按字符数排序的最短代码，从给出的数字卡列表中输出最佳案例的二十一点手。
输入是由空格分隔的1到10（含）之间的数字列表。
输出将是从该卡片列表中形成的最佳二十一点手 - 最接近的可用组合，通过所有卡值的总和达到21而不超过它。卡可以删除但不能添加。
如果需要移除两张或更多张卡以支持一张卡以获得相同的结果（移除5或4,1以获得21），则移除最少的卡。
如果要移除相同数量的牌（删除1,4或3,2），则将删除具有最小值的组（在前面的示例中，1,4将在min(min(3,2), min(1,4))所属的情况下被删除那对）。如果是重复卡，则应删除第一次遭遇。
输出将形成为钻石卡，保留输入顺序：
__________ __________ | || | | || /\ | | || \/ | | /\ || | | \/ || | | || /\ | | || \/ | |__________||__________| __________ __________ | || | | /\ || /\ /\ | | \/ || \/ \/ | | /\ || | | \/ || | | /\ || /\ /\ | | \/ || \/ \/ | |__________||__________| __________ __________ | || | | /\ /\ || /\ /\ | | \/ \/ || \/ \/ | | /\ || /\ /\ | | \/ || \/ \/ | | /\ /\ || /\ /\ | | \/ \/ || \/ \/ | |__________||__________| __________ __________ | || | | /\ /\ || /\ /\ | | \/ /\ \/ || \/ /\ \/ | | /\ \/ /\ || /\ \/ /\ | | \/ \/ || \/ /\ \/ | | /\ /\ || /\ \/ /\ | | \/ \/ || \/ \/ | |__________||__________| __________ __________ | /\ /\ || /\ /\ | | \/ \/ || \/ /\ \/ | | /\ /\ || /\ \/ /\ | | \/ /\ \/ || \/ \/ | | /\ \/ /\ || /\ /\ | | \/ \/ || \/ /\ \/ | | /\ /\ || /\ \/ /\ | |_\/____\/_||_\/____\/_|
Input:
1 5 7 8
Output:
__________ __________ __________ __________
| || || || |
| || /\ /\ || /\ /\ || /\ /\ |
| || \/ \/ || \/ /\ \/ || \/ /\ \/ |
| /\ || /\ || /\ \/ /\ || /\ \/ /\ |
| \/ || \/ || \/ \/ || \/ /\ \/ |
| || /\ /\ || /\ /\ || /\ \/ /\ |
| || \/ \/ || \/ \/ || \/ \/ |
|__________||__________||__________||__________|
Input:
10 3 4 2 6
Output:
__________ __________ __________ __________
| /\ /\ || || || |
| \/ /\ \/ || /\ || /\ || /\ /\ |
| /\ \/ /\ || \/ || \/ || \/ \/ |
| \/ \/ || /\ || || /\ /\ |
| /\ /\ || \/ || || \/ \/ |
| \/ /\ \/ || /\ || /\ || /\ /\ |
| /\ \/ /\ || \/ || \/ || \/ \/ |
|_\/____\/_||__________||__________||__________|
Input
5 10 5 2 3
Output:
__________ __________ __________ __________
| /\ /\ || || || |
| \/ /\ \/ || /\ /\ || /\ || /\ |
| /\ \/ /\ || \/ \/ || \/ || \/ |
| \/ \/ || /\ || || /\ |
| /\ /\ || \/ || || \/ |
| \/ /\ \/ || /\ /\ || /\ || /\ |
| /\ \/ /\ || \/ \/ || \/ || \/ |
|_\/____\/_||__________||__________||__________|
代码计数包括输入/输出（即完整程序）。
答案 0 :(得分：11)
4(1!:2)~,&(10{a.)"1>,.&.>/4 :'''/\/\''(x(<@:+)"1]4 2$1|.4#0 1)}y'&.>/@((<>1 7 1#1&|.@(2 10&#)&.>' _';'| ';'|_'),~>)"0(>({~[:(i.[:>./[*22&>)+/@>)(#:@i.@(2&^)@#<@#"1 _])".1!:1]3){a:,(((1$7);(a 0 _1;7),(a 0 2 _3 _1;7),((a=:,&.>/\)0 2 6 8 _3 _1;4;_5)){&.><,{(2+i.5);2 5 8),(4 5;k);<2 5;6 5;k=:,{2 8;~>:+:i.4
$ echo -n 1 5 7 8 | jconsole test.ijs __________ __________ __________ __________ | || || || | | || /\ /\ || /\ /\ || /\ /\ | | || \/ \/ || \/ /\ \/ || \/ /\ \/ | | /\ || /\ || /\ \/ /\ || /\ \/ /\ | | \/ || \/ || \/ \/ || \/ /\ \/ | | || /\ /\ || /\ /\ || /\ \/ /\ | | || \/ \/ || \/ \/ || \/ \/ | |__________||__________||__________||__________| $ echo -n 10 3 4 2 6 | jconsole test.ijs __________ __________ __________ __________ | /\ /\ || || || | | \/ /\ \/ || /\ || /\ || /\ /\ | | /\ \/ /\ || \/ || \/ || \/ \/ | | \/ \/ || /\ || || /\ /\ | | /\ /\ || \/ || || \/ \/ | | \/ /\ \/ || /\ || /\ || /\ /\ | | /\ \/ /\ || \/ || \/ || \/ \/ | |_\/____\/_||__________||__________||__________| $ echo -n 5 10 5 2 3 | jconsole test.ijs __________ __________ __________ __________ | /\ /\ || || || | | \/ /\ \/ || /\ /\ || /\ || /\ | | /\ \/ /\ || \/ \/ || \/ || \/ | | \/ \/ || /\ || || /\ | | /\ /\ || \/ || || \/ | | \/ /\ \/ || /\ /\ || /\ || /\ | | /\ \/ /\ || \/ \/ || \/ || \/ | |_\/____\/_||__________||__________||__________|
寻找最佳牌局的核心实际上非常简单。生成给定卡片的功率集，并选择值最小值小于22的卡片。
({~[:(i.[:>./[*22&>)+/@>)(#:@i.@(2&^)@#<@#"1 _])
然后我们从一张空白卡开始
<>1 7 1#1&|.@(2 10&#)&.>' _';'| ';'|_'
并在适当的位置镶嵌钻石。
'/\/\'(x(<@:+)"1]4 2$1|.4#0 1)}y
该计划的其余部分由钻石所在的表格占主导地位。
答案 1 :(得分：9)
h=map(int,raw_input().split());H=len(h);R=range
for r in R(9):x=["| "[1>r]];print"".join("_ "[0<r<8].join(x+[" \//\__"[ord(('?'*11+'@'*8+'bb@BBbbbbbQY@AAQQQYYbfH@H@HbffYQD@D@DQQYfb@``bbbbfQY@PPQQQQQbf????????]]')[r*10+c])*2>>i&6:][:2]for i in 0,2,4]+x)for c in max([[h[b]for b in R(H)if n&1<<b]for n in R(1<<H)],key=lambda c:(sum(c)<22,sum(c),len(c),sorted(c),c)))
$ echo 2 4 8 1 8 3| python blackjack.py __________ __________ __________ __________ | || || || | | /\ || /\ /\ || /\ /\ || /\ | | \/ || \/ /\ \/ || \/ /\ \/ || \/ | | || /\ \/ /\ || /\ \/ /\ || /\ | | || \/ /\ \/ || \/ /\ \/ || \/ | | /\ || /\ \/ /\ || /\ \/ /\ || /\ | | \/ || \/ \/ || \/ \/ || \/ | |__________||__________||__________||__________| $ echo 1 4 5 4 7 4 8| python blackjack.py __________ __________ __________ __________ __________ | || || || || | | || /\ /\ || /\ /\ || /\ /\ || /\ /\ | | || \/ \/ || \/ \/ || \/ /\ \/ || \/ \/ | | /\ || /\ || || /\ \/ /\ || | | \/ || \/ || || \/ \/ || | | || /\ /\ || /\ /\ || /\ /\ || /\ /\ | | || \/ \/ || \/ \/ || \/ \/ || \/ \/ | |__________||__________||__________||__________||__________|
369个字符
h=map(int,raw_input().split());H=len(h);R=range;j=[]
for n in R(1<<H):c=[h[b]for b in R(H)if n&1<<b];j+=[(sum(c)<22,sum(c),len(c),sorted(c),c)]
for r in R(9):x=["| "[1>r]];print"".join("_ "[0<r<8].join(x+[" \//\__"[ord(('?'*11+'@'*8+'bb@BBbbbbbQY@AAQQQYYbfH@H@HbffYQD@D@DQQYfb@``bbbbfQY@PPQQQQQbf????????]]')[r*10+c])*2>>i&6:][:2]for i in 0,2,4]+x)for c in max(j)[-1])
377个字符
h=map(int,raw_input().split());H=len(h);R=range;j=[]
for n in R(1<<H):c=[h[b]for b in R(H)if n&1<<b];j+=[(sum(c),len(c),sorted(c),c)]*(sum(c)<22)
T="@@@HD@@?@BA@@`P?@BAHD`P?@bQ@@bQ?@bQHDbQ?@bQbQbQ?@bYfQbQ?@bYfYfQ?bQbYfQb]bYfQbYf]"
for r in R(9):x=["| "[1>r]];print"".join("_ "[0<r<8].join(x+[" \//\__"[ord(('?'+T[c*8-8:])[r])*2>>i&6:][:2]for i in 0,2,4]+x)for c in max(j)[-1])
408个字符
h=map(int,raw_input().split());H=len(h);R=range;j=[]
for n in R(1<<H):c=[h[b]for b in R(H)if n&1<<b];j+=[(sum(c),len(c),sorted(c),c)]*(sum(c)<22)
T="@@@DH@@ @AB@@P` @ABDHP` @Qb@@Qb @QbDHQb @QbQbQb @QfYbQb @QfYfYb QbQfYbQn QfYbQfYn".split();U="\x3f"
for r in R(9):s="_ "[0<r<8];x=["| "[1>r]];print"".join(s.join(x+[(" ","/\\","\/","__")[ord((U+T[c-1]+U)[r])>>i&3]for i in 0,2,4]+x)for c in sorted(j)[-1][-1])
答案 2 :(得分：9)
' ':x/{~}%:h;9,{:r;h,2\?,{:m;h,,{2\?m&},{h\=}%}%1>{:j[{+}*.22<\j,j$j]}$)\;{:c;[x]3,{4\?' \//\__'2/[11'?'*8'@'*'bb@BBbbbbbQY@AAQQQYYbfH@H@HbffYQD@D@DQQYfb@``bbbbfQY@PPQQQQQbf'8'?'*]']]'+10r*c+=@/3&=}%[x]++' _'1/7r&!=*}%'|':x;n}%
$ echo 10 9 8 7 6 5 | ../golfscript.rb black.gs __________ __________ __________ | || || | | /\ /\ || /\ /\ || /\ /\ | | \/ /\ \/ || \/ /\ \/ || \/ \/ | | /\ \/ /\ || /\ \/ /\ || /\ /\ | | \/ /\ \/ || \/ \/ || \/ \/ | | /\ \/ /\ || /\ /\ || /\ /\ | | \/ \/ || \/ \/ || \/ \/ | |__________||__________||__________|
231个字符
' ':x/{~}%:h,2\?,{:m;h,,{2\?m&},{h\=}%}%1>{:j[{+}*.22<\j,j$j]}$-1=:h;9,{:r;h{:c;[x]3,{4\?' \//\__'2/11'?'*8'@'*'bb@BBbbbbbQY@AAQQQYYbfH@H@HbffYQD@D@DQQYfb@``bbbbfQY@PPQQQQQbf'8'?'*']]'++++10r*c+=@/3&=}%[x]++' _'1/7r&!=*}%'|':x;n}%
如何运作
# parse input into a list
' ':x/{~}%:h
# create the powerset
,2\?,{:m;h,,{2\?m&},
# map the powerset onto the cards
{h\=}%}%
# trim the empty set from the powerset
1>
# sort the hands. most preferred hand will be last
{:j[{+}*.22<\j,j$j]}$
# take the preferred hand from the end of the list
-1=:h;
# for r in 0..8
9,{:r
...more to follow
答案 3 :(得分：4)
从gnibbler's Python solution大肆偷走;如果输入是在命令行上给出的，那么结果是在362处缩短了三个字符。
sub p{(local$z=pop)?(@z=&p,map[@$_,$z],@z):[]}
for(p@ARGV){$t=0;$t+=$_ for@$_;($s,@c)=($t,@$_)if$t>=$s&&$t<22}
for$r(0..8){$x=$r?'|':$";for$c(@c){print+join$r*($r<8)?$":'_',$x,
map(join('',((split//,' \//\__')[ord((('?')x11,('@')x8,split//,
'bb@BBbbbbbQY@AAQQQYYbfH@H@HbffYQD@D@DQQYfb@``bbbbfQY@PPQQQQQbf????????]]'
)[$r*10+$c])*2>>$_&6..7])[0,1]),0,2,4),$x}print$/}
要从标准输入读取，请将@ARGV替换为@F并使用perl -an;使用传统的Perl高尔夫球得分，计为362。
当然，也可以复制gnibbler's other other Python trick。
$ perl -pechomp <<'END' \
> >'bb@BBbbbbbQY@AAQQQYYbfH@H@HbffYQD@D@DQQYfb@``bbbbfQY@PPQQQQQbf????????]]'
> sub p{(local$z=pop)?(@z=&p,map[@$_,$z],@z):[]}
> for(p@F){$t=0;$t+=$_ for@$_;($s,@c)=($t,@$_)if$t>=$s&&$t<22}
> for$r(0..8){$x=$r?'|':$";for$c(@c){print+join$r*($r<8)?$":'_',$x,
> map(join('',((split//,' \//\__')[ord((('?')x11,('@')x8,split//,$0)
> [$r*10+$c])*2>>$_&6..7])[0,1]),0,2,4),$x}print$/}
> END
$ wc -c <bb*
287
$ echo 10 5 2 9 3 | perl -an bb*
__________ __________ __________
| /\ /\ || || /\ /\ |
| \/ /\ \/ || /\ || \/ \/ |
| /\ \/ /\ || \/ || /\ /\ |
| \/ \/ || || \/ /\ \/ |
| /\ /\ || || /\ \/ /\ |
| \/ /\ \/ || /\ || \/ \/ |
| /\ \/ /\ || \/ || /\ /\ |
|_\/____\/_||__________||_\/____\/_|
答案 4 :(得分：3)
好的，现在纠正。甚至可以将1分作为王牌。
t=table p=ipairs j=arg x={}r={}o={}b=0 function f()local e,s,n,m=0,0,{}for i,v in p(j)do s=s+v e=e+(v+0==1 and 1 or 0)n[i]=v end t.sort(n)for i,v in p(n)do if o[i]and v~=o[i]then m=v>o[i]break end end while e>0 and s<12 do s=s+10 e=e-1 end if s<=21 and s>b or(s==b and(#j>#r or(#j==#r and m)))then b=s for i=1,#r>#j and #r or #j do r[i]=j[i]o[i]=j[i]end t.sort(o)end if s>0 then for i=1,#j do t.insert(x,t.remove(j,i))f()t.insert(j,i,table.remove(x))end end end f()t={" __________ ","| |","|__________|","| /\\ |","| /\\ |","| /\\ |","| /\\ /\\ |","| \\/ /\\ \\/ |","|_\\/____\\/_|"}c={"12224D223","125E226F3","125E4D6F3","127G227G3","127G4D7G3","127G7G7G3","1278HG7G3","1278H8HG3","17G78HG79","178HG78H9"}for i=1,9 do for j,x in p(r)do v=c[tonumber(x)]:sub(i,i):byte()io.write(v<64 and t[v-48]or(t[v-64]:gsub("[/\\]",{["/"]="\\",["\\"]="/"})))end io.write("\n")end
示例输出：
>lua card.lua 8 9 10 7
__________ __________
| /\ /\ || /\ /\ |
| \/ \/ || \/ /\ \/ |
| /\ /\ || /\ \/ /\ |
| \/ /\ \/ || \/ \/ |
| /\ \/ /\ || /\ /\ |
| \/ \/ || \/ /\ \/ |
| /\ /\ || /\ \/ /\ |
|_\/____\/_||_\/____\/_|
>lua card.lua 1 4 5 4 7 4 8
__________ __________ __________ __________ __________
| || || || || |
| || /\ /\ || /\ /\ || /\ /\ || /\ /\ |
| || \/ \/ || \/ \/ || \/ /\ \/ || \/ \/ |
| /\ || /\ || || /\ \/ /\ || |
| \/ || \/ || || \/ \/ || |
| || /\ /\ || /\ /\ || /\ /\ || /\ /\ |
| || \/ \/ || \/ \/ || \/ \/ || \/ \/ |
|__________||__________||__________||__________||__________|
>lua card.lua 1 9 5 5
__________ __________ __________
| || || |
| || /\ /\ || /\ /\ |
| || \/ \/ || \/ \/ |
| /\ || /\ || /\ |
| \/ || \/ || \/ |
| || /\ /\ || /\ /\ |
| || \/ \/ || \/ \/ |
|__________||__________||__________|
答案 5 :(得分：2)
只需像这样重命名脚本
mv blackjack.py '???????????@@@@@@@@bb@BBbbbbbQY@AAQQQYYbfH@H@HbffYQD@D@DQQYfb@``bbbbfQY@PPQQQQQbf????????]].py'
这是脚本
import sys
h=map(int,raw_input().split());H=len(h);R=range
for r in R(9):x=["| "[1>r]];print"".join("_ "[0<r<8].join(x+[" \//\__"[ord(sys.argv[0][r*10+c])*2>>i&6:][:2]for i in 0,2,4]+x)for c in max([[h[b]for b in R(H)if n&1<<b]for n in R(1<<H)],key=lambda c:(sum(c)<22,sum(c),len(c),sorted(c),c)))
和样本运行
$ echo 1 2 3 4 5 | python '???????????@@@@@@@@bb@BBbbbbbQY@AAQQQYYbfH@H@HbffYQD@D@DQQYfb@``bbbbfQY@PPQQQQQbf????????]].py'
__________ __________ __________ __________ __________ | || || || || | | || /\ || /\ || /\ /\ || /\ /\ | | || \/ || \/ || \/ \/ || \/ \/ | | /\ || || /\ || || /\ | | \/ || || \/ || || \/ | | || /\ || /\ || /\ /\ || /\ /\ | | || \/ || \/ || \/ \/ || \/ \/ | |__________||__________||__________||__________||__________|
答案 6 :(得分：1)
from itertools import*
o=sum;q=range;r=filter
l=map(int,raw_input().split())
s=r(lambda x:o(x)<22,chain.from_iterable(imap(combinations,repeat(l),q(len(l)+1))))
s=r(lambda x:o(x)==max(map(o,s)),s)
t=max(([(min(x),x)for x in s if len(x)==max(map(len,s))]))[1]
a=" __________ "
b="| |"
c="| /\ /\ |";e=c[::-1]
d="| /\ |";l=d[::-1]
g="| \/ |";k=g[::-1]
i="| /\ |";j=i[::-1]
h="| /\ \/ /\ |";f=h[::-1]
n="|_\/____\/_|"
m="|__________|"
y="aaaaaaaaaabbbbbbbbccbddcccccefbggeeeffchibibichhfejbjbjeefhcbkkcccchefblleeeeechmmmmmmmmnn"
for x in q(9):
print''.join([globals()[y[x*10:(x+1)*10][v-1]]for v in t])
示例输出：
10 5 7 4 1 1 __________ __________ __________ __________ __________ | /\ /\ || || || || | | \/ /\ \/ || /\ /\ || /\ /\ || || | | /\ \/ /\ || \/ \/ || \/ \/ || || | | \/ \/ || /\ || || /\ || /\ | | /\ /\ || \/ || || \/ || \/ | | \/ /\ \/ || /\ /\ || /\ /\ || || | | /\ \/ /\ || \/ \/ || \/ \/ || || | |_\/____\/_||__________||__________||__________||__________|
答案 7 :(得分：1)
更新了Haskell版本，现在可以适用于所有情况，但它可能会进行更多测试。
import Data.List --17
--shorteners
m=map --6
r=reverse --10
l=length --9
u=True --7
g=minimum --10
--printing of the cards
p=foldl1(zipWith(++)).m d --26
a=[" __________ ","| |","| /\\ /\\ |",r(a!!2),"| /\\ |",r(a!!4),"| \\/ |",r(a!!6),"| /\\ |",r(a!!8),"| /\\ \\/ /\\ |",r(a!!10),"|_\\/____\\/_|","|__________|"] --190
j(Just x)=x --12
d(n+1)=m(j.(flip$lookup)(zip"abcedlgkijhfnm"a).("aaaaaaaaaabbbbbbbbccbddcccccefbggeeeffchibibichhfejbjbjeefhcbkkcccchefblleeeeechmmmmmmmmnn"!!).(+n))[0,10..80] --160
-- interaction and pipeline
main=interact$unlines.p.f.m read.words --39
f x=head.filter((<22).sum).r.sortBy(o x)$x:z x --47
-- generate all possible hands (power set)
z[]=[] --7
z(x:s)=s:m(x:)(z s)++z s --25
-- sorting logic
c f a b=compare(f a)$f b --25
o x a b|l a==l b=s x a b|u=c l a b --35
s o a b|g a==g b=t o a b|u=c g a b --35
t o a b|all(`elem`b)a=c(n o)a b|u=GT --37
n t=m(j.(`elemIndex`t)) --24
Haskell版尚未完成622个字符
它不会赢得任何选美比赛，也可以改进我确定，并且不完全遵循规范（仍在处理min(min(3,2), min(1,4))问题以及重复过滤（{{ 1}}函数需要更多情况））。尽管它不完整，但确实通过了所有给出的测试用例。
每行都有注释（o）及其中的字符数，包括新行。
--答案 8 :(得分：1)
不知道还有多少改进可能，但是这可能比gnibbler的算法更容易移植到GolfScript：
sub R{for(@_){substr$q[$_],$B+$z,2,'/\\';substr$q[$_+1],$B+$z,2,'\/';
if($B==2){$B=8;R(8-$_);$B=2}}}
for$i(1..1<<(@p=@ARGV)){$s=$n=0;$i&1<<$n++&&($s+=21*$_+1)for@p;
$q=$s,$N=$i if$s<462&&$q<$s}
@q=($/)x9;for$p(grep$N&1<<$_,0..@p){
s/$/| |/for@q;
$B=5;($P=$p[$p])&1&&R(4-($P==7));$P-8||R 5,3;$P>9&&R 6,2;
$B=2;if($P>8){R 1,3,5,7}else{R 2if$P>1;R 6if$P>3;R 4if$P>5}$z+=12}
$q[0]=~y/| / _/;$q[-1]=~y/ /_/;print@q