TRNG THPT TRIU SN 2
IM
KIM TRA H C K è 2
MễN: Tin hc
Thi gian lm bi: 45 phỳt;
H tờn:..................................................Lp: 11....
Ho
I - TRC NGHIM (3 IM)
1. Khai bỏo no sau õy l ỳng cỳ phỏp
A. Type 1mang = array [1..100] of byte;
B. Type mang1c = array [1-100] of byte;
C. Type mang1c = array (1..100) of byte;
D. Type mang1c = array [1100] of byte;
2. on chng trỡnh sau khi thc hin cho kt qu gỡ
Var a:array[1..3] of byte; i:byte;
Begin
For i:=1 to 3 do a[i]:=i;
For i:=1 to 3 do
If a[i] mod 2 = 0 then Write(a[i]);
End.
A. 2 B. 1 C 1 2 3 D. 3
3. Cỏch khai bỏo xõu no l ỳng?
A. s : array[1..255] of char;
B. s : File of String;
C. s : char[256];
D. s : string[30];
4. Mt xõu cú ti a bao nhiờu kớ t?
A. 64 B. 255 C. 128 D. 256
5. Cõu lnh no xúa kớ t cui cựng ca xõu a?
A. delete(a,1,255)
B. delete(a,1,length(a))

chương trình sau, từ câu 1 đến câu 12
program hoadon;
(1)_____muaban = (2)_____
Stt : byte; tenhang : string[20]; dvt :(3)_____[5]; Slng:integer; Dgia, Thtien: real; (4)_____
Var N, i, j: byte; hd:(5)_____[1..30] of muaban; tong: real; vao, ra:text;
begin
assign(vao,'D:\ketoa.txt'); (6)_____(ra,'D:\toahang.txt'); (7)_____(vao); (8)_____(ra);
i:=1; tong:=0;
while not eof(vao) do
Begin
Readln(vao,hd[i].Tenhang); Readln(vao,hd[i].dvt); Readln(vao,hd[i].SLng);
(9)_____(vao,hd[i].dgia); hd[i].STT:= i; hd[i].thtien:= hd[i].Slng*hd[i].dgia;
tong:= tong + hd[i].thtien; i:=i+1; end;
Writeln(ra,'Hoa don ban hang so __ ');
For j:= 1 to i-1 do
(10)_____
Write(ra, hd[j].STT:3, hd[j].Tenhang:20, hd[j].dvt:5, hd[j].Slng:6, hd[j].dgia:8:0);
(11)_____(hd[j].Thtien:10:0); writeln(ra); End;
Write(ra,'Cong tien hang: ',tong:36:0);
(12)_____(vao);close(ra);
end.
1: A. typer B. const C. type D. record
2: A. rercod B. type C. record D. uses
3: A. trings B. integer C. char D. string
4: A. begin B. and C. ent D. end;
5: A. record B. array C. aray D. string
6: A. assign B. asisgn C. select D. atsign
7: A. rewrite B. select C. seret D. reset
8: A. select B. rewrite C. rewite D. reset
9: A. readln B. write C. writeln D. raedln