optimization....help
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Hello all,
Can the below loop be optimized further?
I know that takeing the multiplies out should do it
but am at a lose on this.......
----- begin -----------
count =3D 1=20
across =3D 262
down =3D 236
seq =3D repeat(0,down) =20
=20
for x =3D 1 to down do =20
seq[x] =3D r[count..across * x]=20
count =3D across * x + 1
end for
--- end -------------------
euman at bellsouth.net
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<DIV><FONT face=3DArial size=3D2>Hello all,</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2>Can the below loop be optimized=20
further?</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2>I know that takeing the multiplies out =
should do=20
it</FONT></DIV>
<DIV><FONT face=3DArial size=3D2>but am at a lose on =
this.......</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2>----- begin -----------</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2>count =3D 1 </FONT></DIV>
<DIV><FONT face=3DArial size=3D2>across =3D 262</FONT></DIV>
<DIV><FONT face=3DArial size=3D2>down =3D 236</FONT></DIV>
<DIV><FONT face=3DArial size=3D2> </DIV></FONT>
<DIV><FONT face=3DArial size=3D2>seq =3D=20
<DIV><FONT face=3DArial size=3D2>for x =3D 1 to down =
do =20
<BR> seq[x] =3D r[count..across * x] =
<BR> =20
count =3D across * x + 1</FONT></DIV>
<DIV><FONT face=3DArial size=3D2>end for</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2>--- end =
-------------------</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2><A=20
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