(03-26-2015, 08:31 AM)YinYin Wrote: blerghDid it anyways :P
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PYTHON-Code:
def dlf2(e,d='',a=123,p=0,k="odBearBecauseHeIsVeryGoodSiuHungIsAGo"): return (lambda:d, (lambda:dlf2(e, d+chr(ord(e[a])-ord(k[p])), a+1, p+1),lambda:dlf2(e, d+chr(ord(e[a])-ord(k[0])), a+1, 1))[int(p == len(k))])[a != len(e)]() |
I personally don't like C# either, but it proves to be really useful for when you'd like to do something in a hurry.
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