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