problem_database

Contest problems database on von format (all problems in Portuguese). Base de dados de problemas no formato do von (todos os problemas em português).

In order to the PUIDs be generated correctly the function inferPUID in the puid.py file from von must be modified as shown below. Para que os PUIDs sejam gerados corretamente a função inferPUID no arquivo puid.py do von deve ser modificada conforme mostrado abaixo.

# Regular expressions for Brazilian Olympiads
re_obmep = re.compile(r"OBMEP N(?P<level>[123]) F(?P<phase>[12]) (?P<year>\d{4})/(?P<problem>\d+)")
re_canguru = re.compile(r"Canguru (?P<category>Benjamim|Cadete|Júnior) (?P<year>\d{4})/(?P<problem>\d+)")
re_obm = re.compile(r"OBM (?P<year>\d{4})/(?P<problem>\d+)")
re_obm_banco = re.compile(r"OBM Banco (?P<year>\d{4})/(?P<problem>\d+)")

def inferPUID(source: str) -> str:
    source = source.replace("Finals", "F")
    # Check for the Brazilian Olympiad formats first
    for regex in [re_obmep, re_canguru, re_obm, re_obm_banco]:
        if (m := regex.match(source)) is not None:
            d = m.groupdict()
            # Extract the last two digits of the year
            year_last_two = d['year'][-2:]
            # Construct PUID based on specific format
            if regex == re_obmep:
                contest = "OBMEP"
                level_phase = f"N{d['level']}F{d['phase']}"
            elif regex == re_canguru:
                contest = f"Canguru{d['category']}"
                level_phase = ""
            elif regex == re_obm:
                contest = "OBM"
                level_phase = ""
            elif regex == re_obm_banco:
                contest = "OBMBanco"
                level_phase = ""
            source = f"{year_last_two}{lookup.get(contest, getOnlyAlphanum(contest))}{level_phase}P{getOnlyAlphanum(d['problem'])}"
            thresh = 11
            break
    else:
        # Your existing logic for other formats
        if source[0] == "H" and source[1:].isdigit():
            return source
        else:
            for k in sorted_lookup_keys:
                if source.startswith(k):
                    source = lookup[k] + source[len(k):]
                    thresh = 30
                    break
            else:
                thresh = 8
    source = getOnlyAlphanum(source)
    if len(source) <= thresh:
        return source
    else:
        # still too long, return some sort of hash
        return "Z" + (hashlib.sha256(source.encode("ascii")).hexdigest())[0:7].upper()