VB BaiTap Ver02 - LoiGiai
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Transcript of VB BaiTap Ver02 - LoiGiai
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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
V d: n=5 -> hm tr v kt qu l 9.
Function TinhTong (ByVal n As Integer) As Long
3. Tnh tng S=1-2+3-4+5++(-1)n+1n
V d: n=5 -> hm tr v kt qu l 3.
Function TinhTong (ByVal n As Integer) As Long
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.
V d: k=6 -> hm tr v kt qu l True.
k = 10 -> hm tr v kt qu l 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.
V d: k=4 -> hm tr v kt qu l True.
k = 10 -> hm tr v kt qu l 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