Excel VBA, 816 Bytes
Eine anonyme VBE-Direktfensterfunktion, die Eingaben aus dem Bereich entgegennimmt [A1] und Ausgaben an die Konsole weiterleitet.
Soweit mir bekannt ist, ist dies die erste VBA-Antwort, die die base64Komprimierung verwendet.
For i=1To[Len(A1)]:c=Mid(UCase([A1]),i,1):y=y &IIf(c Like"[0-9A-Z]",c,""):Next:l=Len(y):Set d=New MSXML2.DOMDocument:Set d=d.createElement("b64"):d.DataType="bin.base64":d.Text="HxHxCSEqRkVUjLvGSJSK0cUYIyGEfB8cfFH66Ju0kkHoo3cxRhdnzTHGuuOHEMIouYyYEPI/IeTH+GN8ccIHIYf/Qw6/jzH6ByF8PvroY/zR+fCic9FFh4gI30UPnw8efiG+Mj6c4D90wX9CCHe5Tgc=":b=d.nodeTypedValue:For i=0To 112:k=Right("00000" &Evaluate("=Dec2Bin("&b(i)&")"),8)&k:Next:For i=1To 5:For j=1To l:c=UCase(Mid(y,j,1)):Z=c Like"[0-9]":s=s &IIf(c Like"[A-Z]",Mid(k,IIf(Z,1,25*(Asc(c)-55)+5*i),5)&" ",IIf(Z,Mid(k,25*(Asc(c)-48)+5*i,5)&" ","")):Next:s=Replace(Replace(s,0," "),1,"#") &vbLf:Next:Do:i=InStr(1+(g*l+h)*6+g,s,"#"):p=(p-e)Mod l:e=i<(g*l+h+1)*6+g:s=IIf(e,Left(s,i-1)&Replace(s,"#",Mid(y,p+1,1),i,1),s):g=g-(0=e):h=h-(g>4):g=g Mod 5:Loop While InStr(1,s,"#"):?s
Hinweis: Diese Antwort ist abhängig von der Microsoft XML, v3.0 VBA-Referenz ab
Beispiel I / O
[A1]="'0123456789"
For i=1To[Len(A1)]:c=Mid(UCase([A1]),i,1):y=y &IIf(c Like"[0-9A-Z]",c,""):Next:l=Len(y):Set d=New MSXML2.DOMDocument:Set d=d.createElement("b64"):d.DataType="bin.base64":d.Text="HxHxCSEqRkVUjLvGSJSK0cUYIyGEfB8cfFH66Ju0kkHoo3cxRhdnzTHGuuOHEMIouYyYEPI/IeTH+GN8ccIHIYf/Qw6/jzH6ByF8PvroY/zR+fCic9FFh4gI30UPnw8efiG+Mj6c4D90wX9CCHe5Tgc=":b=d.nodeTypedValue:For i=0To 112:k=Right("00000" &Evaluate("=Dec2Bin("&b(i)&")"),8)&k:Next:For i=1To 5:For j=1To l:c=UCase(Mid(y,j,1)):Z=c Like"[0-9]":s=s &IIf(c Like"[A-Z]",Mid(k,IIf(Z,1,25*(Asc(c)-55)+5*i),5)&" ",IIf(Z,Mid(k,25*(Asc(c)-48)+5*i,5)&" ","")):Next:s=Replace(Replace(s,0," "),1,"#") &vbLf:Next:Do:i=InStr(1+(g*l+h)*6+g,s,"#"):p=(p-e)Mod l:e=i<(g*l+h+1)*6+g:s=IIf(e,Left(s,i-1)&Replace(s,"#",Mid(y,p+1,1),i,1),s):g=g-(0=e):h=h-(g>4):g=g Mod 5:Loop While i<InStrRev(s,"#"):?s
012 567 6789 0123 34 45678 9012 34567 234 567
3 45 8 0 4 5 6 9 3 8 5 6 8 9
6 7 8 9 123 567 78901 0123 4567 9 789 0123
90 1 0 4 8 2 4 8 9 0 0 1 4
234 12345 56789 9012 3 5678 012 1 234 5678
Ungolfed und erklärt
Der größte Teil dieser Lösung speichert die große Schrift als 64er Basiszeichenfolge. Dies geschieht, indem zuerst die Schrift in eine Binärschrift umgewandelt wird, wobei 1ein Ein-Pixel und 0ein Aus-Pixel dargestellt werden. Beispielsweise 0wird dies als dargestellt
### 01110
# ## 10011
0 -> # # # -> 10101 --> 0111010011101011100101110
## # 11001
### 01110
Mit diesem Ansatz kann die alphanumerische Zeichenfolge dann als dargestellt werden
0: 0111010011101011100101110 1: 1110000100001000010011111
2: 1111000001011101000011111 3: 1111000001001110000111110
4: 0011001010111110001000010 5: 1111110000111100000111110
6: 0111110000111101000101110 7: 1111100001000100010001000
8: 0111010001011101000101110 9: 0111010001011110000111110
A: 0111010001111111000110001 B: 1111010001111101000111110
C: 0111110000100001000001111 D: 1111010001100011000111110
E: 1111110000111001000011111 F: 1111110000111001000010000
G: 0111110000100111000101111 H: 1000110001111111000110001
I: 1111100100001000010011111 J: 1111100100001000010011000
K: 1000110010111001001010001 L: 1000010000100001000011111
M: 1000111011101011000110001 N: 1000111001101011001110001
O: 0111010001100011000101110 P: 1111010001111101000010000
Q: 0110010010101101001001101 R: 1111010001111101001010001
S: 0111110000011100000111110 T: 1111100100001000010000100
U: 1000110001100011000101110 V: 1000110001010100101000100
W: 1000110001101011101110001 X: 1000101010001000101010001
Y: 1000101010001000010000100 Z: 1111100010001000100011111
Diese Segmente wurden verkettet und in MSXML Base 64 (Rendering) konvertiert
HxHxCSEqRkVUjLvGSJSK0cUYIyGEfB8cfFH66Ju0kkHoo3cxRhdnzTHGuuOHEMIouYyYEPI/IeTH+GN8ccIHIYf/Qw6/jzH6ByF8PvroY/zR+fCic9FFh4gI30UPnw8efiG+Mj6c4D90wX9CCHe5Tgc=
Die folgende Unterroutine nimmt dies, konvertiert zurück in binär und verwendet diese Referenz, um eine Ausgabezeichenfolge Zeile für Zeile zu erstellen, wobei zuerst die oberen 5 Pixel jedes Zeichens, dann die zweite Zeile usw. abgerufen werden, bis die Zeichenfolge erstellt ist .
Das Unterprogramm durchläuft dann die Ausgabezeichenfolge und ersetzt die Ein-Pixel durch Zeichen aus der Eingabezeichenfolge.
''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''
''
'' Embiggen Function
''
'' @Title : Embiggen
'' @Author : Taylor Scott
'' @Date : 15 June 2018
'' @Desc : Function that takes input, value, and outputs a string in which
'' value has been filtered to alphnumerics only, each char is then
'' scaled up to a 5x5 ASCII art, and each 'pixel' is replaced with
'' a char from value. Replacement occurs letter by letter, line by
'' line
''
''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''
Function EMBIGGEN(ByVal value As String) As String
Dim DOM As New MSXML2.DOMDocument, _
bytes() As Byte
Dim isNum As Boolean, _
found As Boolean, _
index As Integer, _
length As Integer, _
line As Integer, _
letter As Integer, _
pos As Integer, _
alphanum As String, _
char As String, _
filValue As String, _
outValue As String
''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''
''
'' Filter input
''
''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''
For letter = 1 To Len(value) Step 1 '' Iterate Accross `Value`
Let char = Mid$(UCase(value), letter, 1) '' Take the nth char
'' If the char is alphnumeric, append it to a filtered input string
Let filValue = filValue & IIf(char Like "[0-9A-Z]", char, "")
Next letter
Let length = Len(filValue) '' store length of filValue
''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''
''
'' Convert Constant from Base 64 to Byte Array
''
''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''
With DOM.createElement("b64") '' Construct b64 DOM object
Let .DataType = "bin.base64" '' define type of object`
'' Input constructed constant string shown above
Let .Text = "HxHxCSEqRkVUjLvGSJSK0cUYIyGEfB8cfFH66Ju0kkHoo3cxRhdnz" & _
"THGuuOHEMIouYyYEPI/IeTH+GN8ccIHIYf/Qw6/jzH6ByF8PvroY/" & _
"zR+fCic9FFh4gI30UPnw8efiG+Mj6c4D90wX9CCHe5Tgc="
Let bytes = .nodeTypedValue '' Pass resulting bytes to array
End With
''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''
''
'' Convert Byte Array to Byte String
''
''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''
For index = 0 To 112 Step 1
'' convert each byte to binary, fill left with `0`s and prepend
Let alphanum = _
Right("00000" & Evaluate("=Dec2Bin(" & bytes(index) & ")"), 8) & _
alphanum
Next index
''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''
''
'' Construct Embiggened Binary String of Input Value
''
''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''
For line = 1 To 5 Step 1 '' iterate across lines
For letter = 1 To length Step 1 '' iterate across letters
'' take the corresponding letter from
Let char = UCase(Mid(filValue, letter, 1))
If char Like "[0-9]" Then '' if it is a number,
'' Add the 5 bit corresponding to number at line
Let outValue = outValue & _
Mid$(alphanum, 25 * Val(char) + 5 * line, 5) & " "
ElseIf char Like "[A-Z]" Then '' if it is a letter,
'' Add the 5 bits corresponding to letter at line
Let outValue = outValue & _
Mid$(alphanum, 25 * (Asc(char) - 55) + 5 * line, 5) & " "
End If
Next letter
Let outValue = outValue & IIf(line < 5, vbLf, "")
Next line
Let outValue = Replace(Replace(outValue, 0, " "), 1, "#")
''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''
''
'' Replace #s with Input Value
''
''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''
Let pos = 0 '' Reset position in filValue
Let line = 0 '' Reset line index
Let letter = 0 '' Reset letter index
Do
'' Find the index of the first `#` starting at line and letter
Let index = _
InStr(1 + (line * length + letter) * 6 + line, outValue, "#")
'' Iterate position in filValue if a `#` is found in that letter & line
Let pos = (pos - found) Mod length
'' check to see if found index is in the correct letter
Let found = index < (line * length + letter + 1) * 6 + line
'' iff so, replace that # with letter in filValue corresponding to pos
Let outValue = IIf(found, _
Left(outValue, index - 1) & _
Replace(outValue, "#", Mid(filValue, pos + 1, 1), index, 1), _
outValue)
'' if not found, them iterate line
Let line = line - (found = False)
'' iterate letter every five iterations of line
Let letter = letter - (line > 4)
'' Ensure that line between 0 and 4 (inc)
Let line = line Mod 5
'' Loop while there are '#'s in outValue
Loop While InStr(1, outValue, "#")
''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''
''
'' Output
''
''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''
Let EMBIGGEN = outValue
''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''
''
'' Clean Up
''
''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''
Set DOM = Nothoing
End Function
[A-Z\d]- ich glaube nicht, dass das Filtern ungültiger Zeichen der Herausforderung etwas hinzufügt.