x(x*), # 1円 = divisor-1; tail = dividend
( # 2円 = what will be the quotient
 ( # 3円 = the following (after the first iteration has finished),
 # which is always 1
 x # tail -= 1
 (?=
 ( # 4円 = running total
 4円 # recall the previous contents of 4円 (only if it is set)
 1円 # 4円 += divisor-1
 |
 (?!3円) # Match this alternative only if this is the first iteration of
 # the loop, meaning 4円 is unset; 3円 can never not match here if
 # 3円 is set (because 3円==1) unless 1円==0, in which case it
 # doesn't matter if the above alternative is taken instead,
 # because in that case, the value of 4円 isn't changed anyway.
 1円 # 4円 = divisor-1
 )
 )
 )* # Loop the above as many times as possible (zero or more); if
 # it loops zero times, 4円 will be unset (we'll treat that as 0)
)
4円? # tail -= 4,円 or leave tail unchanged if 4円 is unset
(x*) # 5円 = remainder

x(x*), # 1円 = divisor-1; tail = dividend
( # 2円 = what will be the quotient
 ( # 3円 = the following (after the first iteration has finished),
 # which is always 1
 x # tail -= 1
 (?=
 ( # 4円 = running total
 4円 # recall the previous contents of 4円 (only if it is set)
 1円 # 4円 += divisor-1
 |
 (?!3円) # Match this alternative only if this is the first iteration of
 # the loop, meaning 4円 is unset; 3円 can never not match here if
 # 3円 is set (because 3円==1) unless 1円==0, in which case it
 # doesn't matter if this alternative is taken instead of the
 # above – in that case, the value of 4円 isn't changed anyway.
 1円 # 4円 = divisor-1
 )
 )
 )* # Loop the above as many times as possible (zero or more); if
 # it loops zero times, 4円 will be unset (we'll treat that as 0)
)
4円? # tail -= 4,円 or leave tail unchanged if 4円 is unset
(x*) # 5円 = remainder

Regex (Java), (削除) 41 (削除ここまで) 40 bytes

x(x*),((x(?=(4円1円|(?!5円)1円)()))*)4円?(x*)

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This is a port of the Perl/PCRE regex to a flavor that has no conditionals. Emulating a conditional costs 5 bytes here. The quotient and remainder are returned in the capture groups 2円 and 6円, respectively.

x(x*), # 1円 = divisor-1; tail = dividend
( # 2円 = what will be the quotient
 (
 x # tail -= 1
 (?=
 ( # 4円 = running total
 4円 # recall the previous contents of 4円 (only if it is set)
 1円 # 4円 += divisor-1
 |
 (?!5円) # match this alternative only if 4円 is unset
 1円 # 4円 = divisor-1
 )
 () # 5円 = set to indicate that 4円 is set
 )
 )* # Loop the above as many times as possible (zero or more); if
 # it loops zero times, 4円 will be unset (we'll treat that as 0)
)
4円? # tail -= 4,円 or leave tail unchanged if 4円 is unset
(x*) # 6円 = remainder

Regex (Java), (削除) 41 (削除ここまで) (削除) 40 (削除ここまで) 38 bytes

x(x*),((x(?=(4円1円|(?!3円)1円)))*)4円?(x*)

Try it online!

This is a port of the Perl/PCRE regex to a flavor that has no conditionals. Emulating a conditional costs (削除) 5 (削除ここまで) 3 bytes here. The quotient and remainder are returned in the capture groups 2円 and 5円, respectively.

x(x*), # 1円 = divisor-1; tail = dividend
( # 2円 = what will be the quotient
 ( # 3円 = the following (after the first iteration has finished),
  # which is always 1
  x  # tail -= 1
 (?=
 ( # 4円 = running total
 4円 # recall the previous contents of 4円 (only if it is set)
 1円 # 4円 += divisor-1
 |
 (?!3円) # Match this alternative only if this is the first iteration of
 # the loop, meaning 4円 is unset; 3円 can never not match here if
 # 3円 is set (because 3円==1) unless 1円==0, in which case it
  # doesn't matter if the above alternative is taken instead,
 # because in that case, the value of 4円 isn't changed anyway.
 1円 # 4円 = divisor-1
 )
 ) )* # Loop the above as many times as possible (zero or more); if
 # it loops zero times, 4円 will be unset (we'll treat that as 0)
)
4円? # tail -= 4,円 or leave tail unchanged if 4円 is unset
(x*) # 5円 = remainder

Regex (Python^regex / Ruby), 43 bytes

This is a port of the Perl/PCRE regex to flavors that have no support for nested backreferences. Python's built-in re module does not even support forward backreferences, so for Python this requires regex.

Regex (.NET), 14 bytes

Regex (Python^regex / Ruby), 43 bytes

This is a port of the Perl/PCRE regex to flavors that have no support for nested backreferences. Python's built-in re module does not even support forward backreferences, so for Python this requires regex.

Regex 🐘(.NET), 14 bytes