HDC de thi chon HSG THPT cap tinh nam 2008 2009 nam Tin hoc

Bạn đang xem bản rút gọn của tài liệu. Xem và tải ngay bản đầy đủ của tài liệu tại đây (389.48 KB, 6 trang )

(1)<div class='page_container' data-page=1>

S& GD - DT LAO CAI

D$

THI CHINH

THOC

KV

THI HOC SINH

GIOI

CAP

T ~ N H 
L ~ P 12 NAM HOC : 2007-2008

~

N

DAN

G

C H ~ I

-

M ~ N :

Tin hoe

( Hu6ng ddn chc4m gdm 2 trang )

A, TEST: Ph$n thi

l$p

trhh

CSlu 1: ( 7 di6m )

Did.

m

1

Output

DAYSO.0UT

a. So tu trong xau st: 4

b. Ky tu c xuat hien 2 lan, vi

tri

xuat hien dau tien 7
c.

Co

1 tu duoc bat dau boi ky tu t hoac ky

a

T
d. Ki tu N xuat hien nhieu nhat 3

a. So tu trong xau st: 8

b. Ky tu c xuat hien 1 lan, vi tri xuat hien dau tien 17
c. Khong co tu nao duoc bat dau boi ky tu t hoac ky tu

T

d. Ki tu A xuat hien nhieu nhat 4

So

tu trong xau st: 6

b. Ky tu c xuat hien 2 Ian, vi tri xuat hien dau tien 3
c. Co 1 tu duoc bat dau boi ky tu t hoac

ky

tu T
d. Ki tu

H

xuat hien nhieu nhat 3

a. So tu trong xau st: 7

b. Ky tu c xuat hien 1 lan, vi tri xuat hien dau tien 16
c. Co 3 tu duoc bat dau boi ky tu t hoac ky tu T
d. Ki tu 0 xuat hien nhieu nhat 5

a. So tu trong xau st: 3

b. Ky tu c khong xuat hien trong xau

c. Khong co tu nao duoc bat dau boi ky tu t hoac ky tu

T

d. Ki tu N xuat hien nhieu nhat 4

a. So tu trong xau st: 3

b. Ky tu c khong xuat hien trong xau

c. Co 1 tu duoc bat dau boi ky tu t hoac ky tu T
d. Ki tu A xuat hien nhieu nhat 3

So

tu trong xau st: 4

b. Ky tu c khong xuat hien trong xau

c. Co 1 tuduocbatdauboi kytu thoac kytuT
d. Ki tu N xuat hien nhieu nhat 5

Test

5

6

Input 
DAYSO.INP

Tin can ban

'Ong hoi
chu nghia
Narn

Doc

lap tu do hanh
phuc

truong trung hoc
pho thong so 1

nguyen van an

tran ba hoang

</div>
(2)<div class='page_container' data-page=2>

C l u 2: ( 7 di&m ) DCTANG.PAS

C l u 3:( 6 diim ) MUAVE-PAS 
Test

3

6

7

Output 
DAYCON.OUT

Day con tang lon nhat gom 4 pt:

1 2

3 4

Day con tang lon nhat gom 2 pt: 23 31

Day con tang lon nhat gom 3

pt:

23 65

76

Day con tang lon nhat gom

0

pt:

-

Day con tang lon nhat gom 3 pt:

5 6

7

Day con tang lon nhat gom 6 pt: 1 6

7 8

9 23

Day con

tang

Lon nhat gom 5 pt: 2

3

4

5

6

Input 
DAYCON.I[NP

7 4 2 1 2 3 4 1

4 233121209

10 965213423657643

24 27

5

2 2 2 2 2

8

4321 5 6 7 2 2 3

15

2 3 4 5 6 1 6 7 8 9 2 3 4 5

3

13

2 3 4 5 6 1 6 7 8 9 5 3 3

Test

1

2

3

4

5

6

Di6

m

1

Input

Luoi.inp

5

2 5 7 8

4

3

9 10 10

4

3 5 1 7

6 7 1

7

7 6 2 4 1 9 8

12

3 4

12 5

6 3

3

4 5 6 
10 6

6

34 54 23

12

4

3

21 23

45

3 1

13

2

3 8 
6

Output

Luoi.out

17

2

4

7

2

4

22

2 7

10

3

63

2

6

2

Didm

1

</div>
(3)<div class='page_container' data-page=3>

B, Hudng dan ch$m phiin viet chuung trinh va thuat toan.

(

gbm

4 trang

)

Hudng d&n chung.

Thu& t d n hqp

19:

1 didm

Khai Mo bign, h h g hqp

19:

0.5 didm

Khai biio d c chucmg trinh con hqp

19:

0.5 didm

EM

cuc chucmg trinh h q ~

19:

0.5 didm

C l u 1 ( 2.5 diem )
program bail;
uses crt;
Const

fi='c:\xau.inp' ;

fo='c:\xau.outl;
var

st: string;

d: array[1..100] of integer;
f,fl:text;

{

- -- -

-

- -- -

doc du lieu

---

I

Procedure nhap;

Var i,dem j,k,t,dl,d2,il:integer;
ch:char;

vt:array[l..SO] of integer;
luu: string;

Begin

assign(f,fi);
reset(f);
assign(fl,fo);
rewrite(f1);
read(f,st);

Writeln(' Xau st: ',st);
{

---

dem tu

---I

st :=

'

+

st;

dem := 0;

for i := 1 to length(st) do

if (st[i] =

'

') and (st[i+l] o

'

') then dem := dem

+

1;
Writeln(fl,' a. So tu trong xau st: ',dem);

{

-

---

--

-

kiem tra ky tu c

---I

delete(st,l,l);

il:= 0; dl:=O;

for i := 1 to length(st) do
if st[i] = 'c' then

Begin
inc(d1);
inc(i1);
vt[il] := i;
end;
if d l > 0 then

Writeln(f1,' b. Ky tu c xuat hien ',dl,

'

lan, vi tri xuat hien dau tien ',vt[l])
else

Writeln(f1,' b. Ky tu c khong xuat hien trong xau');
d2 :=O; st :=

'

+

st;

for i := 1 to length(st) do

</div>
(4)<div class='page_container' data-page=4>

if d2 > 0 then

Writeln(f1,' c. Co ',d2,' tu duoc bat dau boi ky tu t hoac ky tu T')
else

Writeln(f1,' c. Khong co tu nao duoc bat dau boi ky tu t hoac ky tu T');
{

---

dem ki tu xuat hien nhieu nhat trong xau

---I

k:=O; j:= 0;

for ch := 'A' to

'2'

do
begin

for i := 1 to length(st) do

if upcase(st[i]) = ch then inc(k);
if j < k then

begin
j := k;
luu := ch;
end;
k := 0;
end;

W r i t ! ~ 1 1 l h ' , h l l , h m ~ ~ ' & - - 
--

Close(f);
Close(f1);
End;

{

-

-.

- - -

.

-

-.

1

BEGIN
clrscr ;

nhap;
readln;

END.

C i u 2 ( 2.5 di&m ) DAYCT.PAS
program day-con-tang ;

uses crt ;

Const

fi

= 'c:\dayso.inpt;

fo = 'c:\dayso.out';

type mang = array[l..50] of longint;
var a, a1:mang ;

n:integer ;

f,fl:text ;

{

-

- -

-

--

-

nhap du lieu---

I

p r o c e d u S ~ n p u t ; - - -

var i:integer ;

begin

as&!!n(f,fi) ;

reset(f) ;

Writeln('Doc day so:');
readln(f,n) ;

writeln(n) ;

for i:=l to n do
begin

read(f,a[i]) ;

write(a[i],' ');

</div>
(5)<div class='page_container' data-page=5>

end;

{

---

x u ly

---

----

1

procedure daytang;

Var i j,dem,max,vtl,vt2:integer; 
Begin

dem := 1; max := 0;

for i := 1 to n-1 do
if a[i]ca[i+l] then

begin

j := i;

while (au] c alj+l]) and (j <= n) do
Begin

inc(dem);
inc(j);
end;

if max c dem then
Begin

max := dem;
vtl:= i;

vt2:= j;

end;
dem := 1;
end;
Writeln;

Write(' Day con tang Ion nhat gom: ',max,' phan tu: ');

for i := vtl to vt2 do

Write(a[i]:4);

{

--

---

- -

Ghi file

----

}

assign(f1,fo) ;

Rewrite(f1) ;

Writeln;

Write(f1,'Day con tang Ion nhat gom

',max,'

pt:');
for i := vtl to vt2 do

Write(fl,a[i]:4);
Close(f1);

end;
begin

clrscr ;

input;
daytang;
readln ;

end.

Cilu 3 (2.5 di6m).
Program muave;

Uses

crt;

const inp='c:huave.inp1;
out= 'c:\muave.out';
max = 100;

Var f,t,R: Array[O..max+l] of Integer;

</div>
(6)<div class='page_container' data-page=6>

(

- -

-

- - -

-

Input---

1

Procedure nhap;
Var i:integer;

fl:text;
Begin

Assign(fl,inp);
Reset(fi) ;

Readln(fi,N);
for i := 1 to N do
Read(fl,t[il);

for i := 1 to N

-

1 do
Read(fl,R[i]);

close(fl);
End;

{

-

- - -

x u ly

---

1

Procedure Solve;
Var i : integer;
Begin

R[O]:= 0 ;

no]

:= 0;

fCl] := T[1];

Fillc

har (pass,sizeof(pass),O);
For i

:=

2 to n do

if fli-2]+ R[i-11 c T[i]

+

fli-11 then
Begin

pass[i] := True;
pass[i-1] := False;
f[i] := fli-21

+

R[i-11
end

Else

fCi] := T[i]

+

fCi-11;
end;

(---.-xuat---.--

1

Procedure xuat;
Var

i:integer;
fo:text;
begin

Assign(fo,out);
Rewrite(f0);
Writeln(fo,f[N]);
For i := 1 to N do

if pass[i] then Write(fo,i,' ');

close(f0);
End;

{

- - - - -- - - -- -

-

1

Begin

Clrscr ;

nhap;
Solve;
xuat;
readln ;

</div>

HDC de thi chon HSG THPT cap tinh nam 2008 2009 nam Tin hoc

<b>D$ </b>

<b>THOC </b>

<b>KV </b>

<b>GIOI </b>

<b>CAP </b>

~

N

<b>DAN </b>

G

<b>C H ~ I </b>

-

M ~ N :

<b>Tin hoe </b>

<b>A, TEST: Ph$n thi </b>

<b>l$p </b>

<b>trhh </b>

<b>m </b>

<b>tri </b>

<i><b>Co </b></i>

a

T

<i><b>So </b></i>

<b>ky </b>

H

T

<i><b>So </b></i>

<b>Doc </b>

<b>Day con tang lon nhat gom 4 pt: </b>

<b>1 2 </b>

<b>3 4 </b>

<b>Day con tang lon nhat gom 2 pt: 23 31 </b>

<b>Day con tang lon nhat gom 3 </b>

<b>pt: </b>

<b>23 65 </b>

<b>76 </b>

<b>Day con tang lon nhat gom </b>

<b>0 </b>

pt:

-

<b>Day con tang lon nhat gom 3 pt: </b>

5

<b>6 </b>

<b>7 </b>

<b>Day con tang lon nhat gom 6 pt: 1 6 </b>

<b>7 8 </b>

<b>9 23 </b>

Day con

tang

<b>Lon nhat gom 5 pt: 2 </b>

<b>3 </b>

<b>4 </b>

<b>5 </b>

<b>6 </b>

7

4 2 1 2 3 4 1

4

<b>233121209 </b>

10

<b>965213423657643 </b>

<b>24 27 </b>

<b>5 </b>

<b>2 2 2 2 2 </b>

<b>8 </b>

<b>4321 5 6 7 2 2 3 </b>

<b>15 </b>

<b>2 3 4 5 6 1 6 7 8 9 2 3 4 5 </b>

<b>3 </b>

<b>3 </b>

<b>13 </b>

<b>2 3 4 5 6 1 6 7 8 9 5 3 3 </b>

<b>Di6 </b>

<b>m </b>

<b>1 </b>

<b>1 </b>

<b>1 </b>

<b>5 </b>

<b>2 5 7 8 </b>

<b>9 10 10 </b>

<b>4 </b>