Hah, good job, you found it! Puzzling.SE is proud :D'
This was essentially just a guided solution from the OP's clues!
I had never come across this before, but when he talks about sprinkling pie with irrational sauce he means the reference in the clue he gave:
Index pi according to the unique numbers in it (3,1,4,15,...). pi rational numbers. We use 15 because 1 was already found. This makes the result: 43195195867462520687356193644029372991880
Computationally, this gets hard quickly. But when I used this as in a cipher it returns:
Hah, good gn]3 xxt epsil\x1cgw"$Vwtxnlfg2SC$
You'll note that the first ten characters make sense, and that the OP has a reference to 10 o'clock.
So using this, I
Reused the first ten digits of this pi rational number over and over with the output given.
Previous work for the interested...
Here's a glorious failure for others to avoid (in the spirit of @Excited Raichu's comment to the question...)
So, pie and cipher. Or π and cipher. Mmm. And computer programming. Could it be that we should shift out the cipher letter-wise by the digits of π?
Well, that means shifting:
Ldi5%hxtl&nrc5%zxz(lsxom%j}&(Vy}{unop3[K$lt)usxzl&>G
by
31415926535897932384626433832795028841971693993751058
which is:
Ice4 _vng#ijZ.wvw hmvii"gu#&Opx{sfgl2RD#fk&ljusg%>B
Shifting the other way, or using latter combinations of pi (up to about 30 digits later) still leaves nonsense.
So it isn't that!
Edit: The kind of calculation I did was (in Python):
code="Ldi5%hxtl&nrc5%zxz(lsxom%j}&(Vy}{unop3[K$lt)usxzl&>G"
pi="31415926535897932384626433832795028841971693993751058209749445..."
decode=""
for i,c in enumerate(code):
decode=decode+ chr(ord(c)+int(pi[i])
where chr() converts a decimal to ascii and ord() converts an ascii to decimal.
I also subtracted and did offsets. Also, I assumed that the "I love programming..." is a clue. Since this is baking, we're just going to add in a bunch of ingredients. The trouble is that if we just add stuff together we rapidly get beyond the 128 that is associated with ascii.
So, for example, I can run
> clue.split()
['I', 'love', 'programming,', "it's", 'so', 'fun', 'Just', 'get', 'every', 'computer', 'in', 'the', 'world', 'And', 'just', 'do', 'program', 'as', 'much', 'as', 'you', 'can', 'Because', 'with', 'everyone', 'on', 'your', 'side', 'you', 'can', 'never', 'go', 'wrong', 'with', 'programming', 'And', 'no,', 'this', 'not', 'clue', 'to', 'the', 'puzzle!', "I'm", 'just', 'saying', 'programming', 'is', 'the', 'best', 'Please', 'do', 'program,', 'everyone', 'will', 'help', 'you', 'with', 'your', 'problems', 'Especially', 'StackOverflow,', 'great', 'right?', 'Now', 'go', 'on,', "here's", 'the', 'cipher,', 'ever', 'will', 'you', 'crack', 'it?', 'At', '10', "o'", 'clock', "i'm", 'checking', 'to', 'see', 'if', 'you', 'are', 'done.']
and the use the number of characters in each word. Also, hundred of combinations just down this path: what do we do with carriage returns? What do we do with punctuation? What do we do with numerals? Should "o'" and "clock" be separate words? etc. But running various combinations with this didn't reveal anything promising that could be tweaked.
Another option is to have the clue converted character-by-character into ascii. But then what? You can't add those numbers because they will go above 128. "Add" is implied by the baking clues - average feels uncontrolled (what do you do with the 0.5s). XOR of the binary? OR? AND? Again AND is implied by the "baking", but XOR is much more common in crypto for obvious reasons (that you either switch the bit or not).
And then add the digits of pi (to the ascii decimal? or the ascii hex? or the ord of the digits of pi).
I tried various combinations of the above and, of course, it was all nonsense. Is the "knowledge" a knowledge of the ascii codes? Or something else? Is the "I love programming..." needing to be interpreted in (hard to understand) english? Or read as a string of ascii characters? Are there other ingredients to be added to the pie? Other "irrational sauces"? e? The golden section? Some root?
I'm sure this seemed obvious to the OP, but the number of possibilities is relatively large here. Sorry if I'm missing the "big clues". Gasp, it's a diagonal! Yay! Now I could do that all again but with 10 times the number of permutations!
Ah, well. Maybe I'll guess the next one...