You are given a non-negative integer k. There exists a staircase with an infinite number of stairs, with the lowest stair numbered 0.
Alice has an integer jump, with an initial value of 0. She starts on stair 1 and wants to reach stair k using any number of operations. If she is on stair i, in one operation she can:
i - 1. This operation cannot be used consecutively or on stair 0.i + 2jump. And then, jump becomes jump + 1.Return the total number of ways Alice can reach stair k.
Note that it is possible that Alice reaches the stair k, and performs some operations to reach the stair k again.
\n
Example 1:
\n\nInput: k = 0
\n\nOutput: 2
\n\nExplanation:
\n\nThe 2 possible ways of reaching stair 0 are:
\n\nExample 2:
\n\nInput: k = 1
\n\nOutput: 4
\n\nExplanation:
\n\nThe 4 possible ways of reaching stair 1 are:
\n\n\n
Constraints:
\n\n0 <= k <= 109给你有一个 非负 整数 k 。有一个无限长度的台阶，最低 一层编号为 0 。
虎老师有一个整数 jump ，一开始值为 0 。虎老师从台阶 1 开始，虎老师可以使用 任意 次操作，目标是到达第 k 级台阶。假设虎老师位于台阶 i ，一次 操作 中，虎老师可以：
i - 1 ，但该操作 不能 连续使用，如果在台阶第 0 级也不能使用。i + 2jump 处，然后 jump 变为 jump + 1 。请你返回虎老师到达台阶 k 处的总方案数。
注意 ，虎老师可能到达台阶 k 处后，通过一些操作重新回到台阶 k 处，这视为不同的方案。
\n\n
示例 1：
\n\n输入：k = 0
\n\n输出：2
\n\n解释：
\n\n2 种到达台阶 0 的方案为：
\n\n示例 2：
\n\n输入：k = 1
\n\n输出：4
\n\n解释：
\n\n4 种到达台阶 1 的方案为：
\n\n\n\n
提示：
\n\n0 <= k <= 109x operations of the second type and y operations of the first type, the stair 2x + 1 - y is reached.",
"Since first operations cannot be consecutive, there are exactly x + 1 positions (before and after each power of 2) to perform the second operation.",
"Using combinatorics, we have x + 1Cy number of ways to select the positions of second operations."
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