a) S1= 1+2+3+…+365 b) S2= 1+3+5+…+365 c) S3= 2+4+6+…+364 d)S4= 1^2+2^2+3^2+…+365^2 (Pascal,cấu trúc lặp)

bài chung

program tinhtong;

var i,s1,s2,s3:integer;

       s4:longint;

begin

clrscr;

s1:=0;  s2:=0;  s3:=0;  s4:=0;

for i:=1 to 365 do

 begin

  s1:=s1+i;

  if i mod 2=1 then s2:=s2+i;

  if i mod 2=0 then s3:=s3+i;

  s4:=s4+i*i;

 end;

writeln(‘S1= ‘,s1);

writeln(‘S2= ‘,s2);

writeln(‘S3= ‘,s3);

writeln(‘S4= ‘,s4);

readln

end.

bài 1:

program tinhtong;

uses crt;

var i,t:integer;

begin

clrscr;

t:=0;

for i:=1 to 365 do

 t:=t+i;

write(‘tong la: ‘,t);

readln

end.

Bài 2:

program tinhtong;

uses crt;

var i,t:integer;

begin

clrscr;

t:=0;

for i:=1 to 365 do

 if i mod 2=1 then t:=t+i;

writeln(‘Tong la: ‘,t);

readln

end.

bài 3:

program tinhtong;

uses crt;

var i,t:integer;

begin

clrscr;

t:=0;

for 2:=1 to 364 do

 if i mod 2=0 then t:=t+i;

writeln(‘Tong la: ‘,t);

readln

end.

bài 4:

program tintong;

uses crt;

var i:integer;

      t:longint;

begin

clrscr;

t:=0;

for i:=1 to 365 do

 t:=t+i*i;

write(‘tong la: ‘,t);

readln

end.

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