Python Challenge - Level 0
Problem
Hint: try to change the URL address.
Explore
As the "0" indicates, this is just a warm up.
Loop
The naive way to solve it: multiply by 2 in a loop:
>>> k = 1
>>> for i in range(38):
... k *= 2
...
>>> k
274877906944
Power
Use ** for power:
>>> 2**38
274877906944
In REPL the result will be printed; or you can explicitly print the result by calling print()
>>> print(2**38)
274877906944
Instead of **, you can also use pow()
>>> pow(2,38)
274877906944
or
>>> print(pow(2,38))
274877906944
Shift
Multiply by 2 is equivalent to shifting the binary representation left by one:
>>> 1<<38
274877906944
Numeric Types
Done!
Wait, why 38? what is implied?
If you are coming from C/C++, Java or other languages, you know that there are multiple types just for integers: short, integer, long, and even BigInteger beyond 64-bit. However that is not the case in Python:
>>> type(2**3)
<class 'int'>
>>> type(2**38)
<class 'int'>
>>> type(2**380)
<class 'int'>
So 38 is a good(and random) example to show a int larger than 32-bit.
Similar for float type, in Python it can be arbitrarily large, no double needed
>>> type(2**3.8)
<class 'float'>
>>> type(2.0**38)
<class 'float'>
Complete Solution
print(2**38)
Output:
274877906944
Next Level
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