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

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

Trang 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

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

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

b Ky tu c xuat hien 1 lan, vi tri xuat hien dau tien 16

d Ki tu 0 xuat hien nhieu nhat 5

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 A xuat hien nhieu nhat 3

c Co 1 tuduocbatdauboi kytu thoac kytuT

Test

1

2

3

4

5

6

7

Input DAYSO.INP

chu nghia

Narn

Doc lap tu do hanh

phuc

truong trung hoc

pho thong so 1

nguyen van an

tran ba hoang

nguyen thi hang

nhung

Trang 2

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

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

Test

1

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 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

4

3 5 1 7

6 7 1

7

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

Trang 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;

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;

if ((st[i] = ' ') and (st[i+ll = 't')) or ((st[i] = ' ' ) and (st[i+l] = IT1)) then

inc(d2);

Trang 4

if d2 > 0 then

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

{ - dem ki tu xuat hien nhieu nhat trong xau -I

k:=O; j:= 0;

for ch := 'A' to '2' do

begin

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

uses crt ;

Const fi = 'c:\dayso.inpt;

fo = 'c:\dayso.out';

type mang = array[l 50] of longint;

n:integer ;

f,fl:text ;

{ - - - - - - - nhap du lieu -

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) ;

for i:=l to n do

begin

read(f,a[i]) ;

write(a[i],' ');

end;

close(f) ;

Trang 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;

pass:Array[O max+l] of Boolean;

N:integer;

Trang 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 ;

end

Định dạng
Số trang	6
Dung lượng	389,48 KB