Ky Thuat Xu Li Bit Trong Pascal

7/21/2019 Ky Thuat Xu Li Bit Trong Pascal

1/10

H m nh phn trong my tnhnhng bi ton hay v x l bit

Con ngi thng x l nhng bi ton bng h m thp phn, nhng c l h m nh phn li l h m ca thch hn vi nhng chic my tnh. Trong PC ca bn, cc con s c th hin bng cc bit di dng nhphn (0,1). Khi chng ta nhp cc con s dui dng thp phn, my s x x l v a v h nh phn di dngcc bit m chng ta c th tm hiu l 0 v 1. V ch ti khi xut d liu ra (mn hnh hoc tp vn bn), cc smi c th hin li dng thp phn. Mi tp trong my thc cht l dng cc byte, trong c nhng byte giv tr c bit dng iu khin vic hin th.

Mt s ngi c th hiu t nhiu v bit v cc bi ton x l bit, mt s ngi th khng. Nhng iu khngphi l quan trng, khi m iu quan trng l chng ta s cp n mt vn tng chng nh rt phc tp(i vi nhng ngi cha bit nhiu v n), nhng tht ra li tht l n gin m hiu qu li rt cao trong khi xl nhng bi ton hay v kh. Sau y l nhng thng tin s lc v bit trong PC.

- Trong PC, mi s nguyn u c biu din h nh phn bi mt dy cc bit 0 v 1.

- Cc bit c m s vi nhng s hiu t phi sang tri.

7 6 5 4 3 2 1 0

- 1byte=8bit

- 1word=16bit

- Hm sizeof (x) = s byte ca s nguyn x.

- Cc php ton x l bit:

+ NOT x : php o bit, i cc gi tr trong mi bit ca x t 0->1,1->0.

+ x OR y : Php cng logic trn cc bit.Thc hin trn tng cp bit tng ng ca cc ton hng theo bng sau.

+ x AND y : Php nhn logic trn cc bit. Thc hin trn tng cp bit tng ng ca cc ton hng theo bng sau.

+ x XOR y : Php cng logic trn cc bit.Thc hin trn tng cp bit tng ng ca cc ton hng theo bng sau.

+ x SHR i : Php dch phi, cho gi tr c c t s nguyn x sau khi dch sang phi i bit.


2/10

+ x SHL i : Php dch tri, cho gi tr c c t s nguyn x sau khi dch sang tri i bit.

Trn y l mt s php ton lm vic trn cc bit m ta hay dng, trn c s , ta xy dng c mt s hm,th tc hay dng sau.

- Hm ly bit.

Function LayBit(x:word; i:byte):byte;

Begin

LayBit:=(x SHR i) and 1

End;

Hm tr v gi tr l bit th i ca s nguyn x.

- Th tc bt bit.

Procedure BatBit(Var x:word; i:byte);

Begin

X:=x OR (1 SHL i)

End;

Th tc gn tr 1 cho bit th i trong s nguyn x.

- Th tc tt bit.

Procedure TatBit(Var x:word; i:byte);

Begin

X:=x AND (NOT(1 SHL i))

End;

Th tc gn tr 0 cho bit th i trong s nguyn x.

- Hm Ly s.

Function LaySo(x:word;j,i:byte):byte;

Var

K:byte;

Begin

K:= (8*sizeof(x) - j - 1);

LaySo := (x SHL k) SHR (i+k)

End;

Hm tr v gi tr l s to bi cc bit t tri sang phi, t j -> i ca s x vi iu kin 8*sizeof(x)> j i 0.


3/10

Tip n, chng ta s gii quyt mt s bi ton khhay v x l bit.

Bi ton: Sp xp dy.

Cho tp Data.dat gm cc s nguyn khng m nh hn 500 000 i mt khc nhau. Hy xp tp theo th t tngdn v a ra tp Result.dat.

Bi ton trn khin ta cm thy kh chu v gii hn d liu ca n, nu ta ngh n vic dng mng lu tr vxp xp thun tu th qu l ng ngn bi v b nh 64K chng thm thp vo u v ring vic xp xp chim

mc thi gian k lc ri. Ta cng c th dng phng php xp xp ngoi Merge Sort (dng file sp xp), tuynhin, y khng phi l gii php hu hiu v vn thi gian. Tuy nhin, phng php sp xp nh du bngmt vi th thut dng bit th s li cho ta mt kt qu kh l mmn. Ta ch cn dng mt mng A kiu byte vikhong 63000 phn t l tha x l bi ton ny. Ta tng tng mng A l mt on tu, v mi toa tu lmt phn t c 8 ch ngi, ch ngi c nh du l 1 nu c ngi ngi, l 0 nu ngc li. Banu on tukhng c ngi ngi, ta gn mng bng 0. V vi mi phn t c t file ra, ta nh du bit tng ng trong mngcho ti khi ht file bng th tc batbit. ly kt qu ghi ra file, ta dng function laybit. Sau y l chng trnhth hin thut ton:

{$A+,B-,D+,E+,F-,G-,I+,L+,N-,O-,P-,Q+,R+,S+,T-,V+,X+}

PROGRAM XAP_XEP_DAY_VOI_DU_LIEU_LON;

Const

Datin = 'Data.dat';

Datout = 'Result.dat';

Bt=8;

MAX=63000;

Var

a:array[0..MAX] of byte;

f:text;

Procedure Finish;

Var i,j:longint;

Function LayBit(x,j:longint):byte;

Begin

LayBit:=(x shr j) and 1;

End;

Begin

Assign(f,DATOUT);rewrite(f);

For i:=0 to max do

For j:=0 to bt-1 do

If LayBit(a[i],j)=1 then Write(f,(i*bt+j),' ');


4/10

Close(f)

End;

Procedure Start;

Var i,tg:longint;

Procedure BatBit(x,y:longint);

Begin

A[x]:=a[x] or (1 shl y)

End;

Begin

Fillchar(a, sizeof (a), 0);

Assign(f,DATIN);reset(f);

While not seekeof(f) do

Begin

Read(f,tg);

BatBit(tg div bt,tg mod bt)

End;

Close(f)

End;

BEGIN

Start;

Finish

END.

Mt s bi tp tng t.

Bi s 1:Trn tp.

Cho 2 tp In1.dat v In2.dat cha cc s nguyn khng m c th rt ln (khng ln hn 500 000). Hy to tpOut.dat cha cc s c mt trong 2 tp trn, tuy nhin, khng c trng nhau.

Bi s 2:Ta xu.

Cho file vn bn Str.inp gm nhiu dng, midng khng qu 30 k t, ch cha hai k t * v . Hy loi bbt cc dng ging nhau ri in kt qu ra file Str.out gm cc dng khc nhau.

Bi s 3:M s nhn vin


5/10

Tng gim c mt cng ty X ni ting l ngi k lng. ng ta thc hin vic qun l nhn vinca mnh bng cch gn cho mi nhn vin mt m s. Cng ty c N nhn vin th mi nhn vin i (i=1,2,,N) c mt m s. M s l mt s nguyn dng (hai nhn vin khc nhau phi c m s khc nhau).Do bn i cng tc mt thi gian di nc ngoi nn ng ta giao li quyn qun l cho mt ngi khc.Khi ng tr v, cng ty c thay i s lng nhn vin, khi tip nhn thm nhn vin mi, ng ta yucu mun bit m s nh nht c th gn cho nhn vin mi.

Yu cu:

Cho N m s cacc nhn vin trong cng ty. Hy tm m s nh nht cha xut hin trong N m s cho.

D liu vo t tp vn bn CODE.INP

+ Dng u l s N (1


6/10

Mi s t nhin u c th biu din c di dng nh phn. i vi my tnh, cc gi tr s c lu bi cc

bin di dng chui nh phn c di tng ng vi kiu bin .

Trong ngn ng Pascal, mt s kiu ph bin nh:

byte: di 8 BIT

integer: di 2*8 = 16 BIT

longint: di 4*8 = 32 BIT

Cc BIT ca bin c nh s t phi sang tri bt u t 1.

u im ca x l BIT:

1/ B nh:X l BIT c th c dng lm mng nh du. Thay v s dng 1 bin Boolean ch nh du

c 1 phn t l True hay False, ta c th x l BIT nh du 8 BIT tng ng vi 8 phn t.

2/ Tc :Cc php x l BIT c tc nhanh hn cc php x l khc.V d:Hai php (x div 2)v (x shr

1)l nh nhau nhng php(x shr 1)c tc nhanh hn. Trong cc bi ton i hi vic thay i trng thi

nhiu ln th ngi ta vn hay dng x l BIT ci thin tc chng trnh.

Sau y l mt s php x l BIT c bn:

1/ Cng BIT(or):

Kt qu bng tng gi tr 2 BIT. Trng hp 2 BIT u c gi tr bng 1 th kt qul 1.

V d:

x1 = 9 (00001001)

x2 = 18 (00010010)

x1 or x2 = 27 (00011011)

2/ o BIT (not):

o BIT0thnh BIT 1v BIT 1thnh BIT 0.

V d:

x1 = 13 (00001101)

not (x1) = 242 (11110010)

3/ Nhn BIT (and):

Kt qu bng tch gi tr 2 BIT.

V d:

x1 = 9 (00001001)

x2 = 12 (00001100)

x1 and x2 = 8 (00001000)

4/ Loi tr BIT (xor):

Kt qu l0nu 2 BIT c cng gi tr, l1nu 2 BIT khc gi tr.

V d:

x1 = 9 (00001001)

x2 = 18 (00010010)

x1 xor x2 = 27 (00011011)


7/10

5/ Dch sang tri k BIT (shl):

V d:

x = 27 (00001101)

x shl 2 = 108 (01101100)

6/ Dch sang phi k BIT (shr):

V d:

x = 27 (00001101)

x shr 2 = 6 (00000110)

T cc php ton trn, ta c cc cng thc m rng sau:

7/ Ly gi tr BIT:

Cho bit BIT thkca sxc gi tr bao nhiu:

Function GetBIT(k , x : LongInt) : LongInt;

begin

GetBIT := (x shr (k-1)) and 1;

end;

8/ Gn gi tr BIT:

Gn gi trccho BIT thkca sx:

Procedure SetBIT(c , k : LongInt; var x : LongInt);

begin

if c = 1 then x := x or (1 shl (k-1))

else x := x and (not (1 shl (k-1)));

end;

X l s nguynln

Posted bybasicalgorithmon December 6, 2009

X l s nguyn ln l mt k nng khng th thiu ca mt th sinh tham gia k thi HSGQG. Biton thng lin quan ti vic cng/tr/nhn vi cc s nguyn c nhiu (khong vi trm, vi

nghn) ch s. Bi vit ny xin c cung cp cho cc bn cch thc hin cc php ton vi s nguyn

ln.

1/ tng:Chng ta s thc hin cc php ton ny nh cch lm m hi cp 1 c hc. l thc

hin cc php ton ln lt t phi qua tri v s dng thm mt bin nh.

2/ Khai bo:Cc php ton s c thc hin trn mng, do ta cn xy dng mng 1 chiu c kch thc

l s ch s ti a ca bi ton. Mi phn t ca mng s l 1 ch s. Ngoi ra, cn khai bo binbasel hc s m chng ta dng khi thc hin cc php ton. Trong v d ny, ti tbase = 10 tha mn mi

phn t ch cha 1 ch s.
http://basicalgorithm.wordpress.com/http://basicalgorithm.wordpress.com/http://basicalgorithm.wordpress.com/http://basicalgorithm.wordpress.com/


8/10

Const

maxn = 100; // S ch s ti a

base = 10; // H c s khi s dng

Type

BigNum = array[0..maxn] of LongInt; // Kiu s nguyn ln

3/ Php cng:

Function Plus(x , y : BigNum) : BigNum; //Kt qu l s nguyn ln

var

i , nho : LongInt;

begin

Fillchar(Plus , SizeOf(Plus) , 0); // Khi gn Plus = 0

nho := 0; // Khi gn nho = 0

For i := maxn downto 1 do // Cng ln lt tng ch s t phi qua tri

begin

Plus[i] := x[i] + y[i] + nho; // Cng thc cng nh cp 1

nho := Plus[i] div base; // Tnh li bin nho cho ln cng tip theo

Plus[i] := Plus[i] mod base; // m bo mi phn t ch lu mt ch s

end;

end;

phc tp:O(N)

4/ Php tr:

Mt iu ng lu khi thc hin php tr l bin nh ch nhn1trong 2gi tr:0hoc1. Ngoi ra,

m bo kt qu ng, bn cn ch khi lyX Yth X >= Y(c th vit 1 hm kim tra trc khi thc

hin php tr).

Function Minus(x , y : BigNum) : BigNum;

var

i , nho : LongInt;

begin

Fillchar(Minus , SizeOf(Minus) , 0); // Khi gn Minus = 0

nho := 0;

For i := maxn downto 1 do

begin

Minus[i] := x[i] + base y[i] nho; // Cng thm base m bo Minus[i] >= 0

if x[i] < y[i] + nho then nho := 1 else nho := 0; // Tnh li bin nho

Minus[i] := Minus[i] mod base; // m bo mi phn t ch cha 1 ch s

end;

end;

phc tp:O(N)


9/10

5/ Php nhn:

Chng ta vn thc hin cch lm nh cp 1: Ly tng ch s ca tha s th hai nhn vi tha s th nht,

c bao nhiu cng vo kt qu. Ch sau mi ln nhn 1 ch s ca tha s th hai vi tha s th nht, ta

cn li kt qu sang tri 1 ch s.

Ngoi ra, s ch s ti a ca kt qu s ltng s ch s ca 2 tha s. V th, ta cn m bo ln cho

kt qu trnh bRangeCheck. Tt nht bn hy khai bomaxnln bng 2 ln di ti a ca 2 tha s.

Function Multi(x , y : BigNum) : BigNum;

var

i , j , nho : LongInt;

Temp : BigNum;

begin

Fillchar(Multi , SizeOf(Multi) , 0); // Khi gn Multi = 0

nho := 0;

count := -1; // S ch s phi li vo trc khi nhn l -1For i := maxn downto 1 do

begin

Inc(count); // Tng s lng ch s cn li vo lt nhn th i

Fillchar(Temp , SizeOf(Temp) , 0); // Mng nhn ch s th i ca y vi x

For j := maxn downto (maxn div 2 + 2) do

begin

Temp[j-count] := y[i]*x[j] + nho;

nho := Temp[j-count] div base;

Temp[j-count] := Temp[j-count] mod base;

end;

Minus := Plus(Minus , Temp);

end;

end;

phc tp:O(N^2)

6/ Ci tin:

Khi thc hin php cng/tr, ta nhn thy mi phn t ca kt qu thuc kiuLongInt, c th lu c 9 ch

s. Do , ta c th tbase = 10^9nhm gim i s lng php tnh. S lng phn t ti a ca

mngBigNumcng gim i ~9 ln.

Tuy nhin, khi in ra, do mi phn t lu 9 ch s nn ta cn in ra c 9 ch s ny (tnh c cc ch s 0 v

ngha u).

Procedure Print;

var

i , j : LongInt;

s : string;

begin


10/10

For i := 1 to maxn do

if Res[i] 0 then break; // Tm phn t khc 0 u tin ca kt qu

write(Res[i]);

For j := i + 1 to maxn do

begin

str(Res[j] , s); // Chuyn Res[j] sang xu s

while length(s) < 9 do s := 0 + s;// Thm cc ch s 0 vo u m bo 9 chs

write(s);

end;

end;

Ky Thuat Xu Li Bit Trong Pascal

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