Matematická úloha
Delphi & Pascal (česká wiki)
Kategória: KMP (Klub mladých programátorov)
Autor: Ľuboš Saloky
Program: Prasata.pas, Prasata2.pas, Prasata3.pas, Prasata5.pas
Súbor exe: Prasata.exe, Prasata2.exe, Prasata3.exe, Prasata5.exe
Autor: Ľuboš Saloky
Program: Prasata.pas, Prasata2.pas, Prasata3.pas, Prasata5.pas
Súbor exe: Prasata.exe, Prasata2.exe, Prasata3.exe, Prasata5.exe
Matematická úloha
{ prasata3.pas } { Matematicka uloha: Mame 2 miestnosti a 3 prasiatka v jednej } { z nich. Kazdu minutu prejde jedno prasiatko do druhej miestnosti. } { Za aky cas budu vsetky tri prasiatka v druhej miestnosti? } { } { Author: Ľuboš Saloky } { Datum: 01.01.1996 http://www.trsek.com } {$G+} program Tri_prasiatka_a_assembler; const nc:array[1..10,1..20] of byte= {nahodne cisla} ((8,5,1,4,9,5,7,5,0,4,1,5,0,5,9,3,6,8,4,9), (3,5,2,2,9,7,2,8,3,7,3,6,7,5,2,7,1,6,5,5), (4,0,1,1,7,2,4,3,4,6,3,2,1,8,9,8,7,0,7,4), (6,0,8,7,5,9,7,5,4,5,6,8,6,5,2,9,2,1,6,0), (0,3,7,6,9,3,2,5,5,1,7,4,8,0,1,4,1,6,9,2), (9,4,0,7,7,9,2,2,0,9,3,1,5,0,6,6,9,6,0,6), (8,4,3,7,6,0,0,6,6,6,9,4,1,4,5,2,2,3,9,9), (2,4,3,5,4,1,1,3,4,7,9,0,9,2,0,0,3,8,6,4), (8,1,1,4,9,4,3,2,9,1,1,0,8,1,7,4,4,3,4,8), (1,0,6,3,1,7,8,2,3,8,0,0,1,5,7,4,6,2,5,9)); var ncOfs,tpOfs,x:word; tp:array[1..300] of word; {tabulka pocetnosti} suc:longint; BEGIN {012=1. prasa, 345=2., 678=3., 9 ignoruj} FillChar(tp[1],SizeOf(tp),#0); ncOfs:=Ofs(nc)+200; tpOfs:=Ofs(tp); asm mov di,ncOfs {indexovanie v poli nc} sub di,200 mov cx,10000 {pocet pokusov} @DalsiPokus: mov bx,0 {indexovanie v poli tp a pocet minut} mov al,00000111b {tri jednotky su tri prasiatka} @Znova: cmp al,0 je @Hotovo mov ah,[di] inc di cmp di,ncOfs jl @Dalej sub di,200 {ak je koniec tabulky} @Dalej: cmp ah,9 je @Znova {pre ah=9 ignoruj} cmp ah,6 jl @NieTretie xor al,00000001b {tretie prasa} add bx,2 jmp @Znova @NieTretie: cmp ah,3 jl @NieDruhe xor al,00000010b {druhe prasa} add bx,2 jmp @Znova @NieDruhe: xor al,00000100b add bx,2 jmp @Znova @Hotovo: inc word[tpOfs+bx] loopne @DalsiPokus end; suc:=0; for x:=1 to 300 do suc:=suc+x*tp[x]; writeln('Priemerny cas: ',suc/10000:2:2); END.