DAP AN DE THI TIN HOC TRE KHONG CHUYEN HUONG SON

PROGRAM Bai1;
USES CRT;
VAR dem, n, i, max, maxd, tongc, soduong, tong: integer;
TBC: real;
a: ARRAY[1..100] OF integer;
BEGIN
Clrscr;
Write (‘Moi nhap so nguyen n: ‘); Readln(n);
Writeln(‘Moi nhap cac phan tu cua mang: ‘);
For i:=1 to n do
Begin
Write(‘Moi nhap so thu ‘, i, ‘ cua mang’);
Readln(a[i]);
End;
Max:=-32768; maxd:=1; TBC:=0; Tongc:= 0; Soduong:=0; tong:=0;
For i:=1 to n do
Begin
If max< a[i] then max:=a[i];
If maxd Tong:=tong+a[i];
If (a[i] mod 2 = 0) then dem:=dem+1;
End;
Write(‘So hang lon nhat cua day la:’, max:4);
Writeln;
Write(‘So hang duong lon nhat cua day la:’, maxd:4);
Writeln;
TBC:=tong/n;
Readln;
END.


PROGRAM Bai2;
USES CRT;
VAR
n , i, uoc, tongU: integer;
BEGIN
Clrscr;
Write (‘Moi nhap so nguyen n: ‘); Readln(n);
tongU:=0;
For i:=1 to n-1 do
If (n mod i = 0) then tongU:=tongU+i;
If (tongU=n) then Write (‘So’, n,’ la so hoan thien’)
Else
Write( ‘So ‘, n, ‘khong la so hoan thien’);
Readln;
END.

PROGRAM Bai3;
USES CRT;
VAR
n , i, : integer;
T: real;
BEGIN
Clrscr;
Write (‘Moi nhap so nguyen n: ‘); Readln(n);
T:=0;
For i :=n downto 1 do
T:= SQRT(T+ i);
T:= round(T,6);
Write(‘ Gia tri bieu thuc la: ‘, t:8:3);
Readln;
END.

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Bai 4 toi se gui sau vi quen mat doc tep va ghi tep