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)

0 bình luận về “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)”

1. Cái này mình làm thành 1 bài luôn nha, 4 câu a,b,c,d

   program tong;

   var

            i,S1,S2,S3,S4:longint;

   begin

              S1:=0; S2:=0; S3:=0; S4:=0;

              for i:=1 to 365 do

              begin

                      S1:=S1+i;

                      if i mod 2<>0 then S2:=S2+i else S3:=S3+i;

                      S4:=S4+i*i;

              end;

              writeln(‘a) ‘,S1);

              writeln(‘b) ‘,S2);

              writeln(‘c) ‘,S3);

               writeln(‘d) ‘,S4);

              readln;

   end.

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2. 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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