###### tags: `leetcode` `fibonacci`

# 70. Climbing
It is a fibonacci.
## me, use formula solution

class Solution:
    def climbStairs(self, n: int) -> int:
        x = 5**0.5/5
        c1 = (1+x)
        c2 = (1-x)
        return round((c1*((1+5*x)/2)**n + c2*((1-5*x)/2)**n)/2)

### How to do sqrt() faster?
https://stackoverflow.com/questions/3047012/how-to-perform-square-root-without-using-math-module
> 2**0.5 is pre-calculated, in CPython:

## me, DP

class Solution:
    def climbStairs(self, n: int) -> int:
        if n == 0:
            return 0
        if n == 1:
            return 1
        i = 1
        a = [1, 1]
        while i < n:
            a.append(a[-1] + a[-2])
            i += 1 
        return a[-1]

## Solution: Brute Force
$T: n^2$
$S: n^2$

class Solution:
    def climbStairs(self, n: int) -> int:
        if n==0:
            return 1
        if n==1:
            return 1
        return self.climbStairs(n-1)+self.climbStairs(n-2)

## Solution: Recursion with Memoization

## Solution: Markovian chain, O(lgn)

 public class Solution {
    public int climbStairs(int n) {
        int[][] q = {{1, 1}, {1, 0}};
        int[][] res = pow(q, n);
        return res[0][0];
    }
    public int[][] pow(int[][] a, int n) {
        int[][] ret = {{1, 0}, {0, 1}};
        while (n > 0) {
            if ((n & 1) == 1) {
                ret = multiply(ret, a);
            }
            n >>= 1;
            a = multiply(a, a);
        }
        return ret;
    }
    public int[][] multiply(int[][] a, int[][] b) {
        int[][] c = new int[2][2];
        for (int i = 0; i < 2; i++) {
            for (int j = 0; j < 2; j++) {
                c[i][j] = a[i][0] * b[0][j] + a[i][1] * b[1][j];
            }
        }
        return c;
    }
}