Note that this definition would not work for Double, e.g. 2.25=1.5^2 would be sent to 2^2=4, because 1.5 gets rounded to 2, although 2.25 is closer to 1 than to 4.
Thanks Matthew for naming the language making it more easier healing the issue
It's really weird since 4 Solvers have done it in Haskell Without raising an issue concerning it , But Anyway I've mentioned the Haskell Translator to do the necssary fixing the problem
Hope you all the bestMatthew On/Off CW , Happy to follow you Bro .. Regards .. Zizou
@cliffstamp @MrZizoScream
I am solving in Haskell. The description says (1 ≤ a, b, c ≤ 10), but I am getting input < 1, which makes the problem harder.
Please use spoiler flag, comments are visible from the dashboard. I added it for you this time.
This comment is hidden because it contains spoiler information about the solution
Of course, I was just commenting because for me, as a mathematician, it was on the first glance confusing that this actually works :D
To be fair, the type definition is Int -> Int, not (Num a) => a -> Int, so a Double would never be allowed into the function.
Note that this definition would not work for Double, e.g. 2.25=1.5^2 would be sent to 2^2=4, because 1.5 gets rounded to 2, although 2.25 is closer to 1 than to 4.
Very nice!
That's what all men see, lol
The boobs operator!
This is majestic.
This feels like a 8 kyu kata, not really a 7. Cheers.
Not hard to do by hand, just eta conversion and infix <-> prefix notation.
Definitely not a best practice in terms of code clarity though! :D
Fixed.
@mattmanj17 ..
@cliffstamp @MrZizoScream
I am solving in Haskell. The description says (1 ≤ a, b, c ≤ 10), but I am getting input < 1, which makes the problem harder.
This comment is hidden because it contains spoiler information about the solution
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