if input is
1
4
What should I output?
1
4
OR
1
1
?

and my program:
const
  max = 10000;
var
  num : array [1..max] of integer;
  o1 , o2 : array [0..max-1] of boolean;
  pre : array [0..max-1] of integer;
  i , j , n : integer;
procedure init;
begin
  readln(n);
  for i := 1 to n do
    readln(num[i]);
  fillchar(o1,sizeof(o1),0);
  o1[0] := true;
  fillchar(pre,sizeof(pre),0);
end;
procedure out;
var
  link , tot : integer;
  sz : array [1..max] of integer;
begin
  fillchar(sz,sizeof(sz),0);
  tot := 0;
  link := 0;
  repeat
    inc(tot);
    sz[tot] := pre[link];
    link := (link + n - num[pre[link]] mod n) mod n;
  until link = 0;
  writeln(tot);
  for link := 1 to tot do
    writeln(sz[link]);
  halt;
end;



procedure solve;
begin
  for i := 1 to n do begin
    o2 := o1;
    for j := 0 to n - 1 do
      if (o1[j]) and (pre[(j+num[i] mod n) mod n] = 0) then begin
        o2[(j+num[i] mod n) mod n] := true;
        pre[(j+num[i] mod n) mod n] := i;
      end;
    if (o2[j] = true) and (pre[0]<>0) then out;
    o1 := o2;
  end;
end;


begin
  init;
  solve;
  writeln(0);
end.

I think it's O(n^2),maybe TIME LIMIT EXCEED,but NOW I GET WA!

as they give you N values the next expresion:
Sum(first i elements) mod N
gives you in worst case, N different values, one of them of course is
equal to zero.



> const
>   max = 10000;
> var
>   num : array [1..max] of integer;
>   o1 , o2 : array [0..max-1] of boolean;
>   pre : array [0..max-1] of integer;
>   i , j , n : integer;
> procedure init;
> begin
>   readln(n);
>   for i := 1 to n do
>     readln(num[i]);
>   fillchar(o1,sizeof(o1),0);
>   o1[0] := true;
>   fillchar(pre,sizeof(pre),0);
> end;
> procedure out;
> var
>   link , tot : integer;
>   sz : array [1..max] of integer;
> begin
>   fillchar(sz,sizeof(sz),0);
>   tot := 0;
>   link := 0;
>   repeat
>     inc(tot);
>     sz[tot] := pre[link];
>     link := (link + n - num[pre[link]] mod n) mod n;
>   until link = 0;
>   writeln(tot);
>   for link := 1 to tot do
>     writeln(sz[link]);
>   halt;
> end;
>
>
>
> procedure solve;
> begin
>   for i := 1 to n do begin
>     o2 := o1;
>     for j := 0 to n - 1 do
>       if (o1[j]) and (pre[(j+num[i] mod n) mod n] = 0) then begin
>         o2[(j+num[i] mod n) mod n] := true;
>         pre[(j+num[i] mod n) mod n] := i;
>       end;
>     if (o2[j] = true) and (pre[0]<>0) then out;
>     o1 := o2;
>   end;
> end;
>
>
> begin
>   init;
>   solve;
>   writeln(0);
> end.
>
> I think it's O(n^2),maybe TIME LIMIT EXCEED,but NOW I GET WA!
>