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1TRNG I HC BCH KHOA H NIVIN CNG NGH THNG TIN V TRUYN THNG
TIN HC I CNGBi 5. Mt s thut ton thng dng
Ni dung
5.1. Thut ton s hc
5.2. Thut ton v dy
5.3. Thut ton quy
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5.3. Thut ton s hc
Cc bi ton v s hc
Xc nh mt s nguyn c phi l s nguyn t/hp s hay khng
Tm USCLN, BSCNN ca 2 s nguyn
..
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Bi ton s nguyn t
Cho mt s nguyn dng p. Lm th no bit c p c phi s nguyn t hay
khng?
Input: p nguyn dng
Output: kt lun v tnh nguyn t ca p
tng?
p = 1? Khng phi s nguyn t
p > 1?
Kim tra t 2 n p-1 c phi l c s ca p khng
Nu c th kt lun p khng l s nguyn t, ngc li khng c s no th kt lun
p l s nguyn t
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2Bi ton s nguyn t (2)
Nhp p
if p=1 then begin
Xut: p khng nguyn t;
Dng thut ton;
end
flag := TRUE
for k:=2 to p-1 do
if (k l c s ca p) then begin
flag:=FALSE;
break; { ngt vng lp FOR }
end
if flag=TRUE then
Xut: p l s nguyn t
else
Xut: p khng l s nguyn t 5
Tm UCLN, BCNN ca hai s
B1 : Nhp 2 s a, b
B2 : x a, y b
B3 : Nu y # 0, sang B4
Ngc li, UCLN x, BCNN (a*b)/ UCLN. Kt thc
B4 : r x mod y
x y
y r
Quay li B3.
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V d
1. Tm tt c cc c s ca mt s nguyn dng N
tng : Duyt ln lt cc gi tr t 1 ti N v in ranu gi tr l c s ca
N.
Gii thut
B1 : Nhp N
B2 : Gn i 1
B3 : Nu i
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3V d
3. Kim tra tnh hon ho ca s nguyn dng N.
B1 : Nhp N
B2 : Gn i 1, TongUoc = 0,
B3 : Nu i < N ti B4
Ngc li, ti B6
B4 : Nu N chia ht cho i, gn TongUoc TongUoc + i
B5 : Tng i thm 1 n v. Quay li B3
B6 : Nu TongUoc = N, kt lun N l s hon ho.
Nu TongUoc < N, kt lun N l s khng y .
Nu TongUoc > N, kt lun N l s giu c
B7 : Kt thc
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V d
4. Hin th tt c cc s hon ho trong on [1,10000]
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Ni dung
5.1. Cc cu trc c bn trong lp trnh
5.2. Gi m (pseudocode)
5.3. Thut ton s hc
5.4. Thut ton v dy
5.5. Thut ton quy
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5.4. Thut ton v dy
Lm vic vi mt dy s
Cc bi ton in hnh
Tm s ln nht, nh nht trong dy
Kim tra dy c phi l dy tng hoc dy gim
Sp xp dy tng dn hoc gim dn
Tm trong dy c phn t no bng mt gi tr cho trc
Tnh trung bnh cng ca dy
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4Bi tp
Bi 1. Xy dng thut ton tm phn t c gi tr truyt i ln nht trong dy
gm n phn t.
Bi 2. Xy dng thut ton tm tng ca cc s chn v tng ca cc s l trong
dy gm n phn t c nhp vo t bn phm.
Bi 3. Xy dng thut ton kim tra xem mt dy s gm n phn t c nhp vo t
bn phm c phi l dy s tng (hoc gim) khng.
Bi 4. Xy dng thut ton tnh trung bnh cng ca cc s dng trong dy gm
n s c nhp vo t bn phm.
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Bi 1
tng : Gi s GTT ln nht ca dyMaxAbs l |a1|. Duyt cc phn t ai khc
trongdy, nu c GTT ln hn MaxAbs th gnMaxAbs = |ai|
Gii thut:
B1 : Nhp n v dy s {an}
B2 : Gn MaxAbs |a1|, i2
B3 : Nu i MaxAbs, gn MaxAbs|ai|
B5 : i i +1. Quay li B3
B6 : Xut MaxAbs. Kt thc
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Bi 2
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tng : Gi tng cc s chn v tng cc s l ln lt lTongChan v TongLe.
Duyt ln lt cc phn t ca dy, xt tnhchn l v cng dn vo tng tng ng.
B1 : Nhp n v dy {an}
B2 : Gn TongChan 0, TongLe 0, i 1
B3 : Nu i
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5Ni dung
5.1. Cc cu trc c bn trong lp trnh
5.2. Gi m (pseudocode)
5.3. Thut ton s hc
5.4. Thut ton v dy
5.5. Thut ton quy
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5.5. Thut ton quy
Vi bi ton c th c phn tch v a ti vic gii mt bi ton cng loi nhng
cp thp hn
ln d liu nhp nh hn
gi tr cn tnh ton nh hn
T thc hin li thut ton
V d:
Giai tha: n! = (n-1)! * n
Dy s Fibonacci: 0, 1, 1, 2, 3, 5, 8...
F(n) = F(n-1) + F(n-2)
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5.5. Thut ton quy (2)
xy dng thut ton quy, cn xc nh:
Trng hp c bn: (Cc) trng hp khng cn thc hin li thut ton.
Phn tng qut: C yu cu gi quy
Cn xc nh nguyn l a trng hp tng qut v trng hp c bn
m bo tnh dng ca gii thut quy - chc chn t trng hp tng qut s n c
trng hp c bn
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V d
Tnh giai tha ca n:
Trng hp c bn: 0! = 1
Trng hp tng qut: n! = (n-1)! * n
Xy dng dy Fibonacci
Trng hp c bn: F(0) = F(1) = 1
Trng hp tng qut: F(n) = F(n-1) + F(n-2)
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6Tnh giai tha - Thut ton quy
Input: s t nhin n
Output: GT(n)=n!
Thut gii:
Nhp n
GT:=1;
if n>0 then
GT := GT(n-1)*n;
Xut GT
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Bi tp
Xy dng thut ton cho bi ton tm s Fibonacci F(n)
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Thut gii heuristic
Dng mo
p dng vi nhng bi ton
Cha tm c thut ton v khng bit c tn ti thut ton khng
C thut ton nhng thi gian tnh ton qu lu hoc iu kin ca thut ton kh
p ng
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