#๐ optimization
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@modern anchor
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a represents the number of pigeons a surface can hold
i really dont see how much faster my code can become
What's this?
if the number of pigeons at any time exceeds the number that it can hold, print PLAN REJECTED otherwise print PLAN ACCEPTED
i changed that part this is an earlier version
just pretend i did it without alst
Can you show the current version?
Can you show the full task? Does it tell you how large the input is?
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