minStep

有n层楼梯，每次可以爬1~2步，问能爬的最少的能整除 m 的步数是多少？ n <= 10000, m < 20.

请问怎么把下边的  DFS + Memoization 转换成 DP 数组递推?

class Solution:
    def minStep(self, n, m):
        if n < m:
            return -1
        # DFS + memoization,
        f = [[-1] * (n + 1) for _ in range(n + 1)]
        ans = self.dfs(n, m, 0, f)
        if ans == float("inf"):
            return -1
        return ans

    def dfs(self, n, m, steps, f):
        if n < 0:
            return float("inf")
        if n == 0:
            if steps % m == 0:
                f[n][steps] = steps
                return steps
            return float("inf")
        if f[n][steps] != -1:
            return f[n][steps]
        ans = float("inf")
        ans = min(ans, self.dfs(n - 1, m, steps + 1, f))
        ans = min(ans, self.dfs(n - 2, m, steps + 1, f))
        f[n][steps] = ans
        return ans


n, m = 271, 5
n, m = 10, 2
o = Solution()
ans = o.minStep(n, m)
print(ans)