挑战很简单；输出以下六个2D整数数组：

[[ 1, 11, 21, 31, 41, 51],
 [ 3, 13, 23, 33, 43, 53],
 [ 5, 15, 25, 35, 45, 55],
 [ 7, 17, 27, 37, 47, 57],
 [ 9, 19, 29, 39, 49, 59]]

[[ 2, 11, 22, 31, 42, 51],
 [ 3, 14, 23, 34, 43, 54],
 [ 6, 15, 26, 35, 46, 55],
 [ 7, 18, 27, 38, 47, 58],
 [10, 19, 30, 39, 50, 59]]

[[ 4, 13, 22, 31, 44, 53],
 [ 5, 14, 23, 36, 45, 54],
 [ 6, 15, 28, 37, 46, 55],
 [ 7, 20, 29, 38, 47, 60],
 [12, 21, 30, 39, 52]]

[[ 8, 13, 26, 31, 44, 57],
 [ 9, 14, 27, 40, 45, 58],
 [10, 15, 28, 41, 46, 59],
 [11, 24, 29, 42, 47, 60],
 [12, 25, 30, 43, 56]]

[[16, 21, 26, 31, 52, 57],
 [17, 22, 27, 48, 53, 58],
 [18, 23, 28, 49, 54, 59],
 [19, 24, 29, 50, 55, 60],
 [20, 25, 30, 51, 56]]

[[32, 37, 42, 47, 52, 57],
 [33, 38, 43, 48, 53, 58],
 [34, 39, 44, 49, 54, 59],
 [35, 40, 45, 50, 55, 60],
 [36, 41, 46, 51, 56]]

这些2D整数数组是什么？这些是魔术中使用的包含这些数字的数字卡：

魔术使某人想出一个在[1，60]范围内的数字，并给执行魔术的人提供所有包含该数字的牌。然后，执行魔术技巧的人可以将给定卡片的左上角数字相加（所有2的幂），以得出该人正在考虑的数字。有关此工作原理的其他说明，请参见此处。

挑战规则：

• 您可以以任何合理的格式输出六个2D整数数组。可以打印定界符；可以是包含六个2D整数数组的3D整数数组；可以是行的字符串列表；等等
• 您可以使用范围[-60, -1]或字符中的负值来填充最后四张卡的右下角位置，'*'而不必将其遗漏以使2D整数数组成为矩形矩阵（不，您不能用0或来填充它们-类似于null/的整数undefined，但*因为实际的卡片中也使用了星号）。
• 矩阵中数字的顺序是强制性的。虽然这不要紧，物理魔术，我看到这个挑战主要表现为基质 - 柯尔莫哥洛夫复杂性之一，因此订单的限制。
  矩阵本身在输出列表中的顺序可以是任意顺序，因为从左上角的卡片可以清楚地看出哪个矩阵是哪个。

一般规则：

• 这是代码高尔夫球，因此最短答案以字节为单位。
  不要让代码高尔夫球语言阻止您发布使用非代码高尔夫球语言的答案。尝试针对“任何”编程语言提出尽可能短的答案。
• 标准规则适用于具有默认I / O规则的答案，因此允许您使用STDIN / STDOUT，具有适当参数的函数/方法以及返回类型的完整程序。你的来电。
• 默认漏洞是禁止的。
• 如果可能的话，请添加一个带有测试代码的链接（即TIO）。
• 另外，强烈建议为您的答案添加说明。

MATL，12 11字节

-1字节感谢大师自己：)

60:B"@fQ6eq

说明：

60:           % create a vector [1,2,3,...,60]
   B          % convert to binary matrix (each row corresponds to one number)
    "         % loop over the columns and execute following commands:
     @f       % "find" all the nonzero entries and list their indices
       Q      % increment everything
        6e    % reshape and pad with a zero at the end
          q   % decrement (reverts the increment and makes a -1 out of the zero
              % close loop (]) implicitly
              % display the entries implicitly

在线尝试！

Perl 6的，63 46个字节

say grep(*+&2**$_,^61)[$_,*+5...*for ^5]for ^6

在线尝试！

输出为多行上的2D数组，如有必要，将每个数组的最后一个数组切掉。

Python 2，76个字节

r=range;print[[[i for i in r(61)if i&2**k][j::5]for j in r(5)]for k in r(6)]

在线尝试！

这里的方法是创建所有可能数字的列表，r(61)然后将其缩减为卡的数字列表i&2**k。

然后，通过使用列表切片，将一维数字列表重新排列为正确的6x5卡尺寸[card nums][j::5]for j in r(5)。

然后，仅对6张卡片重复此生成器for k in r(6)。

虽然找不到少于76个字节的解决方案，但这里还有两个也是76个字节的解决方案：

r=range;print[[[i for i in r(61)if i&1<<k][j::5]for j in r(5)]for k in r(6)]

在线尝试！

下一个灵感来自乔纳森·艾伦（Jonathan Allan）。

k=32
while k:print[[i for i in range(61)if i&k][j::5]for j in range(5)];k/=2

在线尝试！

任何意见，不胜感激。

木炭，26字节

Ｅ⁶Ｅ⁵⪫Ｅ⁶§⁺§⪪Φ⁶¹＆πＸ²ι⁵ν⟦*⟧λ 

在线尝试！链接是详细版本的代码。我尝试直接计算条目，但是在调整*右下角的之前已经是27个字节。输出每行之间用空格和卡之间的空白行连接的行。说明：

Ｅ⁶                          Loop over 6 cards
  Ｅ⁵                        Loop over 5 rows
     Ｅ⁶                     Loop over 6 columns
           Φ⁶¹              Filter over 0..60 where
               π            Current value
              ＆             Bitwise And
                 ²          Literal 2
                Ｘ           Raised to power
                  ι         Card index
          ⪪        ⁵        Split into groups of 5
         §          ν       Indexed by column
        ⁺                   Concatenated with
                      *     Literal string `*`
                     ⟦ ⟧    Wrapped in an array
       §                λ   Indexed by row
    ⪫                       Joined with spaces
                            Implicitly print

05AB1E，16个字节

60L2вíƶ0ζε0K5ô®ζ

在线尝试！

说明

60L                 # push [1 ... 60]
   2в               # convert each to a list of binary digits
     í              # reverse each
      ƶ             # multiply each by its 1-based index
       0ζ           # transpose with 0 as filler
         ε          # apply to each list
          0K        # remove zeroes
            5ô      # split into groups of 5
              ®ζ    # zip using -1 as filler

05AB1E，17个字节

6F60ÝNoôāÈϘ5ô®ζ,

在线尝试！

说明

6F                  # for N in [0 ... 5] do
  60Ý               # push [0 ... 60]
     Noô            # split into groups of 2^N numbers
        āÈÏ         # keep every other group
           ˜        # flatten
            5ô      # split into groups of 5
              ®ζ    # transpose with -1 as filler
                ,   # print

稻壳，13个字节

ṠMöTC5Wnünḣ60

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说明

          ḣ60  Range [1..60]
        ü      Uniquify using equality predicate
         n     bitwise AND: [1,2,4,8,16,32]
 M             For each number x in this list,
Ṡ     W        take the indices of elements of [1..60]
       n       that have nonzero bitwise AND with x,
    C5         cut that list into chunks of length 5
  öT           and transpose it.

Python 2中，82个 80 78 74字节

i=1 
exec"print zip(*zip(*[(n for n in range(61)+[-1]if n&i)]*5));i*=2;"*6

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_{-4个字节，感谢Jonathan Allan}

Japt，14个字节

6Æ60õ f&2pX)ó5

试试吧

6Æ              Create a range from 0 to 5 (inclusive) and map each X into
  60õ             Elements in the range [1..60]
      f             Where
       &2pX)          The number bitwise AND with X is not 0
  ó5              Split into 5 arrays, where each array contains every 5th element

-Q flag is just for formatting purposes

JavaScript（ES6）， 90  88字节

_=>[1,2,4,8,16,32].map(n=>(g=i=>i<60?g(++i,i&n?m[y%5]=[...m[y++%5]||[],i]:0):m)(y=m=[]))

在线尝试！

已评论

_ =>                        // anonymous function taking no argument
  [1, 2, 4, 8, 16, 32]      // list of powers of 2, from 2**0 to 2**5
  .map(n =>                 // for each entry n in this list:
    ( g = i =>              //   g = recursive function taking a counter i
      i < 60 ?              //     if i is less than 60:
        g(                  //       recursive call:
          ++i,              //         increment i
          i & n ?           //         if a bitwise AND between i and n is non-zero:
            m[y % 5] =      //           update m[y % 5]:
            [ ...m[y++ % 5] //             prepend all previous values; increment y
              || [],        //             or prepend nothing if it was undefined so far
              i             //             append i
            ]               //           end of update
          :                 //         else:
            0               //           do nothing
        )                   //       end of recursive call
      :                     //     else:
        m                   //       return m[]
    )(y = m = [])           //   initial call to g with i = y = m = []
                            //   (i and y being coerced to 0)
  )                         // end of map()

Python 2，73个字节

灵感来自TFeld和The Matt。

k=32
while k:print zip(*zip(*[(i for i in range(61)+[-1]if i&k)]*5));k/=2

在线尝试！

C（gcc），95字节

i,j,k;f(int o[][5][6]){for(i=6;i;)for(o[--i][4][5]=j=k=-1;j<60;)++j&1<<i?o[i][++k%5][k/5]=j:0;}

在线尝试！

以o中的3D int数组形式返回矩阵。

最后4个矩阵的最后一个值为-1。

感谢Kevin Cruijssen，节省了2个字节。

感谢Arnauld，节省了7 8个字节。

CJam（18字节）

6{61{2A#&},5/zp}fA

在线演示。这是一个输出到stdout的完整程序。

解剖

6{             }fA    # for A = 0 to 5
  61{2A#&},           #   filter [0,61) by whether bit 2^A is set
           5/z        #   break into chunks of 5 and transpose to get 5 lists
              p       #   print

果冻，13 个字节

60&ƇⱮs€5LÐṂZ€

一个尼拉度链接，它产生一个（6）个整数列表的列表。（它使用默认选项不输出*或为负数。）

在线尝试！

怎么样？

每个矩阵按列优先顺序包含最多至 $60$ 与左上（最小）数字共享单个置位。

该程序首先使所有 $60$ 可能的数字顺序列表 $[1,60]$与它们的索引号共享任何设置位。然后将其分成多个块$5$并仅保留长度最小的索引-索引只有一个置位的索引（因此也是其最小值）。最后，它将每个转置以将它们置于列优先顺序中。

60&ƇⱮs€5LÐṂZ€ - Link: no arguments
60            - set the left argument to 60
    Ɱ         - map across ([1..60]) with:  (i.e. [f(60,x) for x in [1..60]])
   Ƈ          -   filter keep if:  (N.B. 0 is falsey, while non-zeros are truthy)
  &           -     bitwise AND
      €       - for each:
     s 5      -   split into chunks of five
         ÐṂ   - keep those with minimal:
        L     -   length
           Z€ - transpose each

许多15s都没有实现“分成五等分时的最小长度”技巧：

5Ż2*Ɱ60&ƇⱮs€5Z€
6µ’2*60&Ƈ)s€5Z€
60&ƇⱮ`LÞḣ6s€5Z€

...并且在尝试寻找更短的时间时，我又获得了13个，而根本不需要花招：

60B€Uz0Ts5ZƊ€

Wolfram语言（Mathematica），88字节

Transpose@Partition[#~Append~-1,5]&/@Last@Reap[Sow[,NumberExpand[,2]]~Do~{,60},Except@0]

Wolfram语言（Mathematica），99字节

Transpose@Partition[#~FromDigits~2&/@Last@GatherBy[{0,1}~Tuples~6,#[[-k]]&],5]~Table~{k,6}/. 61->-1

在线尝试！

果冻，13个字节

60B€Uz0µTs5Z)

在线尝试！

松散地基于 faker的MATL答案。一个尼拉度链接，根据需要输出列表列表。

R，73字节

`!`=as.raw;lapply(0:5,function(i)matrix(c((a=1:60)[(!a&!2^i)>0],-1),5,6))

我不确定我是否满足订购要求，因为默认情况下R会按列填充矩阵，因此实际出现在卡上的顺序与R中分配矩阵的方式相同。

在线尝试！

T-SQL，（1,168 1,139字节）

我只是想知道我能做到。

优化版本

 WITH g AS(SELECT 1 AS n UNION ALL SELECT n+1 FROM g WHERE n+1<61),B as(SELECT cast(cast(n&32 as bit)as CHAR(1))+cast(cast(n&16 as bit)as CHAR(1))+cast(cast(n&8 as bit)as CHAR(1))+cast(cast(n&4 as bit)as CHAR(1))+cast(cast(n&2 as bit)as CHAR(1))+cast(cast(n&1 as bit)as CHAR(1))as b FROM g),P as(SELECT * from (values(1), (2), (4), (8), (16), (32)) as Q(p)),S as(select distinct p,p+(substring(b,6,1)*1)*(case when p=1 then 0 else 1 end)+(substring(b,5,1)*2)*(case when p=2 then 0 else 1 end)+(substring(b,4,1)*4)*(case when p=4 then 0 else 1 end)+(substring(b,3,1)*8)*(case when p=8 then 0 else 1 end)+(substring(b,2,1)*16)*(case when p=16 then 0 else 1 end)+(substring(b,1,1)*32)*(case when p=32 then 0 else 1 end)as e from P cross apply B),D as(select * from S where e>=p and e<61),R as(select p,(row_number()over(partition by p order by cast(e as int)))%5 as r,e from D),H as(select k.p,'['+stuff((select','+cast(l.e as varchar)from R l where l.p=k.p and l.r=k.r for xml path('')),1,1,'')+']'as s from R k group by k.p,k.r)select stuff((select','+cast(x.s as varchar)from H x where x.p=z.p for xml path('')),1,1,'')from H z group by z.p

在线演示

在线尝试！

详细版本-带注释的SQL注释

WITH gen -- numbers 1 to 60
AS (
    SELECT 1 AS num
    UNION ALL
    SELECT num+1 FROM gen WHERE num+1<=60
),
BINARIES -- string representations of binaries 000001 through 111111
as (
SELECT 
    +cast( cast(num & 32 as bit) as CHAR(1))
    +cast( cast(num & 16 as bit)  as CHAR(1))
    +cast( cast(num & 8 as bit)  as CHAR(1))
    +cast( cast(num & 4 as bit)  as CHAR(1))
    +cast( cast(num & 2 as bit)   as CHAR(1))
    +cast(cast(num & 1 as bit)  as CHAR(1)) as binry FROM gen
),
POWERS -- first 6 powers of 2
as (
SELECT * from (values(1), (2), (4), (8), (16), (32)) as Q(powr)
),
SETELEMENTS -- cross apply the six powers of 2 against the binaries
-- returns 2 cols. col 1 = the power of 2 in question.
-- col 2 is calculated as that power of 2 plus the sum of each power of 2 other than the current row's power value, 
-- but only where a given power of 2 is switched "on" in the binary string, 
-- ie. where the first digit in the string represents 32, the second represents 16 and so on. 
-- That is, if the binary is 100100 then the number will be 
-- the sum of (32 x 1) + (16 x 0) + (8 x 0) + (4 x 1) + (2 x 0) + (1 x 0) 
-- but if the current row's power is 32 or 4, then just that number (32 or 4) is excluded from the sum.
-- rows are distinct.
as (
select distinct powr,
powr+
 (substring(binry,6,1) * 1) * (case when powr = 1 then 0 else 1 end)
 +(substring(binry,5,1) * 2) * (case when powr = 2 then 0 else 1 end)
 +(substring(binry,4,1) * 4) * (case when powr = 4 then 0 else 1 end)
 +(substring(binry,3,1) * 8) * (case when powr = 8 then 0 else 1 end)
 +(substring(binry,2,1) * 16) * (case when powr = 16 then 0 else 1 end)
 +(substring(binry,1,1) * 32) * (case when powr = 32 then 0 else 1 end) as elt
from POWERS cross apply BINARIES
),
DISTINCTELEMENTS -- purge calculated numbers smaller than the power of 2 or greater than 60
as (
select * from SETELEMENTS where elt >= powr and elt < 61
)--,
,
ROWNUMBERED -- for each power, number the rows repeatedly from 0 through 5, then back to 0 through 5 again, etc
as (
select powr, (row_number() over (partition by powr order by cast(elt as int)))%5 as r, elt  from DISTINCTELEMENTS
),
GROUPEDSETS -- for each row number, within each power, aggregate the numbers as a comma-delimited list and wrap in square brackets - the inner arrays
as (
select r1.powr, '['+stuff((select ',' + cast(r2.elt as varchar) from ROWNUMBERED r2 where r2.powr = r1.powr and r2.r = r1.r for xml path('')),1,1,'')+']' as s
from ROWNUMBERED r1
group by r1.powr,r1.r
)
select -- now aggregate all the inner arrays per power
stuff((select ',' + cast(g2.s as varchar) from GROUPEDSETS g2 where g2.powr = g1.powr for xml path('')),1,1,'')
from GROUPEDSETS g1
group by g1.powr

瞧！

Note 1: Some of the logic pertains to rendering square brackets and commas.

Note 2: Newer versions of SQLServer have more compact approaches to creating comma-delimited lists. (This was created on SQL Server 2016.)

Note 3: Arrays for a given card are not sorted (which is ok per the spec). Numbers within an array are correctly sorted. In this case, each "card" of the question, renders its arrays on a separate row in the results.

Shorter to hard-code arrays?

Yes.

Byte me.

C# (Visual C# Interactive Compiler), 112 bytes

_=>" ".Select(x=>Enumerable.Range(1,60).Where(l=>(l&x)>0).Select((a,b)=>new{a,b}).GroupBy(i=>i.b%5,i=>i.a))

Try it online!

Red, 108 107 bytes

n: 32 until[b: collect[repeat k 60[if n and k = n[keep k]]]loop 5[print
extract b 5 b: next b]1 > n: n / 2]

Try it online!

APL+WIN, 65 62 bytes

v←∊+\¨n,¨29⍴¨1↓¨(n⍴¨1),¨1+n←2*0,⍳5⋄((v=61)/v)←¯1⋄1 3 2⍉6 6 5⍴v

Try it online! Courtesy of Dyalog Classic

MATLAB, 155 bytes

cellfun(@disp,cellfun(@(x)x-repmat(62,5,6).*(x>60),cellfun(@(x)reshape(find(x,30),[5 6]),mat2cell(dec2bin(1:62)-48,62,ones(1,6)),'Uniform',0),'Uniform',0))

This could be shorter as multiple lines but I wanted to do it in one line of code.

05AB1E, 14 bytes

žOε60LDNo&ĀÏ5ι

Try it online!