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BI TP LP TRNH VISUAL BASIC

I. Tng, t hp, cc dy s ca qui lut nh Fibonacci (10 bi)

1. Tnh tng cc s nguyn dng t 1->n: S=1+2+3+4++n

V d: n=5 -> hm tr v kt qu l 15.

Function TinhTong (ByVal n As Integer) As Long

2. Tnh tng S=1+3+5+7+9++n



3. Tnh tng S=1-2+3-4+5++(-1)n+1n



4. Tnh tng

S = 1 +1

1 + 2+

1

1 + 2 + 3+

1

1 + 2 + 3 + 4++

1

1 + 2 + 3 + 4 ++ n

Function TinhTong (ByVal n As Integer) As Double

5. Tnh tng

+ =

=0

Function TinhTong (ByVal n As Integer, ByVal x As Double, ByVal a As Double) As Double

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6. Tnh tng

S = 12

2!+

4

4!

6

6!++ (1)

2

(2)!

Function TinhTong (ByVal n As Integer, ByVal x As Double) As Double

7. Tnh tng

S = 3

3!+

5

5!

7

7!++ (1)

21

(21)!

Function TinhTong (ByVal n As Integer, ByVal x As Double) As Double

8. Vit hm nhn mt i s l double v tr tr v l double dng tnh tng sau y:

s = 1 +x

1!+

2

2!++

! +

vi sai s |

!| < cho trc.

Function TinhTong (ByVal x As Double) As Double

9. Dy s Fibonacci l dy v hn cc s t nhin bt u bng hai phn t 0 v 1, cc phn t

sau c thit lp theo quy tc mi phn t lun bng tng hai phn t trc n.

Cng thc truy hi ca dy Fibonacci l:

Gii thch thm: v d mt phn ca dy s Fibonacci l:

0 1 1 2 3 5 8 13 21

Vit hm tm s th n trong dy s Fibonacci.

Function Fibonacci (ByVal n As Integer) As Integer

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10. Tnh t hp sau:

=

!

! !

Function ToHop (ByVal n As Integer, ByVal k As Integer) As Double

II.Cc dng bi ton c s (8 bi + 3 bi)

1. Kim tra k c phi l s nguyn t hay khng? Nu l s nguyn t hm tr v kt qu True,

ngc li tr v False.

V d: k=11 -> hm tr v kt qu l True.

k = 10 -> hm tr v kt qu l False.

Function KiemTraNT (ByVal k As Long) As Boolean

2. m cc s nguyn t t 1->n.

V d: n=7 -> hm tr v kt qu l 4 (v c 4 s nguyn t: 2, 3, 5, 7)

Function DemNT (ByVal n As Long) As Integer

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3. Kim tra k c phi l s hon ho hay khng? Nu l s hon ho hm tr v kt qu True,

ngc li tr v False.



Function KiemTraHH (ByVal k As Long) As Boolean

(S hon ho l s nguyn dng m tng cc c s ca n ngoi tr n bng chnh n. V

d: 6 l s hon ho v cc c s ca n l 1,2,3 c tng 1+2+3=6; 10 khng phi l s hon

ho v cc c s ca n l 1,2,5 c tng 1+2+510).

4. m cc s hon ho t 1->n.

V d: n=29 -> hm tr v kt qu l 2 (v c 2 s hon ho 6, 28)

Function DemHH (ByVal n As Long) As Integer

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5. Kim tra k c phi l s chnh phng hay khng? Nu l s chnh phng hm tr v kt qu

True, ngc li tr v False.



Function KiemTraCP (ByVal k As Long) As Boolean

6. Vit chng trnh in ra n s nguyn t u tin, mi s nguyn t cch nhau mt khong trng.

V d: n=4 -> hm tr v kt qu l 2 3 5 7

Function InNT (ByVal n As Integer) As String

7. Tm UCLN, BCNN ca 2 s hoc nhiu s.

Function UCLN (ByVal a As Long, ByVal b As Long) As Long

Function UCLN (ByVal a As Long, ByVal b As Long, ByVal c As Long) As Long

Function BCNN (ByVal a As Long, ByVal b As Long) As Long

Function BCNN (ByVal a As Long, ByVal b As Long, ByVal c As Long) As Long

8. Ti gin phn s vi a l t, b l mu. Hm tr v mt chui c dng phn s t/mu

c ti gin.

V d: a/b = 4/10 -> kt qu 2/5

Function ToiGian (ByVal a As Integer, ByVal b As Integer) As String

III.Cc dng bi ton lin quan ti ch s (10 bi)

1. Vit hm tnh tng cc ch s ca s nguyn dng n.

V d: n=1245 -> hm tr v kt qu: 12

Function TongChuSo (ByVal n As Integer) As Integer

2. Vit hm tnh s lng ch s ca s nguyn dng n.

V d: n=1245 -> hm tr v kt qu: 4

Function DemChuSo (ByVal n As Long) As Integer

3. Vit hm tm ch s u tin ca s nguyn dng n.

V d: n=7245 -> hm tr v kt qu: 7

Function ChuSoDauTien (ByVal n As Long) As Integer

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Cch 1:

Cch 2:

4. Vit hm tm ch s ln nht ca s nguyn dng n.

V d: n=7295 -> hm tr v kt qu: 9

Function ChuSoLonNhat (ByVal n As Long) As Integer

5. Vit hm tm ch s l nh nht ca s nguyn dng n. Nu khng c ch s l nh nht,

hm tr v 0.

V d: n=7295 -> hm tr v kt qu: 5

n=4282 -> hm tr v kt qu: 0

Function LeNhoNhat (ByVal n As Long) As Integer

6*. Tm ch s o ngc ca s nguyn dng n.

V d: n=1352 -> hm tr v kt qu: 2531

Function SoDaoNguoc (ByVal n As Long) As Long

7. Hy kim tra s nguyn dng n c ton ch s l hay khng? Nu ng tr v True, nu

khng tr v False.

V d: n=1733 -> hm tr v kt qu: True

n=1275 -> hm tr v kt qu: False

Function ToanLe (ByVal n As Long) As Boolean

8*. Hy m s lng ch s nh nht ca s nguyn dng n.

V d: n=435364 -> hm tr v kt qu: 2

Function DemNhoNhat (ByVal n As Long) As Integer

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9*. Kim tra mt s c phi l s i xng hay khng? Nu phi tr v True, nu khng tr v

False.

V d: n=14541 -> hm tr v kt qu: True

n=13536 -> hm tr v kt qu: False

n=1551 -> hm tr v kt qu: True

Function SoDoiXung (ByVal n As Long) As Boolean

10. Cho mt s nguyn n, xut ra mt chui c dng sau:

V d: n=3526 -> hm tr v kt qu: 3+5+2+6

Function Xuat (ByVal n As Long) As String

IV.Mng mt chiu (11 bi)

1. Cho mng mt chiu gm n s nguyn, hy vit hm tnh tng ca cc phn t trong mng.

V d: mng: 6 1 100 -5 3 -> hm tr v kt qu: 105

Function TongMang (ByVal a As Integer, ByVal n As Integer) As Long

2. Cho mng mt chiu gm n s nguyn, hy vit hm tnh gi tr trung bnh cng ca cc phn

t trong mng.

V d: mng: 6 1 100 -5 3 -> hm tr v kt qu: 21

Function TrungBinhCong (ByVal a As Variant, ByVal n As Integer) As Long

3.

Cho mng mt chiu cc s nguyn, hy vit hm tnh s ln xut hin ca x.

V d: mng: 6 3 100 -5 3 6 6 3 100 ; x=6 -> hm tr v kt qu: 3

Function TanSuat (ByVal a As Variant, ByVal x As Integer) As Integer

4. Sp xp mng a tng dn. Hm tr v l mt chui cha mng a c sp xp, mi phn t

cch nhau khong trng.

V d: mng a: 6 100 -3 78 34 -> hm tr v kt qu: -3 6 34 78 100

Function SapXep (ByVal a As Variant) As String

5*. Sp xp mng a c cc s l tng dn(s chn gi nguyn). Hm tr v l mt chui cha

mng a c cc s l c sp xp, mi phn t cch nhau khong trng.

V d: mng a: 6 101 -3 78 35 -> hm tr v kt qu: 6 -3 35 78 101

Function SapXep (ByVal a As Variant) As String

6. Kim tra mt mng mt chiu a gm cc phn t nguyn c phi l cp s cng hay khng,

nu phi hm tr v cng sai ca cp s cng, nu khng hm tr v -1.

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V d: mng a: 4 7 10 13 16 19 -> hm tr v kt qu: 3

mng a: 3 4 3 7 5 0 -> hm tr v kt qu: -1

Function CapSoCong (ByVal a As Variant) As Integer

7.Cho mt mng mt chiu a gm cc s thc, kim tra mng a c i xng hay khng, nu i

xng tr v True, nu khng tr v False.

V d: mng a: 0.1 3.4 5 3.4 0.1 -> hm tr v kt qu: True

mng a: 2.3 5 6 7 5 6.7 -> hm tr v kt qu: False

Function DoiXung (ByVal a As Variant) As Boolean

8*. Sp xp mng a tng dn, sau chn thm mt phn t k vo mng sao cho mng vn gi

th t tng dn. Hm tr v kt qu l mt mng.

V d: mng a: 3 4 3 -7 5 0 ; k=1 -> kt qu: -7 0 1 3 3 4 5

Function SapXepChen (ByVal a As Variant, ByVal k As Integer) As Variant

Function SapXepChen (ByVal a As Variant, ByVal k As Integer) As Integer()

khng c.

9. Cho mng mt chiu cc s nguyn dng c n phn t. Hy vit hm tm s nh nh trong

mng. Nu mng c s nh nh th tr v s nh nh, nu mng khng c s nh th tr v 0.

V d: mng 5 5 5 5 5 -> hm tr v kt qu: 0

mng 3 2 4 1 5 -> hm tr v kt qu: 2

Function SoNhoNhi (ByVal a As Variant) As Integer

10. Cho mng mt chiu gm n s nguyn, hy vit hm tm s m ln nht trong mng, nu

mng khng c s m th tr v s 0.

V d: mng 4 2 7 9 3 5 6 -> hm tr v kt qu: 0

mng 4 -5 8 -11 12 18 -2 -> hm tr v kt qu: -2

Function AmLonNhat (ByVal a As Variant) As Integer

11*.Lit k tn sut xut hin ca cc s trong mng a.

V d: mng a: 6 101 -3 -3 35 6 -3

-> hm tr v kt qu:

C 2 s 6, c 1 s 101, c 3 s -3, c 1 s 35

Function TanSuatXuatHien (ByVal a As Integer) As String

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V.Chui (9 bi + 3 bi)

1. Chuyn i gia cc s dng sau:

S thp phn S nh phn

S thp phn S thp lc phn.

Vit hm chuyn s thp phn thnh s nh phn.

Function ThapPhanSangNhiPhan (ByVal s As Long) As String

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Vit hm chuyn s nh phn thnh s thp phn.

Function NhiPhanSangThapPhan (ByVal s As String) As Long

Function ThapPhanSangThapLucPhan (ByVal n As Long) As String

Function ThapLucPhanSangThapPhan (ByVal s As String) As Long

2. Xa khong trng tha trong chui. Ch c php dng hm Len() v hm Mid(), khng

c s dng cc hm v cc th tc x l chui c sn khc ca VB.

V d: Su Pham Ky Thuat TP.HCM -> kt qu Su Pham Ky Thuat TP.HCM

Function XoaKhoangTrangThua (ByVal s As String) As String

3. Vit hm ct b khong trng trong chui, hm ny nhn mt i s l chui c th c khong

trng v tr tr v l chui khng cn khong trng. Lu : Khng c s dng bt k hm hay

th tc x l chui c sn no ca Visual Basic (ngoi tr hm Len v hm Mid).

V d: s= Su Pham Ky Thuat -> kt qu: SuPhamKyThuat.

Function XoaKhoangTrang (ByVal s As String) As String

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4. m s t trong chui.

V d: Su Pham Ky Thuat TP.HCM -> kt qu: 5

Function DemTu (ByVal s As String) As Integer

(Dng hm xa khong trng tha trong chui bi V.2 sau m s khong trng trong

chui. S t = s khong trng + 1)

5. Hm nhn vo mt chui cha cc phn t s nguyn cch nhau khong trng. Kt qu tr v

mt chui sp xp cc phn t s nguyn tng dn, mi phn t cch nhau khong trng.

V d: 3 0 4 -6 7 8 77 -6 -> kt qu -6 -6 0 3 4 7 8 77

Function SXMangChuoi (ByVal s As String) As String

dng nhp mng, khng thi.

6. Hy vit hm o chui. Chui u vo khng c khong trng tha.

V d: I Love You -> kt qu: You Love I

Function DaoChuoi (ByVal s As String) As String

7*. Tm v thay th mt t trong chui bng t khc. Ch c dng hm Mid v Len.

V d: s=DH SP Ky Thuat k1=SP k2=Su Pham -> kt qu DH Su Pham Ky Thuat

Function ThayThe (ByVal s As String, ByVal k1 As String, ByVal k2 As String) As String

8. Vit hm tr v mt chui vit hoa ch u ca cc t trong chui .

V d: s = nhap mon tin hoc -> kt qu: Nhap Mon Tin Hoc

Function VietHoa (ByVal s As String) As String

9. Vit hm chun ha chui dng loi b khong trng tha trong chui, bin ch ci u ca

mt t thnh ch hoa, cc ch ci cn li ca mt t thnh ch thng.

V d: tOi hoC LOP 12 -> hm tr v kt qu Toi Hoc Lop 12

Function ChuanHoaChuoi (ByVal s As String) As String

VI.Mt s bi ton khc (4 bi)

1. Gii phng trnh bc 2 ax2+bx+c=0

2. Tnh chu vi, din tch hnh trn.

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3. Cho n (0n500, bo m iu kin ny, khng cn phi kim tra) l s Kwh in tiu th

trong thng ca mt h gia nh. Vit hm tnh s tin gia nh ny phi tr, bit quy nh v

in nh sau:

48 Kwh u tin gi 600/Kwh

48 Kwh tip theo gi 1004/Kwh

48 Kwh k tip gi 1214/Kwh

T Kwh th 145 tr i, gi 1594/Kwh

V d: n = 155 th hm tr v kt qu l 152798 Function TinhTienDien (ByVal n As Integer) As Long

4. Tnh thu thu nhp (tng t tnh tin in).

Gii THI GIA K:

3:

Cu 4: (2)

Tm v tr u tin ca phn t ln nht trong mng a c n phn t.

V d: 8 99 23 101 -4 101 4 th hm tr v kt qu l 3.

Function VTLN (ByVal a As Variant, ByVal n As Integer) As Integer

lp th ba:

Cho mt s n nguyn dng, vit hm lit k cc ch s ca n theo dng sau: +ch_s _hng_ngn000+ch_s _hng_trm00+ch_s_hng_chc0+ch_s_n_v V d: nu n=8145 th tr v kt qu l mt chui 8000+100+40+5 Function LietKe (ByVal n As Long) As String

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BI TP THM:

1. Lit k nhng s xut hin nhiu nht trong mng a, mi s cch nhau mt khong trng. V d: mng a: 3 100 101 8 8 0 100 3 -> hm tr v kt qu: 3 8 100 mng a: 3 100 100 8 8 0 100 3 -> hm tr v kt qu: 100 Function TanSuatNhieu(ByVal a As Variant) As String

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2. abc

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