Kami memiliki banyak tantangan alfabet. Untuk tantangan ini, Anda melewatkan output dari tantangan alfabet, dan Anda perlu menampilkan pola yang diskalakan ke ukuranN .

Misalnya, jika N=5dan Anda lulus L-phabet :

ABCDEFGHIJKLMNOPQRSTUVWXYZ
BBCDEFGHIJKLMNOPQRSTUVWXYZ
CCCDEFGHIJKLMNOPQRSTUVWXYZ
DDDDEFGHIJKLMNOPQRSTUVWXYZ
EEEEEFGHIJKLMNOPQRSTUVWXYZ
FFFFFFGHIJKLMNOPQRSTUVWXYZ
GGGGGGGHIJKLMNOPQRSTUVWXYZ
HHHHHHHHIJKLMNOPQRSTUVWXYZ
IIIIIIIIIJKLMNOPQRSTUVWXYZ
JJJJJJJJJJKLMNOPQRSTUVWXYZ
KKKKKKKKKKKLMNOPQRSTUVWXYZ
LLLLLLLLLLLLMNOPQRSTUVWXYZ
MMMMMMMMMMMMMNOPQRSTUVWXYZ
NNNNNNNNNNNNNNOPQRSTUVWXYZ
OOOOOOOOOOOOOOOPQRSTUVWXYZ
PPPPPPPPPPPPPPPPQRSTUVWXYZ
QQQQQQQQQQQQQQQQQRSTUVWXYZ
RRRRRRRRRRRRRRRRRRSTUVWXYZ
SSSSSSSSSSSSSSSSSSSTUVWXYZ
TTTTTTTTTTTTTTTTTTTTUVWXYZ
UUUUUUUUUUUUUUUUUUUUUVWXYZ
VVVVVVVVVVVVVVVVVVVVVVWXYZ
WWWWWWWWWWWWWWWWWWWWWWWXYZ
XXXXXXXXXXXXXXXXXXXXXXXXYZ
YYYYYYYYYYYYYYYYYYYYYYYYYZ
ZZZZZZZZZZZZZZZZZZZZZZZZZZ

Anda perlu menampilkan:

ABCDE
BBCDE
CCCDE
DDDDE
EEEEE

Untuk keperluan penjelasan, saya hanya akan menggunakan ABCD, bukan alfabet lengkap. Anda harus dapat mencocokkan L-phabet (di atas), serta pola-pola berikut:

Baris tunggal:

ABCD    or     A
               B
               C
               D

Baris tunggal berulang Nkali

ABCD    or    AAAA
ABCD          BBBB
ABCD          CCCC
ABCD          DDDD

The Tabula Recta :

ABCD
BCDA
CDAB
DABC

Segitiga alfabet ini:

A      or   AAAAAAA
AB           BBBBB
ABC           CCC
ABCD           D
ABC
AB
A

Kami juga memiliki setengah segitiga dalam banyak varietas:

A           AAAA      A        ABCD
BB          BBB       AB       ABC
CCC         CC        ABC      AB
DDDD        D         ABCD     A

Akhirnya, alun-alun:

AAAAAAA
ABBBBBA
ABCCCBA
ABCDCBA
ABCCCBA
ABBBBBA
AAAAAAA

Semua pola di atas berukuran 4. Namun, Anda akan melewati pola ukuran 26, dan juga Nantara 1 dan 26, dan Anda perlu skala pola. Anda tidak perlu menangani pola lainnya.

• Output untuk 1 akan selalu menjadi karakter tunggal A
• Output untuk 26 akan selalu sama, pola ukuran penuh diteruskan.
• Spasi trailing dibolehkan pada akhir setiap baris, dan juga trailing newline di akhir
• Anda dapat menemukan semua pola ukuran 26 di sini

Ini adalah kode-golf , jadi lakukan sesedikit mungkin dalam byte!

PHP, 502 Bytes

<?$c=count($x=explode("\n",$_GET[p]));for($t=$u=$o="",$f="substr",$n=0;$n<$g=$_GET["n"];$n++){if(2651==$l=strlen($s))$o.=$f($x[$n],0,$g).$f($x[$n],1-$g)."\n";elseif($l==1026)echo($t=$f($x[$n],0,$g)).$f(strrev($t),1)."\n";elseif($f($s,-1)=="Y")echo$f($x[$n],0,$g-$n).$f($x[$n],-$n,$n)."\n";elseif($l==376&&$f($s,-2,1)=="\n")echo$f($x[$n],0,$g-$n)."\n";elseif($l==726){$t.=$x[$n]."\n";$n+1==$g?:$u=$x[$n]."\n".$u;}else echo$f($x[$n]??"",0,$g)."\n";}if($o)echo$o.substr(strrev($o),2*$g+1);if($t)echo$t.$u;

Bekerja dengan panjang string pola. Pola memiliki kondisi ini. Surat di mulai akhir. CR dihapus.

Diperluas

foreach($p as$s){ # all patterns in an array
    $c=count($x=explode("\n",$s));
    for($t=$u=$o="",$f="substr",$n=0;$n<$g=$_GET["n"];$n++){
    if(2651==$l=strlen($s))$o.=$f($x[$n],0,$g).$f($x[$n],1-$g)."\n";
# square pattern
    elseif($l==1026)echo($t=$f($x[$n],0,$g)).$f(strrev($t),1)."\n";
#alphabet triangle up down
    elseif($f($s,-1)=="Y")echo$f($x[$n],0,$g-$n).$f($x[$n],-$n,$n)."\n";
# Tabula recta
    elseif($l==376&&$f($s,-2,1)=="\n")echo$f($x[$n],0,$g-$n)."\n";
# two half triangle
    elseif($l==726){$t.=$x[$n]."\n";$n+1==$g?:$u=$x[$n]."\n".$u;}
#alphabet triangle left right
    else echo$f($x[$n]??"",0,$g)."\n";
# all other
    }
    if($o)echo$o.substr(strrev($o),2*$g+1);
    if($t)echo$t.$u;
}

R, 483 412 byte

    x=function(n){d=1:n;a=LETTERS[d];z=rep;f=function(...)cat(...,"\n",sep="");g=c(d,n:2-1);for(i in d)f(a[z(i,i)],if(i!=n)a[(i+1):n]);f(a);for(i in d)f(a[i]);for(i in d)f(a);for(i in d)f(z(a[i],n));for(i in d)f(a[i:n],if(i>1)a[1:i-1]);for(i in g)f(a[1:i]);for(i in d)f(z(" ",i-1),z(a[i],2*(n-i)+1));for(i in d)f(z(a[i],i));for(i in d)f(z(a[i],n-i+1));for(i in g)f(if(i>1)a[2:i-1],z(a[i],2*(n-i)+1),if(i>1)a[i:2-1])}

Ini adalah pertama kalinya saya memposting, saya diberitahu bahwa saya tidak memiliki pengalaman R tahun baru-baru ini, jadi berlatihlah sedikit di sini.

Diperluas

    x=function(n){
    # array of 1 to n
    d=1:n
    # first n capital letters 
    a=LETTERS[d]
    # use z to represent the repeat function
    z=rep
    # use f to represent concatenate, with new line and close the gap
    f=function(...)cat(...,"\n",sep="");
    # use g to represent 1 to n then n to 1
    g=c(d,n:2-1)

    # L-phabet 
    # start and repeat the first letter to i, combine the i+1 to the end
    for(i in d) f(a[z(i,i)],if(i!=n)a[(i+1):n])

    # single line - horizontal
    f(a)

    # single line - vertical
    for(i in d)f(a[i])

    # single line - repeated by row
    for(i in d)f(a)

    # single line - repeated by column
    for(i in d)f(z(a[i],n))

    # Tabula Recta
    # start from i, combine the 1 to i-1
    for(i in d) f(a[i:n],if(i>1)a[2:i-1])

    # alphabet triangle
    for(i in g)f(a[1:i])

    # alphabet triangle - upside down
    for(i in d)f(z(" ",i-1),z(a[i],2*(n-i)+1))

    # half triangles 
    for(i in d)f(z(a[i],i))
    for(i in d)f(z(a[i],n-i+1))

    # the square
    # combine the first part from i to i-1 letters, repeat the i in the middle, combine the last part from i-1 to 1 letters
    for(i in g) f(if(i>1)a[2:i-1],z(a[i],2*(n-i)+1),if(i>1)a[i:2-1])
    } 

JavaScript (ES6), 382 380 370 byte

f=(a,N)=>{X=Y=0
if(a[1]){Y=a[1].trim()[0]=='B'
X=z=a[0][1]==a[1][0]?1:!Y
l="ABCDEFGHIJKLMNOPQRSTUVWXYZ"[s='slice'](0,N)
if(z){if(a[0][1]=='A'){for(a=[],i=0,L=l;i<N;i++)a[N-i]=a[N+i]=L[s](0,-1)+l[N-i-1].repeat(i*2)+[...L].reverse().join``
L=L[s](0,-1)
return a}if(a[1][1]=='C')for(i=0;a[i++]=l,i<N;)l=l[s](1)+l[0]}}else X=1
a=X?a.map(l=>l[s](0,N)):a
return Y?a[s](0,N):a}

Berikan array string ke fungsi f(), seperti ini:

f(
`AAAAAAA
ABBBBBA
ABCCCBA
ABCDCBA
ABCCCBA
ABBBBBA
AAAAAAA`.split('\n'), 3)

Versi yang kurang golf dengan komentar:

f=(a,N)=>{
    // Whether to truncate array horizontally to width N.
    X=0

    // Whether to truncate array vertically to height N.
    Y=0

    // If a second row exists...
    if(I=a[1]){
        // If the first non-whitespace character in the second row == 'B', truncate vertically.
        if(I.trim()[0]=='B')Y=1

        // Truncate horizontally if 2nd character in row 1 == 1st character in row 2; otherwise, if not truncating vertically.
        X=z=a[0][1]==I[0]?1:!Y

        // If 2nd character in row 1 == 1st character in row 2
        if(z){
            // Make an alphabet.
            l="ABCDEFGHIJKLMNOPQRSTUVWXYZ".slice(0,N)

            // If 2nd character in row 1 == 'A', forget everything we just did. Make a new array, generate a Square pattern, then return it.
            if(a[0][1]=='A'){
                for(a=[],i=0,L=l;i<N;i++)
                    a[N-i]=a[N+i]=L.slice(0,-1)+l[N-i-1].repeat(i*2)+[...L].reverse().join``,
                    L=L.slice(0,-1)
                return a
            }

            // If 2nd character in row 2 == 'C', fill array with a Tabula Recta.
            if(I[1]=='C')
                for(i=0;a[i++]=l,i<N;)
                    l=l.slice(1)+l[0]
        }
    }else{
        // If a second row doesn't exist, it's a horizontal line; truncate horizontally.
        X=1
    }

    // Truncate array horizontally if necessary.
    a=X?a.map(l=>l.slice(0,N)):a

    // Truncate array vertically if necessary.
    return Y?a.slice(0,N):a
}