Excel VBA ขนาด 816 ไบต์
ฟังก์ชันหน้าต่าง VBE แบบไม่ระบุชื่อแบบทันทีที่รับอินพุตจากช่วง[A1]และเอาต์พุตไปยังคอนโซล
เท่าที่ฉันทราบนี่เป็นคำตอบ VBA แรกที่ใช้การbase64บีบอัด
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
หมายเหตุ: คำตอบนี้ขึ้นอยู่กับการMicrosoft XML, v3.0อ้างอิง VBA
ตัวอย่าง 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 และอธิบาย
ส่วนที่สำคัญของการแก้ปัญหานี้เก็บตัวอักษรขนาดใหญ่เป็นสตริง 64 ฐาน สิ่งนี้ทำได้โดยการแปลงฟอนต์เป็นไบนารี่เป็นอันดับแรกโดยที่จะ1แสดงเป็นพิกเซลและ0แสดงเป็นพิกเซลแบบปิด ตัวอย่างเช่นสำหรับ0สิ่งนี้ถูกแทนด้วย
### 01110
# ## 10011
0 -> # # # -> 10101 --> 0111010011101011100101110
## # 11001
### 01110
ด้วยวิธีการนี้ตัวอักษรและตัวเลขจึงสามารถแสดงเป็น
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
ส่วนเหล่านี้ถูกต่อกันและแปลงเป็นฐาน MSXML 64 การเรนเดอร์
HxHxCSEqRkVUjLvGSJSK0cUYIyGEfB8cfFH66Ju0kkHoo3cxRhdnzTHGuuOHEMIouYyYEPI/IeTH+GN8ccIHIYf/Qw6/jzH6ByF8PvroY/zR+fCic9FFh4gI30UPnw8efiG+Mj6c4D90wX9CCHe5Tgc=
รูทีนย่อยด้านล่างใช้สิ่งนี้กลับไปแปลงเป็นไบนารี่และใช้การอ้างอิงนี้เพื่อสร้างสตริงเอาท์พุททีละบรรทัดโดยจับแรกสุด 5 พิกเซลของอักขระแต่ละตัวจากนั้นแถวที่สองเป็นต้นจนกว่าจะมีการสร้างสตริง .
รูทีนย่อยจะทำซ้ำผ่านสตริงเอาต์พุตและแทนที่พิกเซล 'on' ด้วยอักขระจากสตริงอินพุต
''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''
''
'' 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]- ฉันไม่คิดว่าการกรองอักขระที่ไม่ถูกต้องจะเป็นการเพิ่มสิ่งใด ๆ ให้กับความท้าทาย