function?X=decode（tempLx_max）
x=0;
for?i=1:L
????x=x+temp（i）*2^（i-1）;
end
x=x/（2^L-1）;
X=（（2*x_max）/（1-0））*（x-0）-x_max;
return

?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----
?????目錄???????????0??2011-08-12?06:11??NBPSO\
?????文件?????????135??2006-03-05?00:12??NBPSO\decode.m
?????文件?????????655??2011-08-11?05:08??NBPSO\display_result.m
?????文件?????????735??2011-08-11?05:40??NBPSO\evaluate.m
?????文件?????????859??2011-08-11?05:49??NBPSO\initialize.m
?????文件????????2439??2011-08-14?04:55??NBPSO\main.m
?????文件?????????509??2011-08-11?05:40??NBPSO\range_func.m
?????文件????????1028??2011-08-14?04:55??NBPSO\Read?me.txt
?????文件????????2494??2011-08-11?00:13??NBPSO\renewg_best.m
?????文件?????????726??2011-08-11?00:05??NBPSO\renewp_best.m
?????文件?????????195??2006-03-13?04:48??NBPSO\Untitled.m
?????文件?????????825??2011-08-12?03:30??NBPSO\update_v_p.m
?????文件????????1346??2011-08-17?07:18??license.txt