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?

有f(n) = f(n - 1) + f(n - 2)。
这是一个斐波那契数列。
方法1，递归，太慢；

方法2，迭代。

方法3，数学公式。斐波那契数列的通项公式为

class Solution {
public:
    int climbStairs(int n) {
        int prev = 0;
        int cur = 1;
        for(int i=1;i<=n;i++){
            int tmp=cur;
            cur+=prev;
            prev=tmp;
        }
        return cur;
    }
};

class Solution {
public:
    int climbStairs(int n) {
        double s = sqrt(5);
        return floor((pow((1+s)/2, n+1) + pow((1-s)/2, n+1))/s + 0.5);
    }
};