Re^2: Faster (but uglier) PWC 350-2 solutions

In the first loop you check for $i being A and in the second for $i being B hence $i / $witness .

That's a different interpretation° of the task B = A * k , the OP was only doing the A part.

Cheers Rolf
_{(addicted to the Perl Programming Language :)

see Wikisyntax for the Monastery}

°) According to Choroba that's the fun part of PWC ;-)

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Re^3: Faster (but uglier) PWC 350-2 solutions by ysth (Canon) on Dec 10, 2025 at 00:46 UTC
None of the examples involve finding A given B, but that just means they work either way. But to me the problem statement is clear: "the number of integers $i in the range $from <= $i <= $to that belong to at least $count different shuffle pairs".	[reply]
Re^4: Faster (but uglier) PWC 350-2 solutions by LanX (Saint) on Dec 10, 2025 at 02:05 UTC
See, different interpretations! All examples list the smallest A first. And B = A * k is explicit about k being an integer. For the other way k would need to be a fraction. Let's agree that only God knows. (and probably Choroba ;-) Cheers Rolf _{(addicted to the Perl Programming Language :) see Wikisyntax for the Monastery}	[reply]