43
Dmitry Khromenko
Last seen 1 hour ago
Member for 3 years, 4 months, 28 days
Difficulty Normal
I study Python myself and help my son and daughter learn programming.
First-PelmenFloopov
1
len(items) // 2 + 1 if len(items) % 2 != 0 else len(items) // 2
you can change to
len(items) // 2 + len(items) % 2
boolean expression is recognized as a number
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Suitable place to use **yield**
def find_divisors(n):
for i in range(1, n):
if n % i == 0:
yield i
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sum([]) can be used without square brackets sum(), its more shorter
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```
0| a
1|<--x-->b *
2| c *
3| d n
4| e<--y-->m
5| f l
6| g k
7| h j
n| i
```
s = side -1
x = abs(s-n)
y = 2s-2abs(s-n)-1
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instead "total_time = datetime(2010, 1, 1, 1, 1, 1) - datetime(2010, 1, 1, 1, 1, 1)"
you can use "total_time = timedelta()"
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len(set(elements)) == 1 or len(set(elements)) == 0
can replace
len(set(elements)) < 2
and the solution could be:
return len(set(elements)) < 2
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