回复:可能是这样
A(0)---a------b-----c----d----e-----B(1000)
I guess
1. take 4 and 5 to b
2. meet 2 at a and pick 3 at when meet him
3. chnage 3 and 4 when meet 4 at c (i.e take 4 leave 3)
4. leave 4 at e
5. go back to meet 3 at d
then all arrive B at same time.
