pip download pippython pip-20.1.1-py2.py3-none-any.whl/pip install pip-20.1.1-py2.py3-none-any.whlpip freeze > r.txt
pip download -r r.txt
import math
a=4.5
x=-1.9
print("y = f(x):\n\t| y = math.tan(a)+x, если x < 0\n\t| \
y = math.(2*a)**0.5-math.sin(x), если 0 < = x < =1\n\t| y = x**3-a**3, если x > 1")
if x < 0:
y = math.tan(a)+x
print(y)
elif 0 <= x <= 1:
y = (2*a)**0.5-math.sin(x)
print(y)
elif x >1:
y = x**3-a**3
print(y)
Как сделать реверс
>>> x
array([[1, 2, 3],
[4, 5, 6],
[7, 8, 9]])
>>> x.T
array([[1, 4, 7],
[2, 5, 8],
[3, 6, 9]])
if(str - 1 >= 0 && mtrx[str-1][col] != 0){ можно невозбранно выкинуть str - 1 >= 0, это условие у вас всегда выполняется, потому что двумя строками выше условие выполнения цикла: str>0;
>>> import itertools
>>> four_numbers_combinations = list(itertools.permutations(eight_numbers_list, 4))
>>> len(four_numbers_combinations)
1680
>>> four_numbers_combinations[:10]
[(1, 2, 3, 4), (1, 2, 3, 5), (1, 2, 3, 6), (1, 2, 3, 7), (1, 2, 3, 8), (1, 2, 4, 3), (1, 2, 4, 5), (1, 2, 4, 6), (1, 2, 4, 7), (1, 2, 4, 8)]
>>> import datetime
>>> from dateutil import relativedelta
>>> past_date = '2017-07-31'
>>> past_date = datetime.datetime.strptime(past_date, '%Y-%m-%d')
>>> today = datetime.datetime.now()
>>> diff = relativedelta.relativedelta(today, past_date)
>>> f'{diff.years} years, {diff.months} months and {diff.days} days'
'2 years, 8 months and 17 days'