给定正整数n，请执行以下操作（并输出每个阶段）：

1. 从包含n副本的列表开始n。
2. 执行以下n次数：
3. 在i步骤th，逐渐减少i列表的th条目，直到到达i

因此，举例来说，如果给定的n是4，那么你下手[4,4,4,4]，然后在你的第一步[3,4,4,4]，[2,4,4,4]，[1,4,4,4]。在第二个步骤，你有[1,3,4,4]，[1,2,4,4]。第三步[1,2,3,4]。第四步什么都没有做。

所以您的输出是[[4,4,4,4],[3,4,4,4],[2,4,4,4],[1,4,4,4],[1,3,4,4],[1,2,4,4],[1,2,3,4]]。

允许使用任何合理的输入/输出格式。

有标准漏洞。这是代码高尔夫：字节数最小的答案胜出。

用于检查目的的Python实现。

果冻，9 个字节

r€⁸Œp»\QṚ

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怎么样？

r€⁸Œp»\QṚ - Link: integer, N    e.g. 4
 €        - for €ach of implicit range of N (i.e. for i in [1,2,3,...N])
  ⁸       -   with the chain's left argument, N on the right:
r         -     inclusive range (for i<=N this yields [i, i+1, ..., N]
          - ...leaving us with a list of lists like the post-fixes of [1,2,3,....,N]
          -                     e.g. [[1,2,3,4],[2,3,4],[3,4],[4]]
   Œp     - Cartesian product* of these N lists
          -                     e.g. [[1,2,3,4],[1,2,4,4],[1,3,3,4],[1,3,4,4],[1,4,3,4],[1,4,4,4],[2,2,3,4],[2,2,4,4],[2,3,3,4],[2,3,4,4],[2,4,3,4],[2,4,4,4],[3,2,3,4],[3,2,4,4],[3,3,3,4],[3,3,4,4],[3,4,3,4],[3,4,4,4],[4,2,3,4],[4,2,4,4],[4,3,3,4],[4,3,4,4],[4,4,3,4],[4,4,4,4]]
      \   - cumulative reduce with:
     »    -   maximum (vectorises)
          -                     e.g. [[1,2,3,4],[1,2,4,4],[1,3,4,4],[1,3,4,4],[1,4,4,4],[1,4,4,4],[2,4,4,4],[2,4,4,4],[2,4,4,4],[2,4,4,4],[2,4,4,4],[2,4,4,4],[3,4,4,4],[3,4,4,4],[3,4,4,4],[3,4,4,4],[3,4,4,4],[3,4,4,4],[4,4,4,4],[4,4,4,4],[4,4,4,4],[4,4,4,4],[4,4,4,4],[4,4,4,4]]
       Q  - de-duplicate        e.g. [[1,2,3,4],[1,2,4,4],[1,3,4,4],[1,4,4,4],[2,4,4,4],[3,4,4,4],[4,4,4,4]]
        Ṛ - reverse             e.g. [[4,4,4,4],[3,4,4,4],[2,4,4,4],[1,4,4,4],[1,3,4,4],[1,2,4,4],[1,2,3,4]]

*使用不同的输入可能更容易了解上面使用的笛卡尔积的情况：

the Cartesian product of [[0,1,2],[3,4],[5]]
is [[0,3,5],[0,4,5],[1,3,5],[1,4,5],[2,3,5],[2,4,5]]

R，83 82 74字节

N=rep(n<-scan(),n);while({print(N);any(K<-N>1:n)})N[x]=N[x<-which(K)[1]]-1

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代替双for循环，在while这里循环就足够了：我们找到列表大于索引的第一个索引，然后在该位置递减。

K有TRUE地方N[i]>i，which(K)回到真实的指数，我们采取先用[1]。

果冻，12字节

R‘®¦<³S©$пṚ

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JavaScript（ES6），75个字节

f=(n,a=Array(n).fill(n))=>[[...a],...a.some(v=>v>++j,j=0)?f(a[j-1]--,a):[]]

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APL + WIN，54个字节

提示屏幕输入整数

((⍴m)⍴n)-+⍀m←0⍪(-0,+\⌽⍳n-1)⊖((+/+/m),n)↑m←⊖(⍳n)∘.>⍳n←⎕

输出矩阵，每行代表每个步骤的结果，例如4：

4 4 4 4
3 4 4 4
2 4 4 4
1 4 4 4
1 3 4 4
1 2 4 4
1 2 3 4

果冻，11字节

x`’Jḟḣ1Ʋ¦ÐĿ

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怎么运行的

x`’Jḟḣ1Ʋ¦ÐĿ  Main link. Argument: n

x`           Repeat self; yield an array of n copies of n.
         ÐĿ  While the results are unique, repeatedly call the link to the left.
             Return the array of all unique results, including the initial value.
  ’     ¦      Decrement the return value at all indices specified by the chain
               in between.
       Ʋ         Combine the four links to the left into a monadic chain.
   J               Indices; yield [1, ..., n].
    ḟ              Filterfalse; remove all indices that belong to the return value.
     ḣ1            Head 1; truncate the result to length 1.

Python 3，91字节

n=int(input())
x=[n]*n;print(x)
for i in range(n):
    for j in[0]*(n-i-1):x[i]-=1;print(x)

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Java（OpenJDK 8），135字节

a->{int r[]=new int[a],i=0;java.util.Arrays x=null;x.fill(r,a);for(r[0]++;i<a;r[i++]++)for(;--r[i]>i;System.out.print(x.toString(r)));}

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说明：

int r[]=new int[a],i=0;    //Initialize array and loop counter
java.util.Arrays x=null;    //reduces the number of of “Arrays” needed from 3 to 1
x.fill(r,a);    //Sets each value in array length n to int n
for(r[0]++;i<a;r[i++]++)    //Increment everything!
  for(;--r[i]>i;    //If decremented array element is larger than element number:
     System.out.print(x.toString(r)));}    //Print the array

信用：

-8个字节感谢Jonathan Frech！

-16个字节感谢Kevin Cruijssen！

-1字节感谢Okx！

Haskell，69 67 65 63字节

递归定义：

f 0=[[]]
f a=map(:(a<$[2..a]))[a,a-1..2]++[1:map(+1)x|x<-f$a-1]

感谢Laikoni 2个字节！

PHP，153字节

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码

function f($n){
$a=array_fill(0,$n,$n);$r=json_encode($a)."\n";$p=0;while($p<$n)
{if($a[$p]!=$p+1){$a[$p]--;$r.=json_encode($a)."\n";}else{$p++;}}echo$r;}

尝试降低字节数或完成递归功能

说明

function f($n){
  $a=array_fill(0,$n,$n);          #start with $nlength array filled with $n
  $r=json_encode($a)."\n";         #pushed to the string to output
  $p=0;                            #first position
  while($p<$n){                    #on position $n ($n-1) we do nothing
    if($a[$p]!=$p+1){              #comparing the position+1 to the value
     $a[$p]--;                     #it gets decreased by 1
     $r.= json_encode($a)."\n";    #and pushed
   } else {
     $p++;                       #when position+1 = the value,
   }                               #position is changed ++
  }
   echo $r;
  }

Python 2中，80 76个字节

i=input();l=[i]*i;print l
for x in range(i):
 while l[x]>x+1:l[x]-=1;print l

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有两个print语句有点浪费，但是我现在想不出更好的方法。

Python 2，70个字节

感谢@LeakyNun
-2字节感谢@JonathanFrech -2字节

i=I=input()
l=[I]*I
exec"exec'print l;l[-i]-=1;'*max(~-i,2);i-=1;"*~-I

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Java (JDK 10), 112 bytes

n->{var s="";for(int i=1,k=n,j;i<=n;k=--k>i?k:n-++i+i)for(j=0;j++<n;)s+=(j<i?j:j>i?n:k)+(j<n?",":";");return s;}

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J, 17 15 bytes

+/\@,(#=)@i.&.-

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Explanation

+/\@,(#=)@i.&.-  Input: n
              -  Negate n
          i.     Reverse of range [0, n)
       =           Identity matrix of order n
      #            Copy each row by the reverse range
              -  Negate
    ,            Prepend n
+/\              Cumulative sum of rows

Retina, 49 bytes

.+
*
_
$`_,$= 
.{*\`_+,(_+)
$.1
0`(\b(_+),\2)_
$1

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.+
*

Convert the input to unary.

_
$`_,$= 

Create a list of n copies of i,n where i is the index of the copy.

.

Don't print anything (when the loop finishes).

{

Loop until the pattern does not change.

*\`_+,(_+)
$.1

Temporarily delete the is and convert the ns to decimal and output.

0`(\b(_+),\2)_
$1

Take the first list entry whose value exceeds its index and decrement it.

Python 3, 70 67 65 bytes

def f(n):
 k=0;a=[n]*n
 while k<n-1:print(a);k+=a[k]==k+1;a[k]-=1

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• (67) Converting to function: -3 bytes
• (65) Removing unneeded parentheses: -2 bytes

Ungolfed version:

def f(n):
    k = 0
    a = [n] * n             # create n-item list with all n's
    while k < n - 1:        # iterate through columns 0..n-1
        print(a)            # print whole list
        if a[k] == k + 1:   # move to the next column when current item reaches k+1
            k += 1
        a[k] -= 1           # decrement current item

C (clang), 131 141 bytes

i,j,k,m[99];p(){for(k=0;m[k];printf("%d ",m[k++]));puts("");}f(n){for(j=k=m[n]=0;k<n;m[k++]=n);p();for(;j<n;j++)for(i=1;i++<n-j;m[j]--,p());}

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This will work for all n upto 99. TIO truncates output. It can support arbitrarily larger n by changing size of array m as memory permits.

Following is limited to n=1..9 but is significantly shorter

C (clang), 89 92 bytes

i,j;char m[12];f(n){j=!puts(memset(m,n+48,n));for(;j<n;j++)for(i=1;i++<n-j;m[j]--,puts(m));}

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Updated: Modified to avoid dependence on static initialization

Clojure, 132 bytes

#(loop[R[(vec(repeat % %))]j(- % 2)i 0](if(> i j)R(recur(conj R(update(last R)i dec))(if(= i j)(- % 2)(dec j))(if(= i j)(inc i)i))))

I was hoping this to be shorter...

Less stateful but longer at 141 bytes:

#(apply map list(for[i(range %)](concat(repeat(nth(cons 0(reductions +(reverse(range %))))i)%)(range % i -1)(if(>(dec %)i)(repeat(inc i))))))

Python 3, 101 bytes

def f(n):
 p=print;m=[n for_ in range(n)];p(m)
 for i in range(n):
    while m[i]>1+i:m[i]-=1;p(m)

I could probably golf more with the print, but I'm away from my computer and am not entirely sure of python 2's rules on setting a variable to print. I'll update later when I get to a computer or if someone clarifies in the comments.

K (ngn/k), 34 32 bytes

{{x[y]-:1;x}\(,x#x),,/(|!x)#'!x}

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