计算使用指定尺寸的瓷砖铺设N长度板的方法

给定一个整数N ，任务是平铺尺寸为N * 1的板，任务是计算使用尺寸为1 * 1和2 * 1的砖平铺板的方法的数量。

例子：

Input: N = 2 
Output: 5 
Explanation: 
Tile the board of size 2 x 1 by placing 1 x 1 tile in a Horizontal way and another 1 x 1 tile in a vertical way. 
Tile the board of size 2 x 1 by placing 1 x 1 tile in a vertical way and another 1 x 1 tile in a Horizontal way. 
Tile the board of size 2 x 1 by placing 1 x 1 tile in a Horizontal way and another 1 x 1 tile in a Horizontal way. 
Tile the board of size 2 x 1 by placing 1 x 1 tile in a vertical way and another 1 x 1 tile in a vertical way. 
Tile the board of size 2 x 1 by placing 2 x 1 tile in a horizontal way. 
Therefore, the required output is 5.

Input: N = 1
Output: 2

方法：可以使用动态编程解决问题。想法是将N x 1板划分为较小的板，然后计算平铺较小的板的方法。最后，打印铺装N x 1板的方法总数。请按照以下步骤解决问题：

• 以下是放置第一个图块的三种方法：
  • 垂直放置1 x 1瓷砖。
  • 将1 x 1瓷砖水平放置。
  • 将2 x 1瓷砖水平放置。
• 因此，解决该问题的递归关系如下：

cntWays(N) = 2 * cntWays(N – 1) + cntWays(N – 2)

• 初始化一个数组dp [] ，其中dp [i]存储平铺ix 1板的方式的数量。
• 迭代范围[2，N]，并使用制表法填充数组dp [] 。
• 最后，打印dp [N]的值。

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
 
#include 
using namespace std;
 
 
// Function to count the ways to tile N * 1
// board using 1 * 1 and 2 * 1 tiles
long countWaysToTileBoard(long N)
{
     
    // dp[i]: Stores count of ways to tile
    // i * 1 board using given tiles
    long dp[N + 1];
 
    // Base Case
    dp[0] = 1;
    dp[1] = 2;
     
    // Iterate over the range [2, N]
    for (int i = 2; i <= N; i++) {
 
        // Fill dp[i] using
        // the recurrence relation
        dp[i] = (2 * dp[i - 1]
                    + dp[i - 2] );
    }
 
    cout << dp[N];
}
 
 
 
// Driver Code
int main()
{
    // Given N
    long N = 2;
 
    // Function Call
    countWaysToTileBoard(N);
 
    return 0;
}

Java

// Java program to implement
// the above approach
class GFG{
     
// Function to count the ways to tile N * 1
// board using 1 * 1 and 2 * 1 tiles
static void countWaysToTileBoard(int N)
{
     
    // dp[i]: Stores count of ways to tile
    // i * 1 board using given tiles
    int dp[] = new int[N + 1];
     
    // Base Case
    dp[0] = 1;
    dp[1] = 2;
     
    // Iterate over the range [2, N]
    for(int i = 2; i <= N; i++)
    {
         
        // Fill dp[i] using
        // the recurrence relation
        dp[i] = (2 * dp[i - 1] +
                     dp[i - 2]);
    }
    System.out.print(dp[N]);
}
 
// Driver Code
public static void main (String[] args)
{
     
    // Given N
    int N = 2;
     
    // Function Call
    countWaysToTileBoard(N);
}
}
 
// This code is contributed by AnkThon

Python3

# Python3 program to implement
# the above approach
 
# Function to count the ways to tile N * 1
# board using 1 * 1 and 2 * 1 tiles
def countWaysToTileBoard( N) :
     
    # dp[i]: Stores count of ways to tile
    # i * 1 board using given tiles
    dp = [0]*(N + 1)
 
    # Base Case
    dp[0] = 1
    dp[1] = 2
     
    # Iterate over the range [2, N]
    for i in range(2, N + 1) :
 
        # Fill dp[i] using
        # the recurrence relation
        dp[i] = (2 * dp[i - 1] + dp[i - 2] )
     
    print(dp[N])
 
# Driver Code
if __name__ == "__main__" :
     
    # Given N
    N = 2
 
    # Function Call
    countWaysToTileBoard(N)
     
  # This code is contributed by jana_sayantan

C#

// C# program to implement
// the above approach
using System;
 
class GFG{
     
// Function to count the ways to tile N * 1
// board using 1 * 1 and 2 * 1 tiles
static void countWaysToTileBoard(int N)
{
     
    // dp[i]: Stores count of ways to tile
    // i * 1 board using given tiles
    int []dp = new int[N + 1];
     
    // Base Case
    dp[0] = 1;
    dp[1] = 2;
     
    // Iterate over the range [2, N]
    for(int i = 2; i <= N; i++)
    {
         
        // Fill dp[i] using
        // the recurrence relation
        dp[i] = (2 * dp[i - 1] +
                     dp[i - 2]);
    }
    Console.Write(dp[N]);
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given N
    int N = 2;
     
    // Function Call
    countWaysToTileBoard(N);
}
}
 
// This code is contributed by shikhasingrajput

输出：

5

时间复杂度： O(N)
辅助空间： O(N)