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ec1:ClrHome :0→P :2→N:1→I :24→T :0→F :Repeat PN=570 or I=8 :real(8) :Ans→T :F+(I≥0)6/(2-(sub(Str1,54abs(I)+N+P,1)="U"))(max(T={4,7,8}))not(F)(sub(Str1,54abs(I)+N+P,1)≠" →F :If I≥1:Then :sub(Str1,54I-53+N+P,1)=" →D :sub(Str1,54I-55+N+P,1)=" →E :Else :1→D :1→E :End :P+(P≠38 and DN=12 and max(T={3,6,8→P :N-(max(T={2,5,7}) and EN>1)+(max(DT={3,6,8}) and ((P=38 and N<15) or N<12→N :F(I≤1)+F(I>1)(sub(Str1,54abs(I)+N+P-108(I>1),1)=" →F :I+(I<0)+not(F)(sub(Str1,54abs(I)+N+P,1)=" ")(I≥0)-not(not(F→I :F-not(not(F→F :Output(1,1,sub(Str1,1+P,16)+sub(Str1,55+P,16)+sub(Str1,109+P,16)+sub(Str1,163+P,16)+sub(Str1,217+P,16)+sub(Str1,271+P,16)+sub(Str1,325+P,16)+sub(Str1,379+P,16 :If 3.5≥abs(4.5-I:Then :Output(I,N,"N :End :End :Return c2 |
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ec1:(sub(Str1,54I-53+N+P,1)=" →D :(sub(Str1,54I-55+N+P,1)=" →Ec2 |
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ec1:P+(P≠38 and DN=12 and max(T={3,6,8→Pc2 |
QUOTE (Radical Pi @ 31 Dec, 2007, 17:24) | ||||
c1-->
ec2 -->I might be mistaken, but on these two lines you could remove the opening parenthesis. Why are they even there? EDIT:c1
ec2 Reordering FTL- of two bytes :P ![]() |
QUOTE (Radical Pi @ 31 Dec, 2007, 17:36) |
= operations, and all other 'tests' for that matter, always produce a binary result. |
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ec1:Repeat I=8 or (P=38 and N=15c2 |
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ec1:Repeat PN=570 or I=8c2 |