Is 0 a decimal literal or an octal literal?

Yes, 0 is an Octal literal in C++.

As per the C++ Standard:

2.14.2 Integer literals [lex.icon]

integer-literal:  
    decimal-literal integer-suffixopt  
    octal-literal integer-suffixopt  
    hexadecimal-literal integer-suffixopt  
decimal-literal:  
    nonzero-digit  
    decimal-literal digit  
octal-literal:  
    0                           <--------------------<Here>
    octal-literal octal-digit

Any integer value prefixed with 0 is an octal value. I.e.: 01 is octal 1, 010 is octal 10, which is decimal 8, and 0 is octal 0 (which is decimal, and any other, 0).

So yes, '0' is an octal.

That's plain English translation of the grammar snippet in @Als's answer :-)

---

An integer prefixed with 0x is not prefixed with 0. 0x is an explicitly different prefix. Apparently there are people who cannot make this distinction.

As per that same standard, if we continue:

 integer-literal:
     decimal-literal integer-suffixopt
     octal-literal integer-suffixopt
     hexadecimal-literal integer-suffixopt
 decimal-literal:
     nonzero-digit                       <<<---- That's the case of no prefix.
     decimal-literal digit-separatoropt digit
 octal-literal:
     0                                    <<<---- '0' prefix defined here.
     octal-literal digit-separatoropt octal-digit <<<---- No 'x' or 'X' is
                                                          allowed here.
 hexadecimal-literal:
     0x hexadecimal-digit                 <<<---- '0x' prefix defined here
     0X hexadecimal-digit                 <<<---- And here.
     hexadecimal-literal digit-separatoropt hexadecimal-digit