Giao Trinh Pascal

GIO TRNH

BI DNG HC SINH GIITIN HC

H tn hc sinh: .............................................................Lp: ...............................................................................

Ti liu bi dng hc sinh gii Tin hc THCS------------------------------------------------------------------------------------------------------------------------------------------------------

------------------------------------------------------------------------------------------------------------------------------------------------------Gio vin: Hong ng Quang in thoi: 0903781593 2

L THUYT



BI 1: GII THIU NGN NG LP TRNH PASCAL

I. Ngn ng lp trnh Pascal:L ngn ng lp trnh c bn gip con ngi giao tip v iu khin my tnh. Ngn ng lp trnh Pascal n gin d hiu nn ph hp vi hc sinh trung hc hoc nhng ai mi lm quen vi lp trnh my tnh.

II. Chng trnh Free Pascal:L phn mm min ph h tr vic to cc chng trnh my tnh bng ngn ngPascal.

III. Cch ci t v s dng:Bc 1: Ti phn mm ti a ch: http://sourceforge.net/projects/freepascal/files/latest/download?source=filesBc 2: Ci t nh cc phn mm khc

IV. Mn hnh giao din phn mm FPC:1) Thanh tiu 2) Thanh menu3) Ca s lp trnh4) Thanh trng thi

V. Cc thao tc trong chng trnh FPC:1) To mi 1 ca s lp trnh: File / New2) ng ca s lp trnh: Alt + F33) Phng to thu nh ca s lp trnh: F54) Chuyn qua li cc ca s lp trnh: Alt + ch s ca ca s5) Lu 1 chng trnh Pascal:

Bc 1: To th mc cha chng trnhBc 2: File / Change Dir i ng dn n th mc cn luBc 3: File / Save t tn cho tp tin chng trnh

Lu :*.pas : Lnh iu khin my tnh ca chng trnh*.bak: L tp tin d phng ca tp tin *.pas*.exe: L chng trnh lm c.

6) M 1 chng trnh Pascal c:Bc 1: M chng trnh FPCBc 2: File / Change Dir i ng dn n th mc cha chng trnh mun mBc 3: File / Open Chn tn chng trnh mun m

7) D li c php t ng: F98) Chy th chng trnh: Ctrl + F9



BI 2: CU TRC CA NGN NG LP TRNH PASCAL

I. Cc thnh phn c bn ca ngn ng Pascal:1) T kha:

Cc t kho l cc t dnh ring ca Pascal m ngi lp trnh c th s dng thit k chng trnh. Khng c dng t kho t cho cc tn ring nh tn bin, tn kiu, tn hm Mt s t kho ca Pascal gm:

AbsoluteAndArrayBeginCaseConstExternalFileForForwardFunctionGotoMod

NilNotObjectOfOrShrStringThenToTypeUnitConstructorDesstructot

DivDoDowntoElseEndIfImplementationInInlineInterfaceInterruptLabelPacked

ProcedureProgramRecordRepeatSetShlUntilUsesVarVirtualWhileWithXor

2) Tn:nh danh l mt dy k t dng t tn cho cc hng, bin, kiu, tn chng trnh con... Khi t tn, ta phi ch mt s im sau:

Khng c t trng tn vi t kho K t u tin ca tn khng c bt u bi cc k t c bit hoc ch s. Khng c t tn vi k t space,cc php ton. Nn t tn ngn gn nhng mang tnh cht gi nh.

3) Cu trc mt chng trnh vit bng Pascal:

Uses crt; Lin kt cc hm th vinType Khai bo kiu d liu PHN KHAI BO CAConst Khai bo hng s CHNG TRNHVar Khai bo bin sChng trnh con

BeginCu lnh 1;Cu lnh 2; PHN THN CHNG TRNHCu lnh 3;Cu lnh n;

End.



4) Mt s lnh c bn:

C php ngha lnhClrscr; Xa mn hnhWriteln; Xung dngReadln; Ngng thc hin lnh. Ch bm phm

enter s thc hin tipWrite(Ni dung ); Hin th phn ni dung ra mn hnhWriteln(Ni dung ); Hin th phn ni dung ra mn hnh sau

xung dng

5) Chng trnh Pascal u tin:Vit chng trnh in ra mn hnh hai dng ch trong khung du * nh hnh bn di

**************************** Chuong trinh Pascal dau tien ** Xin chao ban! ****************************

Bc 1: Khi ng PascalBc 2: Thc hin thao tc luBc 3: Nhp vo cc lnh sau:

Uses crt;Begin

Clrscr;Writeln( *************************** );Writeln( * Chuong trinh Pascal dau tien * );Writeln( * Xin chao ban! * );Writeln( *************************** );Readln;

End.



BI 3: BIN HNG

I. Bin:1) Khi nim:

Bin l 1 thnh phn trong chng trnh dng cha gi tr (gi tr c th l: ch, s, hnh nh, m thanh ). Gi tr cha bn trong bin c th thay i c trong qu trnh chng trnh thc hin.Bin bao gm 2 thnh phn l tn bin, kiu d liu ca bin.Tn bin: do ngi lp trnh t, dng phn bit cc bin.Kiu d liu ca bin: quy nh loi gi tr cha bn trong bin.

2) Bng cc kiu d liu c bn:

TN KIU MIN GI TR B NH HAO TN NGHAByte 0 255 1 byte Kiu s nguyn dngWord 0 65535 2 byte Kiu s nguyn dngInteger -32768..32767 2 byte Kiu s nguynLongint -2147483648 2147483647 4 byte Kiu s nguynReal 2.9*10-39 1.7*1038 6 byte Kiu s thp phnExtended 3.4*10-4932 1.110+4932 10 byte Kiu s thp phnChar 1 k t 1 byte Kiu k tString 255 k t 255 byte Kiu chui k tBoolean True, False 1 byte Kiu lun l

3) Cch khai bo bin: Khai bo trong phn khai bo ca chng trnh.C php khai bo:

VarTn bin : kiu d liu;

V d:So1 : integer;Tong, hieu: longint;Thuong: real;Ho, ten: string;

4) Lnh thao tc vi bin

C php ngha lnhRead(tn bin); Nhp 1 gi tr t bn phm vo bn

trong bin.Readln(tn bin); Nhp 1 gi tr t bn phm vo bn

trong bin sau xung dngTn bin := gi tr Lnh gn gi tr cho bin. Lun ly gi

tr v bn phi b qua v bn tri.



Uses crt;Var

So1, so2, tong : integer;Begin

Clrscr;Writeln( CHUONG TRINH TINH TONG 2 SO );Writeln( Nhap so thu nhat: );Readln(so1);Writeln( Nhap so thu hai: );Readln( so2);Tong := so1 + so2;Writeln( Tong hai so = ,tong);Readln;

End.

V d lnh gn gi tr cho bin: So1 := 10;So2 := 5;Tong := so1 + so2;So3 := so2;

5) Bng cc ton t c bn:Ton t ngha V d Kiu d liu p dng

+ Cng 2 s hoc 2 chui k t S, chui- Tr 2 s S* Nhn 2 s S/ Chia 2 s S thcDiv Chia 2 s ly phn nguyn

V d:6 div 4 110 div 2 57 div 9 00 div 2 0

S nguyn

Mod Chia 2 s ly phn dV d:8 mod 2 07 mod 5 20 mod 7 06 mod 9 6

S nguyn

6) Bi tp v d:Vit chng trnh yu cu ngi s dng nhp vo 2 s nguyn t bn phm ri tin hnh tnh tng 2 s . Sau in kt qu tnh c ra mn hnh.Bc 1: Thc hin lu chng trnhBc 2: Phn tch chng trnh theo ba giai on

- Giai on 1: Nhp d liu (nhp 2 s nguyn t bn phm)- Giai on 2: X l d liu (tnh tng 2 s)- Giai on 3: Xut d liu kt qu (mn hnh)



Uses crt;Const

Pi = 3.1416159;Var

So1, so2, tong : integer;Begin

Clrscr;Writeln( CHUONG TRINH TINH CHU VI DIEN TICH HINH TRON );Writeln( Nhap duong kinh: );Readln(d);R := d/2;P := pi*d;S := pi*r*r;Writeln( Ban kinh = ,r:0:2);Writeln( Chu vi = ,p:0:2);Writeln( Dien tich = ,s:0:2);Readln;

End.

II. Hng:1) Khi nim:

Hng l 1 thnh phn trong chng trnh dng cha gi tr. Gi tr cha trong hng c to trong phn khai bo v gi tr ny lun khng i cho n ht chng trnh.

2) Cch khai bo hng: Khai bo trong phn khai bo ca chng trnh.C php khai bo:

ConstTn hng = gi tr;

V d:Pi = 3.1416159;T = Hong Bang ;

3) Bi tp v d:Vit chng trnh cho nhp vo ng knh hnh trn ri tin hnh tnh bn knh, chu vi v din tch sau in kt qu tnh c ra mn hnh. Yu cu p dng hng s.

III. Mt s hm c bn:

ngha C php Lu Tnh cn bc 2 ca mt s

Bin := Sqrt (gi tr s); Bin kiu s thc

Tnh bnh phngca mt s

Bin := Sqr(gi tr s); Bin kiu s

Ly tr tuyt i ca mt s

Bin := Abs(gi tr s); Bin kiu s

Ly phn nguyn Bin := Trunc(gi tr s); Bin kiu s nguyn



ca s thcLy phn thp phnca s thc

Bin := Frac(gi tr s); Bin kiu s thc

Ly th t ca k t trong bng m ASCII

Bin := Ord (Chrk t); Bin kiu s nguyn

Cho bit k t trong bng m ASCII khi bit th t

Bin := Chr(gi tr s nguyn); Bin kiu k t

i k t thng thnh k t in hoa

Bin := Upcase (k t); Bin kiu k t



BI 4: CU TRC IU KIN CU LNH IF THEN

I. Biu thc iu kin:1) Khi nim:

L nhng biu thc m kt qu ch c 1 trong 2 gi tr l ng (true) hoc sai (false)2) Ton t ca biu thc iu kin:

Ton t ngha V d Kiu d liu p dng> Ln hn S, chui, lun l< B hn S, chui, lun l= Bng S, chui, lun l>= Ln hn hoc bng S, chui, lun l



Uses crt;Var

tb : real;Begin

Clrscr;Writeln( Nhap diem trung binh: );Readln(tb);If (tb >= 8) then writeln( Chuc mung ban! );Readln;

End.

IF (biu thc iu kin) THEN khi lnh;

ngha: Nu biu thc iu kin ng th thc hin khi lnh pha sau THEN ngc li th khng lm g c.

V d: Vit chng trnh nhp im trung bnh. Nu im trung bnh>= 8 th in ra mn hnh cu ni Chuc mung ban

2) Lnh IF THEN dng :S :

C php:

IF (biu thc iu kin) THEN khi lnh ngELSE khi lnh sai;

ngha: Nu biu thc iu kin ng th thc hin cu lnh 1 pha sau THEN ngc li th thc hin cu lnh 2 pha sau ELSE

V d:

Biu thc iu kin

Khi lnh ng

True

False

Khi lnh sai



Uses crt;Var

tb : integer;Begin

Clrscr;Writeln( Nhap diem trung binh: );Readln(tb);If (tb >= 8) then writeln( Chuc mung ban! )Else writeln( Ban can co gang hon );Readln;

End.

Vit chng trnh nhp im trung bnh. Nu im trung bnh>= 8 th in ra mn hnh cu ni Chuc mung ban . Ngc li in ra mn hnh cu ni Ban can co gang hon

III. Cu lnh ghp: ngha:Dng ghp nhiu cu lnh li thnh mt cu lnh ln

C php

BeginCu lnh 1;Cu lnh 2;Cu lnh 3;Cu lnh n;

End;

V d:Vit chng trnh nhp im trung bnh. Nu im trung bnh>= 8 th in ra mn hnh hai cu ni trn 2 dng Chuc mung ban Ban duoc hoc bong 1000000Ngc li in ra mn hnh hai cu ni Ban can co gang hon Ban bi phat 100000



Uses crt;Var

tb : integer;Begin

Clrscr;Writeln( Nhap diem trung binh: );Readln(tb);If (tb >= 8) then begin

writeln( Chuc mung ban! );writeln( Ban duoc hoc bong 1000000 );

endElse begin

writeln( Ban can co gang hon );writeln( Ban bi phat 100000 );

Readln;End.

Uses crt;Var

tb : integer;Begin

Clrscr;Writeln( Nhap diem trung binh: );Readln(tb);If (tb >= 8) then writeln( Gioi )Else if (tb >= 6.5) then writeln( Kha )

Else if (tb >= 5) then writeln( TB )Else if (tb >= 3.5) then writeln( Yeu )

Else writeln( Kem );Readln;

End.

IV. Cu trc lnh if then lng nhau:Dng gii quyt nhng vn c nhiu hn 2 trng hp nh v d sau:Vit chng trnh nhp vo im trung bnh. Sau tin hnh xp loi hc lc theo quy nh:Trung bnh >= 8 Gii8 > Trung bnh >= 6.5 Kh6.5 > Trung bnh >= 5 TB5 > Trung bnh >= 3.5 Yu3.5 > Trung bnh Km



Uses crt;Var

so : integer;Begin

Clrscr;Writeln( Nhap so: );Readln(so);Case so of

1: writeln( Thang gieng );2: writeln( Thang hai );3: writeln( Thang ba );4: writeln( Thang tu );5: writeln( Thang nam );6: writeln( Thang sau );7: writeln( Thang bay );8: writeln( Thang tam );9: writeln( Thang chin );10: writeln( Thang muoi );11: writeln( Thang muoi mot );12: writeln( Thang muoi hai );

End;Readln;

End.

BI 5: CU LNH CHN CASE .. OF

I. Cng dng:Dng chn thc hin 1 trong nhiu trng hp c th xy ra

II. Cu lnh Case .. Of dng thiu:C php

Case (biu thc chn) ofHng 1 : khi lnh 1;Hng 2 : khi lnh 2;Hng 3 : khi lnh 3;Hng n : khi lnh n;

End;

V d: Vit chng trnh nhp vo 1 s. Nu s nm trong khong t 1 n 12 th in ra tn thng tng ng.

III. Cu lnh Case .. Of dng :C php

Case (biu thc chn) ofHng 1 : khi lnh 1;Hng 2 : khi lnh 2;



Uses crt;Var

so : integer;Begin

Clrscr;Writeln( Nhap so: );Readln(so);Case so of

1: writeln( Thang gieng );2: writeln( Thang hai );3: writeln( Thang ba );4: writeln( Thang tu );5: writeln( Thang nam );6: writeln( Thang sau );7: writeln( Thang bay );8: writeln( Thang tam );9: writeln( Thang chin );10: writeln( Thang muoi );11: writeln( Thang muoi mot );12: writeln( Thang muoi hai );Else writeln( Khong co thang nay );

End;Readln;

Hng 3 : khi lnh 3;Hng n : khi lnh n;Elsekhi lnh n+1;

End;

V d: Vit chng trnh nhp vo 1 s. Nu s nm trong khong t 1 n 12 th in ra tn thng tng ng. Nu khng th in ra mn hnh Khong co thang nay

IV. Lu : C th dng lnh ghp thc hin nhiu lnh trong mt nhnh chn Biu thc chn v cc hng phi c cng kiu d liu v ch c 1 trong 2 kiu s

nguyn hoc k t.



BI 6: CC LNH V BN PHM V MN HNH

I. Lnh i mu ch:C php:

Textcolor(tn mu hoc m mu);

V d: Textcolor(red);Textcolor(13);

II. Lnh i mu nn:C php:

Textbackground(tn mu hoc m mu);Clrscr;

V d: Textbackground(10);Clrscr;

III. Lnh di chuyn con tr n mt v tr bt k trn mn hnhC php:

gotoxy(ta ct, ta dng);

V d: gotoxy(5,10);

Lu : mn hnh ca chng trnh vit bng pascal c 80 ct v 25 dng.

IV. Lnh pht sinh ngu nhin:C php:

Randomize;Tn bin := Random(s nguyn);

ngha: My tnh s to ra cc gi tr ngu nhin nm trong khong t 0 gi tr - 1

V. Lnh nhp 1 k t t bn phmC php:

tn bin := readkey;

Lu : bin phi khai bo kiu char



VI. Hm nhn bit c 1 phm no trn bn phm c bm hay khng?

C php:

keypressed;

ngha: Nu c bt k 1 phm no trn bn phm c bm th lnh keypressed; s c gi tr l true, ngc li l false

VII. Lnh lm chm chng trnh:C php:

delay(gi tr);

V d: delay (1000);

VIII.Lnh pht ra m thanh:C php:

sound(gi tr);

V d: sound(530);

IX. Lnh ngng pht ra m thanh:C php:

nosound;



BI 7: CU LNH LP

I. Cu lnh lp vi s ln khng bit trc:

While .. do Repeat .. untilS :

C php:

While (biu thc iu kin) dokhi lnh;

ngha:Trong khi biu thc iu kin ng th lp li khi lnh

S :

C php:

RepeatKhi lnh;

Until (biu thc iu kin);

ngha:Lp li khi lnh cho n khi biu thc iu kin ng

V d: Vit chng trnh yu cu ngi s dng nhp vo t v mu ca mt phn s. Mu s bt buc phi khc 0. Nu nhp sai yu cu th bt nhp li cho n khi ng. Nu nhp ng th tin hnh tnh bnh phng phn s va nhp ri in kt qu ra mn hnh.Lm bng 2 cch

Biu thc iu kin

Khi lnh cn lp

True

false

Thot khi vng lp

Biu thc iu kin

Khi lnh cn lp

False

True

Thot khi vng lp



Cch while .. do Cch Repeat .. until

Uses crt;Var

Tu, mau:integer;Begin

Clrscr;Write( Nhap tu so: );Readln(tu);Write( Nhap mau so khac 0: );Readln(mau);While (mau = 0) doBegin

Write( Nhap mau so khac 0: );Readln(mau);

End;Writeln( Binh phuong phan so vua nhap la: ,tu*tu, / , mau*mau);Readln;

End.

Uses crt;Var

Tu, mau:integer;Begin

Clrscr;Write( Nhap tu so: );Readln(tu);Repeat

Write( Nhap mau so khac 0: );Readln(mau);

Until (mau 0);Writeln( Binh phuong phan so vua nhap la: ,tu*tu, / , mau*mau);Readln;

End.

II. Cu lnh lp vi s ln bit trc:

For .. to .. do For .. downto .. doS : S :

Bin m =gi tr cui

false

True

Khi lnh;Bin m := bin m - 1

Thot khi vng lp



Uses crt;Var

N, s, i : longint;Begin

Clrscr;Writeln( Nhap n: );Readln(n);

For i := 1 to n dos := s + i;

Writeln ( S = ,s);Readln;

C php:

For bin m := gi tr u to gi tr cui doKhi lnh;

ngha:Bin m cha gi tr u, nu gi tr bin m cn nh hn hoc bng gi tr cui th lp li khi lnh pha sau do. Sau mi ln lp li th gi tr ca bin m c t ng tng 1 n v.

C php:

For bin m := gi tr u downto gi tr cui do

Khi lnh;

ngha:Bin m cha gi tr u, nu gi tr bin m cn ln hn hoc bng gi tr cui th lp li khi lnh pha sau do. Sau mi ln lp li th gi tr ca bin m c t ng gim 1 n v.

V d 1:Cho bit kt qu hin th trn mn hnh ca cc cu lnh sau:For i := 1 to 100 do writeln( Chao ban );For j := 1 to 10 do writeln(j);For i := 10 downto 1 do writeln(j);For i := 1 to -10 do writeln( Ban gioi qua! );

V d 2:Vit chng trnh cho ngi s dng nhp vo mt s nguyn dng n sau tin hnh tnh tng s theo cng thc sau:S = 1 + 2 + 3 + + n-1 + nRi in kt qu tnh c ra mn hnh

Lu :- Bin m, gi tr u, gi tr cui ca vng lp For phi l kiu s nguyn hoc

kiu k t.- Khng c t thay i gi tr ca bin m bn trong vng lp For.



BI 9: CHNG TRNH CON

I. ngha:Chng trnh c p dng chng trnh con tun theo nguyn l chia nh vn trong lp trnh. Chng trnh c p dng chng trnh con s c nhng u im sau:

R rng D kim tra sa li Tn dng kh nng s dng li ca chng trnh con

II. V tr ca chng trnh con trong chng trnh:Nm sau phn khai bo v trc phn thn chng trnh.Chng trnh con c chia lm 2 loi l: Procedure (Th tc) vFunction (Hm)

III. Th tc (Procedure):1) Cu trc ca th tc:Procedure tn_th_tc (Khai bo danh sch tham s);Phn khai bo ca th tc;Begin

Cc cu lnh;End; chng trnh con mun thc hin th ta cn c li gi n chng trnh con V d: Vit chng trnh v th cng cc hnh sau, mi hnh lm trong 1 Procedure

ng thng: **********

Hnh ch nht c rut: ****************************************

Hnh vung rng rut: * * * * * ** ** ** ** * * * * *

Hnh tam gic vung: ***************

Hnh tam gic cn: *********

****************



2) Bin ton b v bin cc b:Bin ton b: c khai bo trong phn khai bo ca chng trnh chnh, tn ti vc s dng trong sut chng trnhBin cc b: c khai bo trong phn khai bo ca chng trnh con. Ch tn ti v c s dng trong chng trnh con m n c khai bo.rng hp bin ton b va bin cc b b trng tn th trong chng trnh con s u tin s dng bin cc b.V d: p dng khi nim bin ton b v bin cc b v li 5 hnh ca v d trn theo cch tng qut. Cho ngi s dng nhp k t v hnh, chn hnh mun v. Ty hnh m cho nhp vo chiu cao, chiu rng hoc c hai.

3) Danh sch tham s:L cc bin cc b trong chng trnh con nhng c cng dng trao i d liu gia chng trnh chnh v chng trnh con hoc gia cc chng trnh con vi nhau.Danh sch tham s c chia lm 2 loi

Danh sch tham s tr: a gi tr vo cho chng trnh con x l ri khng s dng na

Danh sch tham s bin: a gi tr vo cho chng trnh con x l ri ly gi tr tr v.

IV. Hm (Function):1) Cu trc ca hm:

Function tn_hm(Khai bo danh sch tham s) : kiu d liu tr v;Phn khai bo ca hm;Begin

Cc cu lnh;Tn_hm := gi tr tr v;

End;



BI 10: CU TRC D LIU MNG (ARRAY)

I. nh ngha:Mng l mt tp hp nhiu bin c cng kiu d liu ghp li vi nhau

II. Cch khai bo mng:1) Khai bo trc tip:

VarTn bin mng : array [1 .. s phn t max] of kiu d liu;Bin qun l s phn t : kiu s nguyn;

V d:a : array [1..100] of integer;na : integer;ho, ten : array [-10..10] of string;nh, nt: integer;

2) Khai bo gin tip:Type (nh ngha 1 kiu d liu mi)

Tn kiu mng = array [1.. s phn t max] of kiu d liu;Var

Tn bin mng : tn kiu mng;Bin qun l s phn t : kiu s nguyn;

V d:

Typemangsothuc = array [1..1000] of real;

Varma,mb,mc : mangsothuc;na,nb,nc:longint;

III. Cch truy xut v thao tc trn mng:1) Php ton gn trn mng:

Ta c th gn 2 mng cng phn t v cng kiu d liuV d: ma := mb;

2) Truy xut n tng phn t trong mng:Tn bin mng [ch s];V d:

Readln(ma[1]);Ma[10] := ma[1] + ma[3];

IV. Cch nhp xut d liu cho 1 mng:1) Nhp mng:

B1: Nhp gi tr cho bin qun l s phn t ca mngB2: Dng vng lp chy t phn t u n phn t cui. Nhp gi tr cho tng phn t trong mngV d:



2) Xut mng:B1: Dng vng lp chy t phn t u n phn t cui ca mngB2: Xut tng phn t ra mn hnh.V d:

BI TP



Bin Hng Cc lnh c bn

Bi 1: Vit chng trnh xut ra mn hnh cu Em s l nim vui ca m .

Bi 2: Vit chng trnh xut ra mn hnh hai cu trn hai dng: ** y l chng trnh Pascal ** Chc cc bn hc tp vui v

Bi 3: Vit chng trnh xut ra mn hnh hnh ch nht v hnh tam gic cn sau:* * * * * ** * * * ** * * * * * * * * *

* * * * * * *Sau tin hnh a 2 hnh ch nht v hnh tam gic ra gia mn hnh.

Bi 4: Vit chng trnh v ngi sao nm cnh.

Bi 5: Vit chng trnh nhp vo hai s cha trong hai bin a, b. Tnh tng, hiu, tch, thng ca chng ri in kt qu ra mn hnh.

Bi 6: Vit chng trnh nhp tn, nm sinh, nm hin ti ca ngi s dng, sau in ra mn hnh tn, tui ca ngi .

Bi 7: Vit chng trnh nhp chiu cao (cm), vng ngc ti a (cm), vng ngc ti thiu (cm), cn nng (kg). Ri tin hnh tnh ch s Pignet bng cng thc:

Pignet = chiu cao * 10 (trung bnh vng ngc + trng lng)Ri in kt qu tnh c ra mn hnh.

Bi 8: Vit chng trnh nhp vo 2 s a, b ri tnh a2, b3.

Bi 9: Vit chng trnh nhp vo 2 s a, b ri thc hin php tnh a MOD b, b DIV a. In kt qura mn hnh.

Bi 10: Vit chng trnh vit chng trnh tnh din tch v chu vi hnh ch nht.

Bi 11: Vit chng trnh nhp vo chu vi ca hnh ch nht ri suy ra din tch ln nht m hnh ch nht c th c c.

Bi 12: Vit chng trnh i inch ra cm bit 1 inch = 2,54 cm.

Bi 13: Vit chng trnh tnh vn tc khi bit qung ng (km) v thi gian (gi). Sau i n v ca vn tc ra m/s.

Bi 14: Vit chng trnh nhp vo 2 s a, b. tin hnh i gi tr ca a cho b v ngc li.

Bi 15: Cho nhp di hai cnh gc vung ca mt tam gic. Hy vit chng trnh tnh: di cnh huyn. Chu vi din tch ca tam gic.



di ng cao xung cnh huyn.

Bi 16: Vit chng trnh tnh din tch v chu vi ca hnh trn khi bit bn knh (s dng const).Bi 17: Vit chng trnh nhp tn v khi lng ca mt ngi sau tnh ra trng lng ca ngi trn: Tri t, Mt Trng, Mt Tri, Sao Thy, Sao Mc. Bit nu khi lng ca tri t l 1 th khi lng ca mt tri l 27,9; mt trng l 0,17; sao thy l 0,37; sao Mc l 2,64.

Bi 18: Vit chng trnh nhp thi gian lm mt cng vic (thi gian tnh bng giy). Sau vit ra thi gian trn di dng bao nhiu gi, bao nhiu pht, bao nhiu giy.

Bi 19: Vit chng trnh nhp vo mt k t ri in ra m ASCII ca n.

Bi 20: Vit chng trnh nhp mt s kiu byte, sau in ra k t tng ng ca s trong bn m ASCII.

Bi 21: Vit chng trnh nhp vo mt s nguyn dng n, sau tnh tng cc s nguyn t 1 n n. (S dng cng thc).

Cu trc iu kin cu lnh ghp

Bi 22: Nhp hai s a, b tnh thng a / b v in ra kt qu.

Bi 23: Vit chng trnh cho nhp hai s a, b. So snh a v b. Nu s no ln hn th gn max bng gi tr .

Bi 24: Vt chng trnh nhp hai s a, b. Kim tra xem a v b c = 0 khng? Nu a v b cng bng khng th in ra cu ni hai so a v b phai khac khong , nu a 0 v b = 0 th tnh a2, nu b 0 v a = 0 th tnh s nghch o ca b, cn nu a v b cng khc 0 th tnh a/b.

Bi 25: Vit chng trnh gii phng trnh bc nht ax + b = 0 vi a, b nhp t bn phm.

Bi 26: Vit chng trnh nhp vo ba s a, b, c. In ra mn hnh s nguyn ln nht In ra mn hnh s nguyn nh nht. Sp xp ba s theo th t tng dn v gim dn.

Bi 27: Vit chng trnh nhp im Ton, Vn, L, Ho, Anh vn (Ton, Vn h s 2). Tnh im trung bnh v in ra hc bng theo iu kin TB >= 8 v khng c mn di 7 l 1.000.000 ng. TB >= 6.5 v khng c mn di 5 l 500.000 ng. Hc bng bng khng cho cc trng hp cn li.

Bi 28: Vit chng trnh nhp vo 3 s a, b, c bt k. Kim tra xem ba s c th l cnh ca mt tam gic khng. Nu khng th in ra thng bo, nu c th kim tra xem l tam gic g trong cc loi tam gic sau (thng, vung, cn, u). Sau tnh din tch tam gic theo cng thc

)).().(( cpbpappS ---= Vi p = (a+b+c) / 2



Bi 29: Mt s t nhin N c gi l Palidrom (i xng) nu cch vit thp phn ca s l i xng (2112, 353). Nhp vo mt s nguyn dng < 10000, hy kim tra xem s c phi l s Palidrom khng?

Bi 30: Gi s thu li tc c nh gi trn thu nhp nh sau:Thu nhp Thu sut

0 400.000400.000 1.000.0001.000.000 2.000.000trn 2.000.000

0%10%20%30%

Vit chng trnh hi thu nhp ca mt ngi, sau tnh thu phi np (s dng const).

Bi 31: Cho nhp qung ng i theo n v km (qung ng c th c gi tr thp phn) hy tnh s tin phi tra cho taxi ri in kt qu ra mn hnh bit: 1 km u tin c gi 9500 ng T km th 2 n km th 10: 9000 ng / 1 km T km th 11 n km th 30: 7000 ng / 1 km T km th 30 tr ln: 5000 ng / 1 km

Bi 32: Vit chng trnh nhp to (x, y) ca im M trn mt phng. Cho bit im M nm gc phn t no ca mt phng (xt 7 trng hp).

Cu lnh chn Case .. of

Bi 33: Cho bit s thng trong nm hy in ra s ngy tng ng (cho thng 2 c 29 ngy).

Bi 34: Vit chng trnh nhp thng, nm, in ra xem thng c bao nhiu ngy bit thng 2 nm thng l 28 ngy, thng 2 nm nhun l 29 ngy. Nm nhun l nm chia ht cho 4 hay 400 nhng khng phi nm u th k.

Bi 35: Vit chng trnh nhp vo mt s n ri in ra gi tr ca s bng ch ( 0 < n



Bi 41: Cho gi bt u cng vic v gi kt thc cng vic. Vit chng trnh tnh tin cng phi tr bit rng mt gi cng bui sng gi 10.000 ng (bui sng c tnh t 7 11 gi); mt gi cng bui chiu gi 20.000 ng (bui chiu c tnh t 12 17 gi); mt gi cng bui ti gi 40.000 ng (bui ti c tnh t 18 21 gi).

Bi 42: Nhp ch s c, ch s mi v mc ch s dng in. Vit chng trnh tnh ho n tin in phi tr bit rng tin in c tnh theo gi bc thang vi cc mc gi v loi gi nh sau: in sinh hot:

o 100 kw u: 550 ng/kw.o 50 kw tip theo: 850 ng/kw.o 151 250 kw: 1000 ng/kw.o 251 kw tr ln: 1500 ng/kw.

in sn xut:o 10000 kw u: 1500 ng/kwo 10001kw tr ln 2000 ng/kw

in nng nghip:o 700 ng/kw

in trng hc v bnh vin:o 800 ng/kw

in vui chi:o 2500 ng/kw.

Bi 43: Vit chng trnh tr chi On t t Hai ngi chi vi nhau. Mt ngi chi vi my.

Vng lp khng bit trc s ln lp

Bi 44: Vit chng trnh tnh tng v tch ca cc s nguyn nhp t bn phm. Chng trnh kt thc khi nhp s 0.

Bi 45: Vit chng trnh nhp t bn phm mt s bt k, kt thc khi nhp s 0. m xem cc s nhp vo c bao nhiu s chn, bao nhiu s l.

Bi 46: Vit chng trnh lp li vic nhp mt s thc bt k, cho bit phn nguyn ca s ny c my ch s. Chng trnh kt thc khi ngi s dng nhn ESC.

Bi 47: Vit chng trnh tnh tng S = 1/1 + 1/2 + 1/3 + (iu kin dng ca chng trnh l 2 S < 0,0001)

Bi 48: Vit chng trnh in ra mn hnh thng nguyn v d ca php chia s nguyn dng cho 5.

Vng lp bit trc s ln lp



Bi 49: Vit chng trnh tnh cc gi tr sau (vi n nhp t bn phm): S1 = 1 + 3 + 5 + 7 + + n. S2 = 1*1 + 2*2 + 3*3 + + n*n. S3 = 1 * 2 * 3 * * n S4 = 1/(1*2) + 1/(2*3) + 1/(3*4) + + 1/(n*(n+1)) S5 = (1 + 1/12)*(1 + 1/22) * * (1+1/n2) S6 = xn/n!

Bi 50: Vit chng trnh in ra cn bc hai ca cc s chn t 0 100.

Bi 51: Vit chng trnh kim tra xem c bao nhiu s chia ht cho 3 m khng chia ht cho 6 trong khong t 120 650.

Chn loi vng lp ph hp nht vit chng trnh

Bi 52: Vit chng trnh in ra mn hnh cc s c 3 ch s vi iu kin s chia cho 6 d 2 v tng ca cc ch s n l 20.

Bi 53: Vit chng trnh nhp chiu cao v k t sau cho hin ln mn hnh kt qu sau:a) *

********************

b) $$$$$$$$$$$$$$$$$$$$$$$$$$$$

c) @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

Bi 54: Vit chng trnh nhp vo chiu cao sau in ra mn hnh nh sau:PASCAL

PASCALPASCAL

PASCAL

Bi 55: Vit chng trnh cho to bng s nh sau:0 1 2 3 4 5 6 7 8 910 11 12 13 14 15 16 17 18 19 90 91 92 93 94 95 96 97 98 99

Bi 56: Nhp vo mt s nguyn dng, vit chng trnh in ra s theo chiu ngc li. V d: 12345 54321.

Bi 57: Vit chng trnh cho nhp s tin gi, li xut mt thng (tnh theo %), s thng gi. Tnh tin li v s tin c c sau n thng gi ngn hng.Bi 58: Vit chng trnh cho nhp s tin gi (a), li xut thng (x). Hi mun c s tin (b) th phi gi bao nhiu thng.



Bi 59: Vit chng trnh cho nhp ngy / thng / nm. In ra xem ngy ny l ngy th my. Bit ngy 1 / 1 / 1 l th Hai.

Bi 60: Vit chng trnh nhp vo mt s n (n >=2). Kim tra xem s n c phi l s nguyn thay khng?

Bi 61: Vit chng trnh nhp vo mt s n > 0, in ra tt c cc s nguyn t nm gia 2 v n, ng thi m xem c bao nhiu s nguyn t.

Bi 62: Vit chng trnh cho nhp mt s nguyn dng a, hy vit li s a bng cch phn tch thnh cc tha s nguyn t. V d: a = 40 in ra mn hnh a = 2, 2, 2, 5.

Bi 63: Vit chng trnh tm c chung ln nht ca hai s nguyn dng m, n.

Bi 64: Vit chng trnh tm bi chung nh nht ca hai s nguyn dng m, n.

Bi 65: Vit chng trnh nhp vo mt s nguyn dng ri kim tra xem s c phi l mt s hon ho hay khng? (s hon ho l s c tng cc ch s bng tch cc ch s).

Bi 66: Vit chng trnh in ra bn cu chng n vi n nhp t bn phm

Bi 67: Vit chng trnh in bng cu chng t 2 9. Trnh by 9 bng cu chng trn 1 hng ngang

Bi 68: Vit chng trnh tm tt c cc s nguyn a, b ( 1



Bi 72: Vit chng trnh in ra gi tr bng m ASCII ca 1 k t c nhp t bn phm (tt ccc phm trn bn phm).

Bi 73:Vit chng trnh nhp chiu cao ca tam gic, xut ra mn hnh tam gic cn c bng cc du *

V d: n = 4*

********

*******Bi 74: Vit chng trnh nhp chiu cao ca tam gic, xut ra mn hnh tam gic c bng cc du * v du [

V d: n = 4*

*[**[*[*

*[*[*[*

Bi 75: Vit chng trnh tnh phn t th n ca dy Fibonacci vi n nhp t bn phm. Bit dy Fibonacci c xc nh nh sau:

u0 = u1 = 1un+2 = un+1 + un (n > 0)

V d: n = 8 1 1 2 3 5 8 13 21.

Bi 76: Vit chng trnh nhp mt s t nhin ln hn 7. sau phn tch s ny thnh tng tuyn tnh ca hai s 3 v 5. V d: Nhp 14 14 = 3*3 + 1*5.

Bi 77: Vit chng trnh in ra mn hnh cc s nguyn c 4 ch s sao cho: Tng ch s hng trm + hng ngn = tng ch s hng chc + hng n v.

Bi 78: Nhp cc s nguyn dng a, b, c. Tm cc nghim nguyn dng ca phng trnh ax + by = c.

Bi 79: Nhp cc s nguyn dng e, a, b, c, d. Tm cc nghim nguyn dng ca h phng trnh.

x + y + z = eax + by + cz = d

Bi 80: Va g va ch, b li cho trn, ba mi su con, mt trm chn chn . Lp trnh tnh s g, s ch.

Bi 81: Vit chng trnh gii bi ton sau: 100 tru, 100 c, tru ng n 5, tru nm n 3, lkh tru gi, 3 con 1 b.Bi 82: C 3 loi giy bc 500, 200 v 50 ng. Tm tt c cc t hp c th c ca 3 loi giy bc trn cho s tin l 5750 ng.



Bi 83: Vit chng trnh nhp v kim tra gi tr 2 s nguyn m, n vi 1



b:=b+1;c:=c+1;d:=a+b+c;

End;

Procedure Q (var a: integer; b: integer; var c: integer);Begin

a:=a+1;b:=b+1;P (b,c,a);c:=c+1;d:=d+1;

End;

Begin {main}Clrscr;a:=1;b:=1;c:=1;d:=1;P(a,b,c);Writeln ( a= ,a, b= ,b, c= ,c, d= ,d);Q(b,c,a);Writeln ( a= ,a, b= ,b, c= ,c, d= ,d);Readln;

End.

Bi 85: Nhp vo 2 s nguyn m, n tho iu kin 0 < m < n < maxlongint. Vit chng trnh cho bit (mt cu lm trong 1 procedure hoc Function): Procedure nhp n, m n c chia ht cho m khng? (function) m c l s nguyn t khng? (function) Xut ra cc s nguyn t nh hn n v ln hn m. (Procedure)

Bi 86: Vit chng trnh nhp vo 2 phn s. Thc hin cc php ton +, -, *, / ca hai phn s ri in ra kt qu di dng ti gin.



Chuyn mngBi 87: Vit chng trnh theo dng chng trnh con thc hin cc yu cu sau

a) Nhp mng s nguyn c n phn t t bn phm (0 < n < 1000)b) Pht sinh ngu nhin mng s nguyn c n phn t (0 < n < 1000)c) Xut mng ra mn hnhd) Tnh trung bnh cng cc phn t m trong mng. Tnh trung bnh nhn cc phn t

dng trong mng.e) Tm phn t ln nht v nh nht trong mngf) Tm v xut ra mn hnh cc phn t l s nguyn t trong mngg) Sp xp mng c th t tng dnh) Sp xp mng c th t gim dni) Nhp vo s nguyn x, tm xem x c trong mng hay khng, nu c th bo tt c cc v

tr ca x trong mng.j) Nhp vo s nguyn x, xa phn t c gi tr bng x u tin tm thy trong mng. Bo v

tr va xa v in mng sau khi xa ra mn hnh.k) Nhp vo s nguyn x, xa tt c phn t c gi tr bng x tm thy trong mng. Bo cc

v tr va xa v in mng sau khi xa ra mn hnh.l) Nhp vo s nguyn x, xa t v tr x cho n cui mng, in mng sau khi xa ra mn

hnh.m) Nhp vo s nguyn x, xa t v tr x cho n u mng, in mng sau khi xa ra mnn) Nhp v tr x v s phn t cn xa y, xa i y phn t t v tr x v cui mng, in mng

sau khi xa ra mn hnh.o) Nhp vo v tr x v phn t y, chn phn t y vo v tr x trong mngp) Sp xp mng gim dn, Nhp vo phn t x, tin hnh chn phn t x vo mng sao

cho khng lm mng thay i tnh cht gim dnq) Nhp vo s nguyn p, tm v tr k tha mn ng thc sau:

[ ] [ ] [ ] [ ] [ ] [ ]( )nAkAkApkAAA +++++=+++ ...21*...21Trong A l mng theo bi

r) Xt tnh i xng, tnh n iu, tnh tun hon ca mng. Nu c th thng bo, nu khng c th sp xp mng gim dn. (S dng d liu mng gc ban u nhp vo)

s) T mng A ban u, tch thnh 2 mng con B v C. Mng B cha cc s hon ho, mng C cha cc s khng hon ho.

t) Nhp thm mng B, gp 2 mng A v B thnh mng C sao cho cc phn t ca mng B nm trc ri mi n mng A.

u) Nhp thm mng B, gp 2 mng A v B thnh mng C sao cho cc phn t ca mng B v B nm xen k nhau.

v) Sp xp mng A theo dng s l tng dn ri n s chn gim dn.w) Nhp vo mng B, kim tra xem mng B c phi l mng con ca mng A hay khng?x) Xc nh s mng con tng v s mng con gim trong mng A. In ra mn hnh mng con

tng di nht v mng con gim t nht.



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