


import pytest

from datasets.splits import SplitDict, SplitInfo
from datasets.utils.py_utils import asdict







@pytest.mark.parametrize(
    '''split_dict'''	,[
        SplitDict(),
        SplitDict({'''train''': SplitInfo(name='''train'''	,num_bytes=1337	,num_examples=42	,dataset_name='''my_dataset''')}),
        SplitDict({'''train''': SplitInfo(name='''train'''	,num_bytes=1337	,num_examples=42)}),
        SplitDict({'''train''': SplitInfo()}),
    ]	,)
def 						lowerCamelCase(   a__):
				_SCREAMING_SNAKE_CASE			  =split_dict._to_yaml_list()
				assert len(a__) == len(a__)
				_SCREAMING_SNAKE_CASE			  =SplitDict._from_yaml_list(a__)
				for split_name, split_info in split_dict.items():
								# dataset_name field is deprecated, and is therefore not part of the YAML dump
								_SCREAMING_SNAKE_CASE			  =None
								# the split name of split_dict takes over the name of the split info object
								_SCREAMING_SNAKE_CASE			  =split_name
				assert split_dict == reloaded







@pytest.mark.parametrize(
    '''split_info'''	,[SplitInfo(), SplitInfo(dataset_name=a__), SplitInfo(dataset_name='''my_dataset''')])
def 						lowerCamelCase(   a__):
				# For backward compatibility, we need asdict(split_dict) to return split info dictrionaries with the "dataset_name"
				# field even if it's deprecated. This way old versionso of `datasets` can still reload dataset_infos.json files
				_SCREAMING_SNAKE_CASE			  =asdict(SplitDict({'''train''': split_info}))
				assert "dataset_name" in split_dict_asdict["train"]
				assert split_dict_asdict["train"]["dataset_name"] == split_info.dataset_name



snake_case_						:			dict[tuple[int, int, int], int]													=							{}







def 						lowerCamelCase(   a__	,a__	,a__):

				# if we are absent twice, or late 3 consecutive days,
				# no further prize strings are possible
				if late == 3 or absent == 2:
								return 0

				# if we have no days left, and have not failed any other rules,
				# we have a prize string
				if days == 0:
								return 1

				# No easy solution, so now we need to do the recursive calculation

				# First, check if the combination is already in the cache, and
				# if yes, return the stored value from there since we already
				# know the number of possible prize strings from this point on
				_SCREAMING_SNAKE_CASE			  =(days, absent, late)
				if key in cache:
								return cache[key]

				# now we calculate the three possible ways that can unfold from
				# this point on, depending on our attendance today

				# 1) if we are late (but not absent), the "absent" counter stays as
				# it is, but the "late" counter increases by one
				_SCREAMING_SNAKE_CASE			  =_calculate(days - 1	,a__	,late + 1)

				# 2) if we are absent, the "absent" counter increases by 1, and the
				# "late" counter resets to 0
				_SCREAMING_SNAKE_CASE			  =_calculate(days - 1	,absent + 1	,0)

				# 3) if we are on time, this resets the "late" counter and keeps the
				# absent counter
				_SCREAMING_SNAKE_CASE			  =_calculate(days - 1	,a__	,0)

				_SCREAMING_SNAKE_CASE			  =state_late + state_absent + state_ontime

				_SCREAMING_SNAKE_CASE			  =prizestrings
				return prizestrings







def 						lowerCamelCase(   a__ = 30):
				return _calculate(a__	,absent=0	,late=0)


if __name__ == "__main__":
							print(solution())