GIO TRNH

BI DNG HC SINH GIITIN HC

H tn hc sinh:Lp:

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 ng Pascal.

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 trnhLu :*.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 m7) 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:

Ti liu bi dng hc sinh gii Tin hc THCS

Gio vin: Hong ng Quang in thoi: 090378159333

AbsoluteAndArrayBeginCaseConstExternalFileForForwardFunctionGotoModNilNotObjectOfOrShrStringThenToTypeUnitConstructorDesstructotDivDoDowntoElseEndIfImplementationInInlineInterfaceInterruptLabelPackedProcedureProgramRecordRepeatSetShlUntilUsesVarVirtualWhileWithXor

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 vinTypeKhai bo kiu d liuPHN KHAI BO CAConstKhai bo hng sCHNG TRNHVarKhai 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 lnh

Clrscr;Xa mn hnh

Writeln;Xung dng

Readln;Ngng thc hin lnh. Ch bm phm enter s thc hin tip

Write(Ni dung);Hin th phn ni dung ra mn hnh

Writeln(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 lu

Uses crt;BeginClrscr;Writeln(***************************);Writeln(* Chuong trinh Pascal dau tien *);Writeln(* Xin chao ban! *);Writeln(***************************);Readln;End.Bc 3: Nhp vo cc lnh sau:

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 KIUMIN GI TRB NH HAO TN NGHA

Byte0 2551 byteKiu s nguyn dng

Word0 655352 byteKiu s nguyn dng

Integer-32768..327672 byteKiu s nguyn

Longint-2147483648 21474836474 byteKiu s nguyn

Real2.9*10-39 1.7*10386 byteKiu s thp phn

Extended3.4*10-4932 1.110+493210 byteKiu s thp phn

Char1 k t1 byteKiu k t

String255 k t255 byteKiu chui k t

BooleanTrue, False1 byteKiu 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 lnh

Read(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 dng

Tn bin := gi trLnh gn gi tr cho bin. Lun ly gi tr v bn phi b qua v bn tri.

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 dKiu d liu p dng

+Cng 2 s hoc 2 chui k tS, chui

-Tr 2 sS

*Nhn 2 sS

/Chia 2 sS thc

DivChia 2 s ly phn nguynV d:6 div 4 110 div 2 57 div 9 00 div 2 0S nguyn

ModChia 2 s ly phn dV d:8 mod 2 07 mod 5 20 mod 7 06 mod 9 6S 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;VarSo1, so2, tong : integer;BeginClrscr;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.

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:

Uses crt;ConstPi = 3.1416159;VarSo1, so2, tong : integer;BeginClrscr;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.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:

nghaC phpLu

Tnh cn bc 2 ca mt sBin := Sqrt (gi tr s);Bin kiu s thc

Tnh bnh phng ca mt sBin := Sqr(gi tr s);Bin kiu s

Ly tr tuyt i ca mt sBin := Abs(gi tr s);Bin kiu s

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

Ly phn thp phn ca s thcBin := Frac(gi tr s);Bin kiu s thc

Ly th t ca k t trong bng m ASCIIBin := Ord (Chrk t);Bin kiu s nguyn

Cho bit k t trong bng m ASCII khi bit th tBin := Chr(gi tr s nguyn);Bin kiu k t

i k t thng thnh k t in hoaBin := Upcase (k t);Bin kiu k t

BAI 4: CU TRUC IU KIN CU LNH IF THEN

I. Biu thc iu kin:1) Khi nim:La nhng biu thc ma kt qua chi co 1 trong 2 gia tri la ung (true) hoc sai (false)2) Ton t ca biu thc iu kin:

Ton t ngha V dKiu d liu p dng

>Ln hnS, chui, lun l

=Ln hn hoc bngS, chui, lun l

= 8 thi in ra man hinh cu noi Chuc mung ban

Uses crt;Vartb : real;BeginClrscr;Writeln(Nhap diem trung binh: );Readln(tb);If (tb >= 8) then writeln(Chuc mung ban!);Readln;End.

2) Lnh IF THEN dang u:

Biu thc iu kinKhi lnh ngTrueFalseKhi lnh saiS :

Cu phap:

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

Y nghia: Nu biu thc iu kin ung thi thc hin cu lnh 1 phia sau THEN ngc lai thi thc hin cu lnh 2 phia sau ELSE

Vi du: Vit chng trinh nhp im trung bnh. Nu im trung bnh>= 8 thi in ra man hinh cu noi Chuc mung ban. Ngc li in ra mn hnh cu ni Ban can co gang hon

Uses crt;Vartb : integer;BeginClrscr;Writeln(Nhap diem trung binh: );Readln(tb);If (tb >= 8) then writeln(Chuc mung ban!)Else writeln(Ban can co gang hon);Readln;End.

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

Cu phap

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

V d:Vit chng trinh nhp im trung bnh. Nu im trung bnh>= 8 thi in ra man hinh hai cu noi trn 2 dng Chuc mung banBan duoc hoc bong 1000000Ngc li in ra mn hnh hai cu ni Ban can co gang honBan bi phat 100000

Uses crt;Vartb : integer;BeginClrscr;Writeln(Nhap diem trung binh: );Readln(tb);If (tb >= 8) then beginwriteln(Chuc mung ban!);writeln(Ban duoc hoc bong 1000000);endElse beginwriteln(Ban can co gang hon);writeln(Ban bi phat 100000);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;Vartb : integer;BeginClrscr;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.

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.

Uses crt;Varso : integer;BeginClrscr;Writeln(Nhap so: );Readln(so);Case so of1: 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.

III. Cu lnh Case .. Of dng :C phpCase (biu thc chn) ofHng 1 : khi lnh 1;Hng 2 : khi lnh 2;Hng 3 : khi lnh 3;Hng n : khi lnh n;Elsekhi lnh n+1;End;

Uses crt;Varso : integer;BeginClrscr;Writeln(Nhap so: );Readln(so);Case so of1: 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;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 .. doRepeat .. until

S :

Biu thc iu kinKhi lnh cn lpTruefalseThot khi vng lp

C php:

While (biu thc iu kin) dokhi lnh;

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

S :

Biu thc iu kinKhi lnh cn lpFalseTrueThot khi vng lp

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

Cch while .. doCch Repeat .. until

Uses crt;VarTu, mau:integer;BeginClrscr;Write(Nhap tu so: );Readln(tu);Write(Nhap mau so khac 0: );Readln(mau);While (mau = 0) doBeginWrite(Nhap mau so khac 0: );Readln(mau);End;Writeln(Binh phuong phan so vua nhap la: ,tu*tu, / , mau*mau);Readln;End.

Uses crt;VarTu, mau:integer;BeginClrscr;Write(Nhap tu so: );Readln(tu);RepeatWrite(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 .. doFor .. downto .. do

S :

Bin m = gi tr cuifalseTrueKhi lnh;Bin m := bin m - 1Thot khi vng lp

C php:

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

Uses crt;VarN, s, i : longint;BeginClrscr;Writeln(Nhap n: );Readln(n);For i := 1 to n dos := s + i;Writeln (S = ,s);Readln;End.

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;BeginCc 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 Procedureng 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 v c 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;BeginCc 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;VarTn 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 qu ra 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

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 nhpThu sut

0 400.000400.000 1.000.0001.000.000 2.000.000trn 2.000.0000%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 =2). Kim tra xem s n c phi l s nguyn t hay 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