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
1 var a:array[0..3000010] of longint; 2 n,k,x,i,m,j,t:longint; 3 4 procedure sort(l,r: longint); 5 var i,j,x,y:longint; 6 begin 7 i:=l; 8 j:=r; 9 x:=a[(l+r) shr 1]; 10 repeat 11 while a[i]<x do inc(i); 12 while x<a[j] do dec(j); 13 if not(i>j) then 14 begin 15 y:=a[i]; 16 a[i]:=a[j]; 17 a[j]:=y; 18 inc(i); 19 j:=j-1; 20 end; 21 until i>j; 22 if l<j then sort(l,j); 23 if i<r then sort(i,r); 24 end; 25 26 begin 27 readln(n,k); 28 for i:=1 to n do 29 begin 30 readln(x); 31 m:=trunc(sqrt(x)); 32 for j:=1 to m do 33 if x mod j=0 then 34 begin 35 inc(t); 36 a[t]:=j; 37 if j*j<>x then 38 begin 39 inc(t); 40 a[t]:=x div j; 41 end; 42 end; 43 end; 44 sort(1,t); 45 m:=1; 46 for i:=t-1 downto 0 do 47 begin 48 if a[i]=a[i+1] then inc(m) 49 else begin 50 if m>=k then 51 begin 52 writeln(a[i+1]); 53 halt; 54 end; 55 m:=1; 56 end; 57 end; 58 end.
