Những thách thức nhỏ bị hạn chế Cướp chủ đề


15

Cops chủ đề

Nhiệm vụ của bạn là những tên cướp là tìm các giải pháp cảnh sát và viết một chương trình bằng ngôn ngữ được cung cấp để tính toán thuật ngữ thứ n của chuỗi chỉ sử dụng các byte trong tập hợp được cung cấp.

Mục tiêu sẽ là crack càng nhiều câu trả lời của cảnh sát càng tốt. Với mỗi vết nứt trao cho bạn một điểm duy nhất.

Các vết nứt không cần phải là giải pháp dự định của cảnh sát miễn là chúng hoạt động.


1
Một bảng xếp hạng cướp sẽ là tốt đẹp.
Ông Xcoder

@ Mr.Xcoder Nếu bạn muốn biến một người thành khách của mình, tôi không làm JavaScript vì vậy tôi sẽ không tự mình thêm một người.
Đăng Rock Garf Hunter

Câu trả lời:


13

Haskell, xnor

s=show s
ss=show[[s]]
w h=[ss!!h]<show[h]

Hãy thử trực tuyến!

Dòng đầu tiên định nghĩa slà điểm cố định của showhàm, là chuỗi vô hạn

"\"\\\"\\\\\\\"\\\\\\\\\\\\\\\"\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\...

Nó có "s tại các chỉ số 0, 2, 6, 14, 30 Sức mạnh của hai, trừ hai.

Dòng thứ hai định nghĩa sslà chuỗi

[["\"\\\"\\\\\\\"\\\\\\\\\\\\\\\"\\\\\\\\\\\\\\\\\\\\\\\\\\\\\...

Nó không có các \chỉ số 0, 1, 2, 4, 8, 16, 32 Quyền hạn của hai, bỏ qua 0.

Chúng tôi đang ở may mắn: trong ASCII, "< [< \, vì vậy chúng tôi có thể viết một chức năng chỉ cho backslashes ít hơn trong chuỗi này, và chúng tôi đã hoàn tất!

w hlà câu trả lời của chúng tôi: nó kiểm tra xem hphần tử thứ của ssít hơn dấu gạch chéo ngược. Chà, chúng ta không thể xây dựng một chuỗi chỉ chứa dấu gạch chéo ngược, vì vậy chúng ta xây dựng một chuỗi khác luôn lớn hơn "[", cụ thể là show[h].



4

JavaScript (ES6), Arnauld

$=>eval(`${$}${atob(`ICo=`)}${$}`)

Đầu ra 0, 1, 4, 9, 16, 25, ....

Đặt tên đầu vào $làm cho nó trông phức tạp hơn một chút so với nó. atob('ICo=')đánh giá *với không gian hàng đầu (tìm thấy bằng tay). Điều này làm cho ${$}${atob(`ICo=`)}${$}n *nnơi nđầu vào. evaling cho hình vuông.


1
Làm tốt! (Giải pháp dự định của tôi ngắn hơn một chút, nhưng đó là ý tưởng.)
Arnauld

@Arnauld Cảm ơn! Bây giờ tôi tò mò :)
LarsW

1
Tôi đã có a=>eval(`a${atob`ICo`}a`). Tất nhiên, bất kỳ chữ cái nào khác có thể được sử dụng làm biến đầu vào. (Ngay cả khi bạn muốn sử dụng $, bạn không cần phải làm ${$}.)
Arnauld

@Arnauld Oh wow, ${$}quả thực là khá ngu ngốc.
LarsW

3

Con trăn 2, que

Rất tối ưu mặc dù. Mã số:

a=sum((()or()))
s=g=int(not(a))
for m in range(input()):a,s,g=s,g,sum((s,g))
print(a)

Hãy thử trực tuyến!


Tôi đã rất gần, không tìm ra andnotvới ()làm 10. :(
Stephen

tốt, nhưng không phải là giải pháp dự định mặc dù
Rod

Tôi đã thông báo importlà có sẵn, cũng nhưfrom
Stephen

@Rod Đã sửa lỗi d, nhưng có cần thiết phải sử dụng tất cả các byte ít nhất một lần không?
Ad Nam

Bạn không phải sử dụng tất cả, vì vậy câu trả lời này có giá trị một lần nữa c :, nếu muốn thử thách khác, có phiên bản mới không cho phépsum
Rod

3

Haskell, Christian Sievers

one=product(map(pred)mempty)
zero=pred(one)
add(a)(b)|a==zero=b|True=add(pred(a))(addone(b))
addone(x)=ao(x)(enumFrom(x))
ao(x)(y)|x==pred(head(y))=head(y)|True=ao(x)(drop(product(map(pred)mempty))y)
t(x)|x==zero=zero|x==one=one|True=add(t(pred(x)))(t(pred(pred(x))))

Hãy thử trực tuyến!

Lưu ý: product(map(pred)mempty)trong dòng 5 khác với dòng 1, bởi vì dòng trước là loại Inttrong khi dòng sau là loại Integer.


Ôi mempty! Lần sau tôi nói Haskell 98... Bạn có thể làm mà không có y? Nhưng sử dụng productlà ý tưởng chính, vì vậy tôi nghĩ rằng tôi không bắt đầu một phiên bản mới của thử thách này. Và tôi ước có (hoặc tôi có thể đặt) giới hạn thời gian. BTW, có aovẻ phức tạp không cần thiết.
Christian Sievers

@ChristianSievers: vâng, aolà thừa, bởi vì addone(x)=head(drop(product(map(pred)mempty))(enumFrom(x))). Và one(x)=product(enumFromTo(x)(pred(x)))đâu xlà số đầu vào, nhưng nó không hoạt động cho việc 1sử dụng với drop.
nimi


3

Haskell, Christian Sievers

head.maybe[]id.head[head[flip lookup.zip[head[pred.pred.pred.floor]pi..]].map pure][head[pred.floor]pi..]

Try it online!


Wow, so many functions in the Prelude! I was beginning to believe this might stay uncracked... BTW, my solution is slightly shorter
Christian Sievers

@ChristianSievers: I've found your challenge 5 days after you've posted it and it took some time to get the all the head in the right place. Like all your other challenges, it was quite some fun.
nimi


2

R, Jarko

I don't know R at all, so this is just a guess.

identity

Try it online!


You got it! Maybe that was too easy...
JAD

1
@JarkoDubbledam, if a guy who never touched R before this got it the, yeah, it was definitely too easy! :p
Shaggy

Eh, its just a matter of finding the right function. Now if only R didn't give their functions a name related to what theyre doing....
JAD

2

R, Jarko

f=function(n)`if`(n,n*f(n-n**(n-n)),n**n)

Try it online!

Definitely took me a good five minutes to try and figure out how to get 1 from the letters, but then I remembered that ** is ^ so that worked out nicely! (and 0^0=1 in R)


Well whaddaya know, **=^. Didn't know that.
JAD

1
I intended n==n btw.
JAD

f=function(n,c=n==n)'if'(n,f(n-(n==n),c*n),c*(n==n))
JAD

@JarkoDubbeldam well nice to know there are two solutions here
Giuseppe

1
As an R noob, I did the reverse. Also I was stumped because I didn't know about backtick notation:f=function(n)'if'(n-n==n,n**n,n*f(n-(n**(n-n))))
geokavel


2

C (gcc), dj0wns

??=import<stdlib.h>
int main(int a,char**f)<%int h=""==" ",o=""=="",c=h,p=o,m=(int)atof(*(f+o));for(;p<=m;p++)c+=m%p==h;printf("%i",c);%>

Try it online! Digraphs, digraphs everywhere!


z is not in the bytespace but you have plenty of other letters you can use. Well done!
dj0wns

@dj0wns Ah! True enough, thanks
Conor O'Brien

2

Ruby, Value Ink

With -n flag,

p$./$$

This is my guess for the intended solution. Ignores the input and just outputs the floor of 1/the process id. Since the process id generally can't be 0 or 1, this should always be 0.


Yep, there it is :D
Value Ink




2

Haskell, Christian Sievers

u(n)|negate(n)==(n)=negate(pred(n))|True=head(drop(length(enumFrom()))(enumFrom(n)))
g(a)(x)|null(x)=a|True=g(u(a))(drop(length(enumFrom()))x)
p(a)(b)=pred(g(rem(b)(b))(enumFromTo(negate(b))(a)))
a(n)|negate(n)==n||u(negate(pred(n)))==n=n|True=p(a(pred(pred(n))))(a(pred(n)))

Try it online!

Finally fixed it for Integers rather than Ints. Very slow.


The task challenge demands that "Answers must be capable of calculating every term in the b-files of the sequence", and the b-file for A000045 goes up to 2000th Fibonacci number, which is much more than maxBound::Int, so I think this is not correct.
Christian Sievers

Good point, didn't think of that.
H.PWiz

Well done! Now, can you do it without g? - BTW, you could have used u(n)=negate(pred(negate(n)))
Christian Sievers

I knew there was a better way to write u, I wish I thought of that. I don't know how to do it without g, I found it really hard to define addition for Integers with your byteset.
H.PWiz


2

R, Jarko, again

cat(diag(diag(('i'%in%'i'):(x=scan()),x%x%x)),sep='')

This differs from Jarko Dubbeldam's intended solution, but the general idea is the same: to generate 1:n repeated 1:n times, it leverages diag in two different ways:

  • diag(matrix) returns the diagonal of a matrix.
  • diag(vector, nrow) generates an nrowxnrow matrix with vector along the diagonal, recycling as necessary.

%x% is the Kronecker matrix product which when applied to two numbers returns their usual product.

Finally, cat prints the diagonal out with separator '' which results in the sequence.

Try it online!


Đó là (gần như) nó, nhưng đủ gần để tính nó như dự định :)
JAD

2

C, Yimin Rong

long long t(long long n, long o){
  return o ? t(n,--o)-(-n) : o;
}
long long r(long n, long o){
  return --o ? t(r(n,o),n) : n;
}
long long g(long n){
  return n ? r(n,n) : -(--n);
}

tlà phép nhân, rlà lũy thừa, glà hàm thực. Các dòng mới được thêm vào để dễ đọc, chúng không cần thiết.



2

R bởi Jarko Dubbeldam

Mất một thời gian kể từ khi tôi không biết R và tôi đã không quản lý để sử dụng thai lần:

cos(pi*(t=scan())/2)/abs(cos(pi*t/2))

Hãy thử trực tuyến!

Giải trình

Nó sử dụng thực tế cos(t*pi/2)có một khoảng thời gian 4trong tvà chuỗi bắt đầu bằng: [1, 6.123234e-17, -1, -1.83697e-16]. Như bạn có thể thấy các dấu hiệu là chính xác, vì vậy chúng ta chỉ cần bình thường hóa các giá trị nhưng vẫn giữ dấu hiệu - đó là những gì x/abs(x)để làm x != 0.

Kiểm tra nguồn!


bây giờ tôi không hiểu làm thế nào R liên kết các biến nữa. Ngoài ra, thật buồn cười là cách chính xác của dấu phẩy động làm cho điều này khả thi!
Giuseppe

In C an assignment returns the value that gets assigned, so I guess that's what R is doing here too.
ბიმო

1
@Giuseppe This works, unless you're in a function call. cos(t=scan()) wouldn't work because R can't distinguish between = for assignment or for argument specification. In braces however, this is not a problem: cos((t=scan())), but this is longer than cos(t<-scan()), so not used when golfing.
JAD

Cũng là một công việc tốt, chính xác là những gì tôi đã nghĩ: D
JAD

@JarkoDubbeldam cuộc gọi chức năng là điều gây rối cho tôi. CSONG mặc dù đã cố gắng hết sức nhưng tôi không thể giải quyết được vì tôi rất chắc chắn rằng đó *là cá trích đỏ và cospithực sự tính toán chính xác cos(pi/2)=0... và điều này sẽ được lập chỉ mục 1 sinthay vìcos
Giuseppe

2

CPython 3.6, bởi wizzwizz4 (A000002)

t=int().__sub__(int()==int())
t=t.__sub__(int().__sub__(t));t=t.__sub__(int().__sub__(t));t=t.__sub__(int().__sub__(t));t=t.__sub__(int().__sub__(t));t=t.__sub__(int().__sub__(t));
t=int().__sub__(t).__sub__(int()==int())
d=__builtins__.__dict__;l=list(d);
c=d[l[t]]((d,d,d,d,d,d,d,d,d,d,d,d,d))
_=d[l[t]]((d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d))
dd=d[l[_]].__dict__
ll=list(dd)
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d,d,d,d))),d[l[c]](d[l[t]]((d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d))),d[l[c]](d[l[t]]((d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d))),d[l[c]](d[l[t]]((d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d))),d[l[c]](d[l[t]]((d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d))),d[l[c]](d[l[t]]((d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d))),d[l[c]](d[l[t]]((d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d))),d[l[c]](d[l[t]]((d,d,d,d,d,d,d,d,d,d))),d[l[c]](d[l[t]]((d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d))),d[l[c]](d[l[t]]((d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d))),d[l[c]](d[l[t]]((d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d))),d[l[c]](d[l[t]]((d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d))),d[l[c]](d[l[t]]((d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d))),d[l[c]](d[l[t]]((d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d))),d[l[c]](d[l[t]]((d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d))),d[l[c]](d[l[t]]((d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d))),d[l[c]](d[l[t]]((d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d))),d[l[c]](d[l[t]]((d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d))),d[l[c]](d[l[t]]((d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d))),d[l[c]](d[l[t]]((d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d))),d[l[c]](d[l[t]]((d,d,d,d,d,d,d,d,d,d))),d[l[c]](d[l[t]]((d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d))),d[l[c]](d[l[t]]((d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d))),d[l[c]](d[l[t]]((d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d))),d[l[c]](d[l[t]]((d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d))),d[l[c]](d[l[t]]((d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d))),d[l[c]](d[l[t]]((d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d))),d[l[c]](d[l[t]]((d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d))),d[l[c]](d[l[t]]((d,d,d,d,d,d,d,d,d,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Tôi đã tự viết các dòng cho đến khi ll=list(dd), nhưng phần còn lại của mã được tạo bởi chương trình python này:

def construct_number(n):
    return "d[l[t]](("+",".join(("d",)*n)+"))"
def construct_character(n):
    return "d[l[c]]("+construct_number(ord(n))+")"
def construct_string(n):
    return "dd[ll[%s]](d[l[_]](),(%s))"%(construct_number(24),",".join(map(construct_character, n)))
def construct_code(code):
    return "\nd[l[%s]](%s)"%(construct_number(19),construct_string(code))
code = construct_code("""def k(n):
 v=1;s=[1,1,2];c=1;r=1
 for _ in range(n):
  for _ in range(s[c]):
   if r>=len(s):s.append(v)
   r+=1
  c+=1;v=(v%2)+1
 return s[n]
print(k(int(input())))""")

Mã được tạo gần như chắc chắn có thể được đánh golf tốt hơn, nhưng điều đó sẽ làm cho mã máy phát phức tạp hơn.


Chào! Bạn đã tự động hóa vấn đề. :-p Tôi sẽ đánh dấu nó sau khi tôi đã thử nghiệm.
wizzwizz4

Có một số lý do tôi không thể sử dụng tự động hóa?
pppery

Không có lý do nào cả. Tôi cũng làm như vậy - mặc dù người chơi golf tôi cảm thấy rằng giải pháp của tôi có phần táo bạo và kém thanh lịch hơn nhiều so với của bạn. Một sự kết hợp của cả hai sẽ thực sự golf.
wizzwizz4







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