You are climbing a stair case. It takes n steps to reach to the top.

Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?

分析：这道题就相当于青蛙跳，就是一个费波那契数列。如果只有1级台阶，那显然只有一种跳法，如果有2级台阶，那么有两种跳的方法了:一种分两次条，每次跳一个；另外就是一次跳2级。我们把n级台阶的跳法看成是n的函数记为f(n)。当n>2时，f(n)=f(n-1)+f(n-2);

class Solution {public:    int climbStairs(int n) {        int result[2]={
   1,1};        if(n<2)            return result[n];        int fibN=0;        int fibOne=1;        int fibTwo=1;        for(unsigned int i=2;i<=n;++i)        {            fibN=fibOne+fibTwo;            fibOne=fibTwo;            fibTwo=fibN;        }        return fibN;    }};