我在浏览＃1，看到这个问题有关平铺显示M×N的矩形，我认为这将是伟大的高尔夫球场。这是任务。

给定尺寸M和N，编写一个程序，输出在给定这些约束的情况下可以平铺MxN矩形（N是行数，而不是列数。这并不重要）有多少种独特方式。

1. 所有图块均为2x1或3x1
2. 所有图块都留在其行内（即它们都是水平的）
3. 每两个相邻的行之间，除了两端之外，磁贴不应对齐
4. M和N保证至少为1

例如，有效的8x3矩阵切片将是

  2    3     3
  |    |     |
  v    v     v
 _______________
|___|_____|_____| 
|_____|_____|___|
|___|_____|_____|

但是以下内容将无效，因为行对齐

  2    3     3
  |    |     |
  v    v     v
 _______________
|___|_____|_____| 
|_____|___|_____|
|_____|_____|___|

测试用例：

8x3：4

3x1：1

1x1：0

9x4：10

编码高尔夫，最短的答案将获胜。

果冻，20字节

2*ḃ€2‘ÄṪ⁼¥Ƈ⁸ṗfƝẸ$€ċ0

在线尝试！

JavaScript（ES6）， 119 110 106 96  91字节

将输入作为。$(N,M)$

f=(n,m,p=0,g=(w,h=x=>g(p[g[w-=x]=1,w]||w)*g[w]--)=>w>3?h(2)+h(1):w>1&&f(n,m-1,g))=>m?g(n):1

在线尝试！

已评论

注意：此代码使用3个互不相同的函数。这使得跟踪变量的范围有点困难。请记住，在的范围内定义，而在的范围内定义。$g$$f$$h$$g$

f = (                    // f is a recursive function taking:
  n,                     //   n = number of columns
  m,                     //   m = number of rows
  p = 0,                 //   p = object holding the previous row
  g = (                  //   g = recursive function taking:
    w,                   //     w = remaining width that needs to be filled in the
                         //         current row
    h = x =>             //     h = helper function taking x
                         // h body:
      g(                 //   recursive call to g:
        p[g[w -= x] = 1, //     subtract either 2 or 1 from w and mark this width as used
          w              //     test p[w]
        ]                //     pass p[w] if p[w] = 1 (which will force the next iteration
                         //     to fail immediately)
        || w             //     otherwise, pass w
      )                  //   end of recursive call
      * g[w]--           //   then restore g[w] to 0
  ) =>                   // g body:
    w > 3 ?              //   if w > 3, we need to insert at least 2 more bricks:
      h(2) + h(1)        //     invoke h with x = 2 and x = 1
    :                    //   else:
      w > 1              //     this is the last brick; we just check if it can be inserted
      &&                 //     abort if w is equal to 1 (a brick does not fit in there)
      f(                 //     otherwise, do a recursive call to f:
        n,               //       n is unchanged
        m - 1,           //       decrement m
        g                //       pass g as the new reference row
      )                  //     end of recursive call
) =>                     // f body:
  m ? g(n) : 1           //   yield 1 if we made it to the last row or call g otherwise

R，243231字节

function(m,n,i=2*0:(m%/%6)+m%%2,j=i+(m-3*i)/2,M=Map)`if`(m<2,0,sum((e=eigen(lengths(outer(p<-unlist(M(M,list(function(x,y)cumsum(2+1:y%in%x)),M(combn,j,i,s=F),j),F),p,Vectorize(intersect)))<2))$ve%*%diag(e$va^(n-1))%*%solve(e$ve)))

在线尝试！

带换行符的版本：

function(m,n,i=2*0:(m%/%6)+m%%2,j=i+(m-3*i)/2,M=Map)`if`(m<2,0,
sum((e=eigen(lengths(outer(p<-unlist(M(M,list(function(x,y)cumsum(2+1:y%in%x)),
M(combn,j,i,s=F),j),F),p,Vectorize(intersect)))<2))$ve%*%diag(e$va^(n-1))%*%solve(e$ve)))

注意没有递归，并且处理相当大的m和n值（例如24x20-> 3.3e19）

这是一个评论的答案，其作用与上面的大致相同，但我没有嵌套所有功能，因此它实际上是可读的：

f <- function(m,n) {
  # First work out what potential combinations of 2s and 3s add up to m
  i <- 2*0:(m %/% 6) + m %% 2 # Vector with numbers of possible 3s
  j <- i + (m - 3 * i) / 2 # Vector with total number of 2s and 3s
  if (m < 2) {
    0 # If wall less than 2 wide, no point in continuing because answer is 0
  } else {
    # Work out all possible positions of threes for each set
    positions_of_threes <- Map(combn, j, i, simplify = FALSE)
    # Function to work out the cumulative distance along the wall for a given
    # Set of three positions and number of bricks
    make_cumulative_bricks <- function(pos_threes, n_bricks) {
      bricks <- 1:n_bricks %in% pos_threes
      cumsum(2 + bricks)
    }
    # Find all possible rows with cumulative width of wall
    # Note because this is a `Map` with depth two that needs to be vectorised
    # for both `positions_of_threes` and `j`, and we're using base R, the
    # function `make_cumulative_bricks` needs to be placed in a list
    cum_bricks <- Map(Map, list(make_cumulative_bricks), positions_of_threes, j)
    # Finally we have the list of possible rows of bricks as a flat list
    cum_bricks_unlisted <- unlist(cum_bricks, recursive = FALSE)
    # Vectorise the intersect function
    intersect_v <- Vectorize(intersect, SIMPLIFY = FALSE)
    # Find the length of all possible intersects between rows
    intersections <- outer(cum_bricks_unlisted, cum_bricks_unlisted, intersect_v)
    n_intersections <- lengths(intersections)
    # The ones not lined up will only have a single intersect at `m`
    not_lined_up <- n_intersections == 1
    # Now use method described at /programming//a/9459540/4998761
    # to calculate the (matrix of TRUE/FALSE for lined-up) to the power of `n`
    eigen_nlu <- eigen(not_lined_up)
    final_mat <- eigen_nlu$vectors %*%
      diag(eigen_nlu$values ^ (n - 1)) %*%
      solve(eigen_nlu$vectors)
    # The sum of this matrix is what we're looking for
    sum(final_mat)
  }
}
f(20,20)

取矩阵并将其自身重复乘以的方法来自于stackoverflow的问题。这种方法之所以在这里起作用，是因为它有效地计算了通过砖的不同可能行的累计分支数。

如果允许使用外部软件包，我可以将其降低到192：

function(m,n,i=2*0:(m%/%6)+m%%2,j=i+(m-3*i)/2,M=purrr::map2)`if`(m<2,0,sum(expm::`%^%`(lengths(outer(p<-unlist(M(M(j,i,combn,s=F),j,M,~cumsum(2+1:.y%in%.)),F),p,Vectorize(intersect)))<2,n-1)))

果冻，26个字节

2*ḃ€2‘ÄṪ⁼¥Ƈ⁸œ&L¬ɗþ`æ*⁴’¤SS

在线尝试！

细分：

生成可能的墙的列表作为累积总和，并去除末端：

2*ḃ€2‘ÄṪ⁼¥Ƈ⁸

找到所有可能没有相交的墙的外表：

œ&L¬ɗþ`

将这个矩阵乘以（N-1）的幂，然后将其总和：

æ*⁴’¤SS

使用@EriktheOutgolfer的答案的第一位生成可能的墙的列表，然后使用我的R答案的矩阵相交和矩阵求幂方法。这样，即使在大N情况下也能很好地工作。这是我的第一个果冻答案，我怀疑它可以打更多的高尔夫球。理想情况下，我也想更改第一部分，以便时间和内存需求不会随M呈指数增长。

05AB1E，42 个字节

Åœʒ23yåP}€œ€`Ùε.¥¦¨}IиI.ÆÙεøyíø‚€€üQOO_P}O

我几乎不敢发表这个文章，而且绝对可以用一种不同的方法让A LOT打高尔夫球，但是由于花了一段时间才完成，所以我决定还是张贴它，然后从这里开始打高尔夫球。挑战似乎比imo容易，但是我在这里肯定使用了错误的方法，我感到05AB1E可以完成大约25个字节。

在线尝试。注意：它不仅很长，而且效率很低，因为9x4测试用例在TIO上运行大约40秒。

说明：

Ŝ             # Get all possible ways to sum to the (first) implicit input
               #  i.e. 8 → [[1,1,1,1,1,1,1,1],[1,1,1,1,1,1,2],[1,1,1,1,1,3],[1,1,1,1,2,2],[1,1,1,1,4],[1,1,1,2,3],[1,1,1,5],[1,1,2,2,2],[1,1,2,4],[1,1,3,3],[1,1,6],[1,2,2,3],[1,2,5],[1,3,4],[1,7],[2,2,2,2],[2,2,4],[2,3,3],[2,6],[3,5],[4,4],[8]]
  ʒ23yåP}      # Only leave those consisting of 2s and/or 3s
               #  → [[2,2,2,2],[2,3,3]]
         €œ    # For each: get all permutations
           €`  # Flatten this list of lists once
             Ù # And uniquify it (leaving all possible distinct rows of bricks)
               #  → [[2,2,2,2],[3,3,2],[3,2,3],[2,3,3]]
ε    }         # For each:
 .¥            #  Get the cumulative sum
   ¦¨          #  With the leading 0 and trailing first input removed
               #   → [[2,4,6],[3,6],[3,5],[2,5]]
      Iи       # Repeat this list the second input amount of times
               #  i.e. 3 → [[2,4,6],[3,6],[3,5],[2,5],[2,4,6],[3,6],[3,5],[2,5],[2,4,6],[3,6],[3,5],[2,5]]
        I.Æ    # Get all combinations of lists the size of the second input
           Ù   # And uniquify the result (leaving all possible distinct walls)
               #  → [[[2,4,6],[3,6],[3,5]],[[2,4,6],[3,6],[2,5]],[[2,4,6],[3,6],[2,4,6]],[[2,4,6],[3,6],[3,6]],[[2,4,6],[3,5],[2,5]],[[2,4,6],[3,5],[2,4,6]],[[2,4,6],[3,5],[3,6]],[[2,4,6],[3,5],[3,5]],[[2,4,6],[2,5],[2,4,6]],[[2,4,6],[2,5],[3,6]],[[2,4,6],[2,5],[3,5]],[[2,4,6],[2,5],[2,5]],[[2,4,6],[2,4,6],[3,6]],[[2,4,6],[2,4,6],[3,5]],[[2,4,6],[2,4,6],[2,5]],[[2,4,6],[2,4,6],[2,4,6]],[[3,6],[3,5],[2,5]],[[3,6],[3,5],[2,4,6]],[[3,6],[3,5],[3,6]],[[3,6],[3,5],[3,5]],[[3,6],[2,5],[2,4,6]],[[3,6],[2,5],[3,6]],[[3,6],[2,5],[3,5]],[[3,6],[2,5],[2,5]],[[3,6],[2,4,6],[3,6]],[[3,6],[2,4,6],[3,5]],[[3,6],[2,4,6],[2,5]],[[3,6],[2,4,6],[2,4,6]],[[3,6],[3,6],[3,5]],[[3,6],[3,6],[2,5]],[[3,6],[3,6],[2,4,6]],[[3,6],[3,6],[3,6]],[[3,5],[2,5],[2,4,6]],[[3,5],[2,5],[3,6]],[[3,5],[2,5],[3,5]],[[3,5],[2,5],[2,5]],[[3,5],[2,4,6],[3,6]],[[3,5],[2,4,6],[3,5]],[[3,5],[2,4,6],[2,5]],[[3,5],[2,4,6],[2,4,6]],[[3,5],[3,6],[3,5]],[[3,5],[3,6],[2,5]],[[3,5],[3,6],[2,4,6]],[[3,5],[3,6],[3,6]],[[3,5],[3,5],[2,5]],[[3,5],[3,5],[2,4,6]],[[3,5],[3,5],[3,6]],[[3,5],[3,5],[3,5]],[[2,5],[2,4,6],[3,6]],[[2,5],[2,4,6],[3,5]],[[2,5],[2,4,6],[2,5]],[[2,5],[2,4,6],[2,4,6]],[[2,5],[3,6],[3,5]],[[2,5],[3,6],[2,5]],[[2,5],[3,6],[2,4,6]],[[2,5],[3,6],[3,6]],[[2,5],[3,5],[2,5]],[[2,5],[3,5],[2,4,6]],[[2,5],[3,5],[3,6]],[[2,5],[3,5],[3,5]],[[2,5],[2,5],[2,4,6]],[[2,5],[2,5],[3,6]],[[2,5],[2,5],[3,5]],[[2,5],[2,5],[2,5]]]
ε              # Map all walls `y` to:
 ø             #  Zip/transpose; swapping rows and columns
 yí            #  Reverse each row in a wall `y`
   ø           #  Also zip/transpose those; swapping rows and columns
 ‚             #  Pair both
  €            #  For both:
   €           #   For each column:
    ü          #    For each pair of bricks in a column:
     Q         #     Check if they are equal to each other (1 if truthy; 0 if falsey)
    O          #    Then take the sum of these checked pairs for each column
   O           #   Take the sum of that entire column
   _           #   Then check which sums are exactly 0 (1 if 0; 0 if anything else)
   P           #   And check for which walls this is only truthy by taking the product
}O             # After the map: sum the resulting list
               # (and output it implicitly as result)

木炭，89字节

Ｎθ⊞υ⟦⟧≔⟦⟧ηＦυＦ⟦²¦³⟧«≧⁺∧Ｌι§ι⁰κ¿⁼κθ⊞ηι¿‹κθ⊞υ⁺⟦κ⟧ι»≔Ｅη⟦ι⟧ζＦ⊖Ｎ«≔ζι≔⟦⟧ζＦιＦη¿¬⊙§κ⁰№λμ⊞ζ⁺⟦λ⟧κ»ＩＬζ

在线尝试！链接是详细版本的代码。可在TIO上处理大小最大为约12的矩形，但可以通过使用位旋转而不是列表交集，以2字节的成本将速度提高大约三倍。说明：

Ｎθ

输入宽度。

⊞υ⟦⟧

从无砖块开始。

≔⟦⟧η

从没有完成的行开始。

Ｆυ

循环遍历行。

Ｆ⟦²¦³⟧«

循环在砖头上。

≧⁺∧Ｌι§ι⁰κ

将砖块宽度添加到当前行宽度。

¿⁼κθ⊞ηι

如果这导致输入宽度，则将此行添加到已完成行的列表中。

¿‹κθ⊞υ⁺⟦κ⟧ι»

否则，如果它仍然小于输入宽度，则将新行添加到行列表中，从而使它在以后的迭代中被拾取。

≔Ｅη⟦ι⟧ζ

列出一排墙。

Ｆ⊖Ｎ«

比高度低一圈。

≔ζι

保存墙壁列表。

≔⟦⟧ζ

清除墙壁列表。

Ｆι

循环浏览已保存的墙列表。

Ｆη

循环完成的行。

¿¬⊙§κ⁰№λμ⊞ζ⁺⟦λ⟧κ»

如果可以将行添加到此墙，则将其添加到墙列表。

ＩＬζ

输出墙的最终列表的长度。