Chuong 1 Giói Thiệu R

Nguyn Quang Ngc

Chng 1 : Thut ton

1.1 Mn hc lp trnh :

Gii mt bi ton bng my tnh.

- my tnh gii c mt bi ton cn c cc ch

dn.

- Mi ch dn c vit bng mt chui k hiu m

my tnh thc hin c.

- Mi ch dn c gi l mt thao tc.

- Ngi a ra mt dy cc thao tc my tnh gii

mt bi ton (ngi hng dn my tnh) c

gi l ngi lp trnh.

1.2 Thut ton :

- L cc ch dn cho my tnh my tnh gii quyt

mt bi ton no .

- Cc ch dn c th c vit bng m gi, lu ,

bng ngn ng t nhin. Sau cc ch dn ny s

c vit li di dng ngn ng m my tnh thc

hin c, gi l ngn ng lp trnh.

Cc ngn ng lp trnh thng dng: VB, C, C++, Java, ...

Thut ton c trnh by nh sau :

Thut ton :

B1. Ch dn 1

B2. Ch dn 2

. . .

Bn Ch dn n

Kt thc .

Cc ch dn s c my tnh thc hin theo th t B1, B2, . . , Bn

1.3 Biu thc i s trong ton (nhc li) :

1.3.1 Mt biu thc i s (trong ton hc) c xy dng t :

a) cc s nguyn ,hu t ,thc . . . m ta gi l hng s.

b) cc bin x, y, . . . c th ly gi tr l cc hng s.

c) cc php ton (+, -, * (nhn) , / , , logab, . . .) thao tc trn cc hng s v cc bin theo mt th t nht nh.

V d :

a) 5 l mt biu thc i s.

b) x l mt biu thc i s.

c) x + 5 l mt biu thc i s.

d) x2 + 3y - || - 2 l mt biu thc i s.

1.3.2 Gi tr ca mt biu thc i s (trong ton hc)

c xy dng t :

Khi thay th cc bin trong mt biu thc i s bi cc

hng s th kt qu thc hin cc php ton trong biu thc

s l mt hng s no , c gi l gi tr ca biu thc.

V d : Khi thay x bi 7, y bi 2 ta c :

a) gi tr biu thc 5 l 5.

b) gi tr biu thc x l 7.

c) gi tr biu thc x + 5 l 7 + 5=12.

d) gi tr biu thc x2 + 3y - || - 2 l 72 + 3*2 - |7| - 2.

Ta cng ni biu thc x + 5 cho gi tr l 7 + 5=12.

1.4 Bin trong Visual basic:

Khi ni bin x, y , ... Ta hiu bin nh trong 1.3.

1.4.1 Kiu gi tr :

- Double , Single : s thc (R). Cch vit s thc trong

VB : 3,14 3.14

- Integer, Long : s nguyn (Z).

- Byte : 0, 1, 2, . . . , 255.

1.4.2 Khai bo bin : Dng qui nh kiu gi tr cho

bin.

Dim x As Kiu gi tr

Dim y As Kiu gi tr

Dim z As Kiu gi tr

- x, y, z : bin.

- Kiu gi tr : Double, Byte, . . .

Gi tr ban u ca x, y , z l 0.

1.5 Biu thc i s trong VB :

1.5.1 Mt biu thc i s c xy dng t :

a) cc s nguyn ,hu t ,thc . . . m ta gi l hng s.

b) cc bin x, y, . . . c th ly gi tr l cc hng s.

c) cc php ton

c1) +, -, *, /

c2) \ , mod

c3) cc biu thc : , logab, . . . c trnh by

1.5.2.

Cc php ton ny thao tc trn cc hng s v cc bin

theo mt th t nht nh.

1.5.2 Biu thc (VB) :

1) Sqr(x)

2) Log(x) Lnx

3) Abs(x) |x|

4) Exp(x) ex

5) x^y xy

6) -x -x

7) Sin(x), Cos(x) Sinx, Cosx , x tnh theo radiant.

8) Int(x), Fix(x)

9) Round(x, n) (v nh)

10) Rnd

V d 1:

Trong ton ta vit

2 + .

Trong VB ta vit

Sqr(x^2 + Exp(x)).

Bi tp ti lp: Cc bin ly gi tr s thc. Vit cc biu

thc sau bng VB :

1. 2 + + xy

2. sin(x+y) + lnx

3. logab + 3x + cos(z)

1.6 Thao tc gn (php gn / lnh gn) trong trong VB :

x = v

ngha : bin x ly gi tr v. Nu v l biu thc th bin x

ly gi tr ca biu thc v.

Ch : bin x ly gi tr v cn c ni l :

- gn gi tr v cho bin x;

- gi tr v c gn cho bin x;

- x c gi tr v;

V d 2: Cc bin x , y ly gi tr l s nguyn. Cc ch

dn c cho:

Thut ton:

B1. x = 5

B2. y = x2 + 1

Kt thc thut ton.

Duyt thut ton :

- B1 ga tr 5 c gn cho bin x (x ly gi tr 5).

- B2 gi tr 52 + 1 = 26 c gn cho bin y (y ly gi

tr 26).

Qui c trnh by duyt thut ton nh sau :

Thut ton:

B1. x = 5

B2. y = x2 + 1

Kt thc thut ton.

Duyt thut ton :

B1. x 5.

B2. y 52 + 1 (= 26)

Vit V d 2 bng ngn ng VB:

A1. Dim x As Long ( hoc Integer)

A2. Dim y As Long ( hoc Integer)

B1. x = 5

B2. y = x^2 + 1 (Hoc y = x*x+1)

Duyt chng trnh :

A1. x 0 , bin kiu s nguyn

A2. y 0 , bin kiu s nguyn

B1. x 5.

B2. y 52 + 1 (= 26)

V d 3:

Thut ton:

B1. x = 5

B2. x = x2 + 1

B3. y = x + 1

Kt thc thut ton.

Duyt thut ton:

B1. x 5

B2. x 52 + 1 (= 26)

B3. y 26+1 (=27)

BT ti lp :

1. Vit V d 3 bng ngn ng VB. Duyt chng trnh.

2. Hy vit thut ton sau bng VB. Duyt chng trnh. Cc bin kiu nguyn.

Thut ton:

B1. x = 5

B2. y = 2

B3. x = x2 + x + 1

B4. y = y + x + 1

Kt thc thut ton.

1.7 Cc ch dn c bn cho my tnh :

(Cc thut ton c bn)

Ch dn 1: (Ch dn cho my tnh)

Nhn n gi tr v gn n gi tr ny cho n bin x1, x2, . . ., xn .

( Nhp x1, x2, . . ., xn )

Ch dn 1 c vit trong VB:

x1 = InputBox()

x2 = InputBox()

. . .

xn = InputBox()

Ch dn 2:

Vit cc gi tr v1, v2, . . ., vn ra mn hnh.

( Vit v1, v2, . . ., vn )

Vi : vi l cc biu thc.

Ch dn 2 c vit trong VB:

MsgBox v1

MsgBox v2

. . .

MsgBox vn

Ch : Nu cn vit mt cu thng bo ta

thc hin :

MsgBox Cu thng bo

V d : MsgBox Chao cc ban !

V d 4: Tnh tng 2 s thc.

Thut ton:

B1. Nhn 2 gi tr v gn vo 2 bin a, b

B2. Tnh tng a+b v gn vo T

B3. Vit T ra mn hnh

Kt thc.

Trong VB ta vit :

B1. a = InputBox()

b = InputBox()

B2. T = a + b

B3. MsgBox T

Bi tp ti lp:

1) Hy vit cc thut ton sau bng VB v cho bit kt qu c vit ra mn hnh khi my tnh thc hin chng trnh , gii thch tng lnh (Duyt chng trnh). Cc bin c kiu s nguyn. Gi s B1 a v b c nhp l 9 v 4.

a) Thut ton:

B1. Nhn 2 gi tr v gn vo 2 bin a, b

B2. T= a mod b

B3. Vit T ra mn hnh

Kt thc.

b) Thut ton:

B1. Nhn 2 gi tr v gn vo 2 bin a, b

B2. T= a \ b

B3. Vit T ra mn hnh

Kt thc.

Bi tp ti lp:

2) Vit chng trnh (bng VB) cho cc bi ton sau :

a) Tnh din tch ca mt hnh ch nht.

b) Tnh din tch hnh trn.

c) Tnh modul ca vetor v=(x, y). Modul ca v l

2 + 2

Kim tra qu trnh (1):

a) Cc bi tp : 1 5 (10 im, h s 1).

b) Yu cu duyt chng trnh. Gi tr cc bin sinh

vin t cho.

c) Thi gian : 15.

d) c xem ti liu.