给定数字n，计算使用1 x 4瓦片填充anx 4网格的方法数。
例子：
Input : n = 1
Output : 1
Input : n = 2
Output : 1
We can only place both tiles horizontally
Input : n = 3
Output : 1
We can only place all tiles horizontally.
Input : n = 4
Output : 2
The two ways are :
1) Place all tiles horizontally
2) Place all tiles vertically.
Input : n = 5
Output : 3
We can fill a 5 x 4 grid in following ways :
1) Place all 5 tiles horizontally
2) Place first 4 vertically and 1 horizontally.
3) Place first 1 horizontally and 4 horizontally.我们强烈建议您单击此处并进行实践，然后再继续解决方案。
此问题主要是此拼贴问题的扩展
假设“ count(n)”是在“ nx 4”网格上放置图块的方法计数，当我们放置第一个图块时会出现以下两种情况。
- 水平放置第一个图块：如果我们水平放置第一个图块，问题将减少为“ count(n-1)”
- 垂直放置第一个图块：如果我们垂直放置第一个图块，则必须再垂直放置3个图块。因此问题减少到“ count(n-4)”
因此，count(n)可以写成如下形式。
count(n) = 1 if n = 1 or n = 2 or n = 3
count(n) = 2 if n = 4
count(n) = count(n-1) + count(n-4) 这种重复类似于斐波纳契数，可以使用动态编程解决。
C++
// C++ program to count of ways to place 1 x 4 tiles
// on n x 4 grid.
#include
using namespace std;
// Returns count of count of ways to place 1 x 4 tiles
// on n x 4 grid.
int count(int n)
{
// Create a table to store results of subproblems
// dp[i] stores count of ways for i x 4 grid.
int dp[n+1];
dp[0] = 0;
// Fill the table from d[1] to dp[n]
for (int i=1; i<=n; i++)
{
// Base cases
if (i >= 1 && i <= 3)
dp[i] = 1;
else if (i==4)
dp[i] = 2 ;
else
// dp(i-1) : Place first tile horizontally
// dp(n-4) : Place first tile vertically
// which means 3 more tiles have
// to be placed vertically.
dp[i] = dp[i-1] + dp[i-4];
}
return dp[n];
}
// Driver program to test above
int main()
{
int n = 5;
cout << "Count of ways is " << count(n);
return 0;
} Java
// Java program to count of ways to place 1 x 4 tiles
// on n x 4 grid
import java.io.*;
class Grid
{
// Function that count the number of ways to place 1 x 4 tiles
// on n x 4 grid.
static int count(int n)
{
// Create a table to store results of sub-problems
// dp[i] stores count of ways for i x 4 grid.
int[] dp = new int[n+1];
dp[0] = 0;
// Fill the table from d[1] to dp[n]
for(int i=1;i<=n;i++)
{
// Base cases
if (i >= 1 && i <= 3)
dp[i] = 1;
else if (i==4)
dp[i] = 2 ;
else
{
// dp(i-1) : Place first tile horizontally
// dp(i-4) : Place first tile vertically
// which means 3 more tiles have
// to be placed vertically.
dp[i] = dp[i-1] + dp[i-4];
}
}
return dp[n];
}
// Driver program
public static void main (String[] args)
{
int n = 5;
System.out.println("Count of ways is: " + count(n));
}
}
// Contributed by Pramod KumarPython
# Python program to count of ways to place 1 x 4 tiles
# on n x 4 grid.
# Returns count of count of ways to place 1 x 4 tiles
# on n x 4 grid.
def count(n):
# Create a table to store results of subproblems
# dp[i] stores count of ways for i x 4 grid.
dp = [0 for _ in range(n+1)]
# Fill the table from d[1] to dp[n]
for i in range(1,n+1):
# Base cases
if i <= 3:
dp[i] = 1
elif i == 4:
dp[i] = 2
else:
# dp(i-1) : Place first tile horizontally
# dp(n-4) : Place first tile vertically
# which means 3 more tiles have
# to be placed vertically.
dp[i] = dp[i-1] + dp[i-4]
return dp[n]
# Driver code to test above
n = 5
print ("Count of ways is"),
print (count(n))C#
// C# program to count of ways
// to place 1 x 4 tiles on
// n x 4 grid
using System;
class GFG
{
// Function that count the number
// of ways to place 1 x 4 tiles
// on n x 4 grid.
static int count(int n)
{
// Create a table to store results
// of sub-problems dp[i] stores
// count of ways for i x 4 grid.
int[] dp = new int[n + 1];
dp[0] = 0;
// Fill the table from d[1]
// to dp[n]
for(int i = 1; i <= n; i++)
{
// Base cases
if (i >= 1 && i <= 3)
dp[i] = 1;
else if (i == 4)
dp[i] = 2 ;
else
{
// dp(i-1) : Place first tile
// horizontally dp(i-4) :
// Place first tile vertically
// which means 3 more tiles have
// to be placed vertically.
dp[i] = dp[i - 1] +
dp[i - 4];
}
}
return dp[n];
}
// Driver Code
public static void Main ()
{
int n = 5;
Console.WriteLine("Count of ways is: "
+ count(n));
}
}
// This code is contributed by Sam007PHP
= 1 && $i <= 3)
$dp[$i] = 1;
else if ($i == 4)
$dp[$i] = 2 ;
else
// dp(i-1) : Place first tile horizontally
// dp(n-4) : Place first tile vertically
// which means 3 more tiles have
// to be placed vertically.
$dp[$i] = $dp[$i - 1] + $dp[$i - 4];
}
return $dp[$n];
}
// Driver Code
$n = 5;
echo "Count of ways is " , countt($n);
// This code is contributed by nitin mittal.
?>Javascript
输出 ：
Count of ways is 3