You are given two integers n and k.
For any positive integer x, define the following sequence:
p0 = xpi+1 = popcount(pi) for all i >= 0, where popcount(y) is the number of set bits (1's) in the binary representation of y.This sequence will eventually reach the value 1.
\n\nThe popcount-depth of x is defined as the smallest integer d >= 0 such that pd = 1.
For example, if x = 7 (binary representation "111"). Then, the sequence is: 7 → 3 → 2 → 1, so the popcount-depth of 7 is 3.
Your task is to determine the number of integers in the range [1, n] whose popcount-depth is exactly equal to k.
Return the number of such integers.
\n\n\n
Example 1:
\n\nInput: n = 4, k = 1
\n\nOutput: 2
\n\nExplanation:
\n\nThe following integers in the range [1, 4] have popcount-depth exactly equal to 1:
| x | \n\t\t\tBinary | \n\t\t\tSequence | \n\t\t
|---|---|---|
| 2 | \n\t\t\t"10" | \n\t\t\t2 → 1 | \n\t\t
| 4 | \n\t\t\t"100" | \n\t\t\t4 → 1 | \n\t\t
Thus, the answer is 2.
\nExample 2:
\n\nInput: n = 7, k = 2
\n\nOutput: 3
\n\nExplanation:
\n\nThe following integers in the range [1, 7] have popcount-depth exactly equal to 2:
| x | \n\t\t\tBinary | \n\t\t\tSequence | \n\t\t
|---|---|---|
| 3 | \n\t\t\t"11" | \n\t\t\t3 → 2 → 1 | \n\t\t
| 5 | \n\t\t\t"101" | \n\t\t\t5 → 2 → 1 | \n\t\t
| 6 | \n\t\t\t"110" | \n\t\t\t6 → 2 → 1 | \n\t\t
Thus, the answer is 3.
\n\n
Constraints:
\n\n1 <= n <= 10150 <= k <= 5给你两个整数 n 和 k。
对于任意正整数 x，定义以下序列：
p0 = xpi+1 = popcount(pi)，对于所有 i >= 0，其中 popcount(y) 是 y 的二进制表示中 1 的数量。这个序列最终会达到值 1。
\n\nx 的 popcount-depth （位计数深度）定义为使得 pd = 1 的 最小 整数 d >= 0。
例如，如果 x = 7（二进制表示 \"111\"）。那么，序列是：7 → 3 → 2 → 1，所以 7 的 popcount-depth 是 3。
你的任务是确定范围 [1, n] 中 popcount-depth 恰好 等于 k 的整数数量。
返回这些整数的数量。
\n\n\n\n
示例 1:
\n\n输入: n = 4, k = 1
\n\n输出: 2
\n\n解释:
\n\n在范围 [1, 4] 中，以下整数的 popcount-depth 恰好等于 1：
| x | \n\t\t\t二进制 | \n\t\t\t序列 | \n\t\t
|---|---|---|
| 2 | \n\t\t\t\"10\" | \n\t\t\t2 → 1 | \n\t\t
| 4 | \n\t\t\t\"100\" | \n\t\t\t4 → 1 | \n\t\t
因此，答案是 2。
\n示例 2:
\n\n输入: n = 7, k = 2
\n\n输出: 3
\n\n解释:
\n\n在范围 [1, 7] 中，以下整数的 popcount-depth 恰好等于 2：
| x | \n\t\t\t二进制 | \n\t\t\t序列 | \n\t\t
|---|---|---|
| 3 | \n\t\t\t\"11\" | \n\t\t\t3 → 2 → 1 | \n\t\t
| 5 | \n\t\t\t\"101\" | \n\t\t\t5 → 2 → 1 | \n\t\t
| 6 | \n\t\t\t\"110\" | \n\t\t\t6 → 2 → 1 | \n\t\t
因此，答案是 3。
\n\n\n
提示:
\n\n1 <= n <= 10150 <= k <= 5n: let dp[pos][ones][tight] = number of ways to choose bits from the most significant down to position pos with exactly ones ones so far, where tight indicates whether you're still matching the prefix of n.",
"Precompute depth[j] for all j from 0 to 64 by repeatedly applying popcount(j) until you reach 1.",
"After your DP, let dp_final[j] be the count of numbers <= n that have exactly j ones; the answer is the sum of all dp_final[j] for which depth[j] == k."
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