[ DjQb @ 06.06.2009. 16:25 ] @
| Ljudi ako moze neko da pomogne bio bih zahvalan. k=3, p = 1, r =3 ... trazi se kranji rezultat K ? var i,j,k,n,p,r:integer; begin read(k,p,r); n:=0 for j:=p to r do if j mod 2 = 0 then n:= n+p else n:=n+2*p; i=o; repeat k:= k+ sqr(i-1) + n; i:=i+1 until i>3; i:=1; while (i<5) do begin k:= 2*(k div i) +n; i:=i+2; end writeln(k) end |