There is a stair, the length of which is n. The person is at No.0 step at first, and his destination is the step NO.n. He has two kind of way forward: 1. One step forward(x->x+1). 2. Two steps forward(x->x+2). But there is an evil trap in one step, and the person has to skip it. That means if the number of that step is k, the person has to a two steps forward at the step k-1. In addition, due to a two steps forward very arduous. So he can\'t often use it. To simplify the problem, in the process of climbing stairs, the person had used the number of times one step forward must be not less than the number of times two steps forward. Now the person wants to know, how many methods exist that he can finish the climbing?
3 3 2 4 3 6 3
Case 1: 1 Case 2: 1 Case 3: 2