Hi, haukex.
*while (/(?=($re))./g) { ... }*
"Look for a fib anywhere ahead, then if found, consume its leftmost digit, then repeat" probably makes no sense, even for this assignment/homework. The `%fib` was built but not actually used. The "3 element" part of assignment (not addressed in your code) could be either consecutive, or just ascending fibs. But, the way `%fib` was constructed, it may be to help looking for a fib whose initial digits are a fib too, whose initial digits, ..., etc. While leftmost digit of any 2-digit fib is a fib, -- I wonder if any value of `%fib` could be an array with more than 3 elements? :)
P.S. Please, I'm not appealing to rewrite anything, don't waste your time, the task is too much unclear.
P.P.S. And, I suspect the "PERFECT SQUARE" in OP, in C, won't work for 6-digit numbers and `int` type, because of overflow?
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*The *`%fib` was built but not actually used.
Well not quite, it is used for what you said:
*to help looking for a fib whose initial digits are a fib too*
This is what I was using it for, since I didn't feel like getting too fancy with the regex. The code I wrote was mostly just an implementation of my interpretation of the description the OP gave:
*"Check 4 is fibonacci number , no , go ahaed 49 is fibonacci number , no, next 496 is fibonacci ,no, 4969 after 44693 [sic] ,after 496934 no.There isnt any fibonacci , go to next digit and do it again 9 , 96,969,9693,96934."*
*I wonder if any value of *`%fib` could be an array with more than 3 elements?
Good question, I didn't try cranking up `$DIGITS` past 60 yet...
*the task is too much unclear*
Agreed, which is why I just implemented the part I (thought I) understood ;-)
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