFG
{
     
static int getNoOfWays(int n)
  {
  
    // Base case
    if (n <= 2) {
      return n;
    }
    return getNoOfWays(n - 1) + getNoOfWays(n - 2);
  }
 
// Driver Code
public static void Main()
{
    Console.WriteLine(getNoOfWays(4));
    Console.WriteLine(getNoOfWays(3));
}
}
 
// This code is contributed by code_hunt.

<script>
// JavaScript program to count the
// no. of ways to place 2*1 size
// tiles in 2*n size board.
 
function getNoOfWays(n)
{
    // Base case
    if (n <= 2)
      return n;
       
    return getNoOfWays(n - 1) + getNoOfWays(n - 2);
}
 
// Driver Function
document.write(getNoOfWays(4));
document.write(getNoOfWays(3));
 
// This code is contributed by shinjanpatra
</script>