bzoj2257

不难发现其实这个捣鼓过程就是一个辗转相减
所以最小燃料是瓶容量的最小公约数
然后穷举约数即可

var a:array[0..3000010] of longint;
    n,k,x,i,m,j,t:longint;

procedure sort(l,r: longint);
  var i,j,x,y:longint;
  begin
    i:=l;
    j:=r;
    x:=a[(l+r) shr 1];
    repeat
      while a[i]<x do inc(i);
      while x<a[j] do dec(j);
      if not(i>j) then
      begin
        y:=a[i];
        a[i]:=a[j];
        a[j]:=y;
        inc(i);
        j:=j-1;
      end;
    until i>j;
    if l<j then sort(l,j);
    if i<r then sort(i,r);
  end;

begin
  readln(n,k);
  for i:=1 to n do
  begin
    readln(x);
    m:=trunc(sqrt(x));
    for j:=1 to m do
      if x mod j=0 then
      begin
        inc(t);
        a[t]:=j;
        if j*j<>x then
        begin
          inc(t);
          a[t]:=x div j;
        end;
      end;
  end;
  sort(1,t);
  m:=1;
  for i:=t-1 downto 0 do
  begin
    if a[i]=a[i+1] then inc(m)
    else begin
      if m>=k then
      begin
        writeln(a[i+1]);
        halt;
      end;
      m:=1;
    end;
  end;
end.