Dap an de cuong on tap Tin Hoc 8(Duong Gia Thinh 8/4)

Baitap1

uses crt;
var a:real;
n,i,dem0,dem1,dem2:integer;
begin
clrscr;
write('Nhap so so hang trong mang: ');readln(n);
for i:=1 to n do
begin
write('Nhap so hang thu ',i,': ');readln(a);
if a>0 then
dem1:=dem1+1;
if a<0 then
dem2:=dem2+1;
if a=0 then
dem0:=dem0+1;
end;
a:=0;
end;
writeln('Co ',dem1,' so duong, ',dem2,' so am va ',dem0,' so khong.');
readln;
end.

Baitap2

uses crt;
var a:array [1..100] of integer;
b:array [1..100] of integer;
x,n,dem,i,j:integer;
begin
clrscr;
write('Nhap so x: ');readln(x);
write('Nhap so so hang trong mang: ');readln(n);
for i:=1 to n do
begin
write('Nhao so thu ',i,': ');readln(a[i]);
if a[i]=x then
begin
dem:=dem+1;
j:=j+1;
b[j]:=i;
end;
end;
write('Co ',dem,' phan tu giong x va o vi tri: ');
for i:=1 to j do
write(b[i]);
readln;
end.

Baitap3

uses crt;
var a:array [1..100] of integer;
n,min,i,j:integer;
begin
clrscr;
write('Nhap so so hang trong mang: ');readln(n);
for i:=1 to n do
begin
write('Nhap so thu ',i,': ');readln(a[i]);
end;
min:=a[1];
for i:=2 to n do
begin
if a[i]<min then
begin
min:=a[i];
j:=i;
end;
end;
if j>0 then
begin
a[1]:=a[1]+a[j];
a[j]:=a[1]-a[j];
a[1]:=a[1]-a[j];
end;
write('Mang sau khi duoc sap xep la: ');
for i:=1 to n do
write(a[i]:5);
readln;
end.

Baitap4

uses crt;
var a,i,j,dem,dem1,n:integer;
begin
clrscr;
write('Nhap so so hang: ');readln(n);
for i:=1 to n do
begin
write('Nhap so thu ',i,': ');readln(a);
for j:=1 to a do
if a mod j=0 then
dem:=dem+1;
if dem=2 then
dem1:=dem1+1;
end;
write('Co ',dem1,' so nguyen to trong day.');
readln;
end.

Baitap5

uses crt;
var a,b,c,i,dem,n:integer;
begin
clrscr;
write('Nhap so so hang: ');readln(n);
for i:=1 to n do
begin
write('Nhap so thu ',i,': ');readln(a);
b:=1;
c:=1;
if a>1 then
begin
repeat
b:=b+c;
c:=b+c;
until (a<=b)or(a<=c);
if (a=b)or(a=c) then
dem:=dem+1;
end;
if a=1 then
dem:=dem+1;
end;
write('Co ',dem,' so fibonaci.');
readln;
end.

Baitap6

uses crt;
var a:array[1..50]of string;
n,i:integer;
begin
clrscr;
write('Nhap so hoc sinh trong to: ');readln(n);
for i:=1 to n do
begin
write('Nhap ten hoc sinh thu ',i,': ');readln(a[i]);
end;
clrscr;
writeln('Danh sach to: ');
for i:=1 to n do
writeln(a[i]);
readln;
end.

Baitap7

uses crt;
var a:array[1..100]of integer;
i,n:integer;
begin
clrscr;
write('Nhap so so hang: ');readln(n);
for i:=1 to n do
begin
write('Nhap so thu ',i,': ');readln(a[i]);
end;
for i:=1 to n do
if abs(a[i])>1 then
write(a[i]:5);
readln;
end.

Baitap8

uses crt;
var a:array[1..100]of integer;
n,i,j,k,dem:integer;
begin
clrscr;
write('Nhap so so hang: ');readln(N);
for i:=1 to n do
begin
write('Nhap so thu ',i,': ');readln(a[i]);
end;
for i:=1 to n do
begin
for j:=1 to a[i] do
if a[i] mod j=0 then
dem:=dem+1;
if dem=2 then
begin
if k=0 then
writeln('Cac so nguyen to la: ');
k:=k+1;
write(a[i]:5);
end;
dem:=0;
end;
if k=0 then
write(‘Khong co so nguyen to trong mang.’);
readln;
end.

Baitap9

uses crt;
var a:array[1..100]of integer;
b,c,n,i,j:integer;
begin
clrscr;
write('Nhap so so hang: ');readln(n);
for i:=1 to n do
begin
write('Nhap so thu ',i,': ');readln(a[i]);
end;
for i:=1 to n do
begin
b:=1;
c:=1;
if a[i]>1 then
begin
repeat
b:=b+c;
c:=b+c;
until (a[i]<=b)or(a[i]<=c);
if (a[i]=b)or(a[i]=c) then
begin
j:=j+1;
if j=1 then writeln('Cac so fibonaci la: ');
write(a[i]:5);
end;
end;
if a[i]=1 then
begin
j:=j+1;
if j=1 then
writeln('Cac so fibonaci la: ');
write(a[i]:5);
end;
end;
if j=0 then writeln(‘Khong co so fibonaci nao trong mang. ‘);
readln;
end.

Baitap10

uses crt;
var a:array [1..100] of integer;
min:array[1..100] of integer;
n,i,j,k:integer;
begin
clrscr;
write('Nhap so so hang trong mang: ');readln(n);
for i:=1 to n do
begin
write('Nhap so thu ',i,': ');readln(a[i]);
end;
min[1]:=a[1];
for i:=2 to n do
begin
for j:=1 to i-1 do
if a[i]<=min[j] then
begin
for k:=i downto j+1 do
min[k]:=min[k-1];
min[j]:=a[i];
break;
end;
if min[j]<>a[i] then
min[j+1]:=a[i];
end;
write('Mang sau khi duoc sap xep la: ');
for i:=1 to n do
write(min[i]:5);
readln;
end.

Mỏi mắt nhỉ....