Let $ f(x) $ denote the number of 1-bits in the binary representation of $ x $, i.e., the population count (popcount).
Define $ g(x) $ as the minimum number of applications of $ f $ required to reduce $ x $ to 1:
- $ g(1) = 0 $
- For $ x > 1 $, $ g(x) = 1 + g(f(x)) $
A number $ x $ is **special** if $ g(x) = k $.
Given:
- $ n $: a positive integer given in binary string representation (without leading zeros), with $ 1 \leq n < 2^{1000} $
- $ k $: an integer, $ 0 \leq k \leq 1000 $
**Objective:** Count the number of integers $ x \in [1, n] $ such that $ g(x) = k $, modulo $ 10^9 + 7 $.
---
### Formal Statement:
Let $ S = \{ x \in \mathbb{Z}^+ \mid 1 \leq x \leq n \text{ and } g(x) = k \} $
Compute $ |S| \mod (10^9 + 7) $
---
### Notes on $ g(x) $:
- $ f(x) = \text{popcount}(x) \leq \lfloor \log_2 x \rfloor + 1 $
- Since $ f(x) < x $ for all $ x > 1 $, the sequence $ x, f(x), f(f(x)), \dots $ strictly decreases until reaching 1.
- Thus, $ g(x) $ is well-defined and finite for all $ x \geq 1 $.
- $ g(x) = k $ means that after exactly $ k $ applications of $ f $, we reach 1, and not earlier.
Let $ h(m) = g(m) $. Then $ h $ can be precomputed for small values (since $ f(x) \leq 1000 $ for $ x < 2^{1000} $, and $ f(x) $ reduces the value drastically).
Let $ A_k = \{ m \in \mathbb{Z}^+ \mid g(m) = k \} $
We are to count:
$$
\left| \left\{ x \in [1, n] \mid \text{popcount}^{(k)}(x) = 1 \text{ and } \text{popcount}^{(k-1)}(x) > 1 \right\} \right|
$$
Equivalently, use dynamic programming over the binary digits of $ n $, tracking:
- Position in the binary string
- Tight constraint (whether prefix equals prefix of $ n $)
- Current popcount value (which is at most the number of bits so far, i.e., ≤ 1000)
- The number of operations applied so far (i.e., how many times we have "simulated" $ f $, but we cannot simulate full $ g $ without knowing the entire number)
But note: we cannot compute $ g(x) $ for arbitrary $ x $ without knowing $ x $, and $ x $ can be up to $ 2^{1000} $.
However, observe:
- $ f(x) = \text{popcount}(x) \leq \text{bit-length}(x) \leq 1000 $
- So $ f(x) \in [1, 1000] $
- Thus, $ g(x) = 1 + g(f(x)) $, and $ g $ only depends on the value of $ f(x) $, which is ≤ 1000.
So we can precompute $ g(m) $ for all $ m \in [1, 1000] $.
Let $ G[m] = g(m) $ for $ m = 1 $ to $ 1000 $
Then, for any $ x \leq n $, we have $ f(x) = \text{popcount}(x) \in [1, 1000] $, so:
> $ g(x) = k $ if and only if $ G[\text{popcount}(x)] = k - 1 $
Wait — correction:
By definition:
- $ g(x) = 1 + g(f(x)) $
- So $ g(x) = k \iff g(f(x)) = k - 1 \iff G[\text{popcount}(x)] = k - 1 $
Thus:
> $ x $ is special ⇔ $ G[\text{popcount}(x)] = k - 1 $
But note: if $ k = 0 $, then $ g(x) = 0 \Rightarrow x = 1 $
So:
- If $ k = 0 $: count numbers $ x \in [1, n] $ such that $ x = 1 $
- If $ k \geq 1 $: count numbers $ x \in [1, n] $ such that $ \text{popcount}(x) = m $, where $ G[m] = k - 1 $
Therefore, define:
Let $ T = \{ m \in [1, 1000] \mid G[m] = k - 1 \} $
Then the answer is:
$$
\sum_{m \in T} \left( \text{number of integers } x \in [1, n] \text{ with } \text{popcount}(x) = m \right)
$$
But note: if $ k = 1 $, then we require $ G[m] = 0 \Rightarrow m = 1 $, so we count numbers with exactly one 1-bit — i.e., powers of two.
If $ k = 0 $, then we only count $ x = 1 $, provided $ 1 \leq n $
Also, if $ T = \emptyset $, then answer is 0.
Thus, the problem reduces to:
> Given binary string $ n $, and a set $ T \subseteq [1, 1000] $, count the number of integers $ x \in [1, n] $ such that the number of 1-bits in $ x $ is in $ T $.
This is a classic digit DP problem.
---
### Final Formal Statement:
Let $ n $ be a binary string of length $ L \leq 1000 $, representing an integer $ N $.
Let $ G: [1, 1000] \to \mathbb{N} $ be defined recursively:
- $ G(1) = 0 $
- For $ m > 1 $, $ G(m) = 1 + G(\text{popcount}(m)) $
Let $ T = \{ m \in [1, 1000] \mid G(m) = k - 1 \} $
If $ k = 0 $, then $ T = \{1\} $ if $ k = 0 $, but wait — correction:
Actually:
- $ g(x) = 0 \iff x = 1 $ → so for $ k = 0 $, we want $ x = 1 $
- For $ k \geq 1 $, $ g(x) = k \iff g(f(x)) = k - 1 \iff G(\text{popcount}(x)) = k - 1 $
So define:
- If $ k = 0 $: answer = 1 if $ n \geq 1 $, else 0 → but $ n \geq 1 $ always, so answer = 1 if $ 1 \leq n $, which it is.
- But wait: what if $ k = 0 $ and $ n = 1 $? Then only $ x = 1 $, so 1 number.
- What if $ k = 0 $ and $ n > 1 $? Still only $ x = 1 $ qualifies → answer = 1.
But what if $ k = 0 $ and $ n = 0 $? But $ n \geq 1 $ per input.
So:
- If $ k = 0 $: answer = 1 (since $ x = 1 $ is always ≤ $ n $, and $ n \geq 1 $)
- If $ k \geq 1 $: let $ T = \{ m \in [1, 1000] \mid G(m) = k - 1 \} $, and compute
$$
\sum_{m \in T} \text{count}(m)
$$
where $ \text{count}(m) $ = number of integers $ x \in [1, n] $ with exactly $ m $ bits set to 1.
But note: $ x = 0 $ is not allowed (positive integers), and $ \text{popcount}(x) \geq 1 $, so we are safe.
Also, if $ m > L $, then $ \text{count}(m) = 0 $, since $ n $ has $ L $ bits, so no number ≤ $ n $ can have more than $ L $ bits set.
So we can compute $ \text{count}(m) $ for $ m = 1 $ to $ \min(L, 1000) $ using digit DP.
---
### Final Answer Formulation:
Let $ L = \text{len}(n) $, where $ n $ is given as a binary string.
Precompute $ G[1..1000] $:
- $ G[1] = 0 $
- For $ i = 2 $ to $ 1000 $: $ G[i] = 1 + G[\text{popcount}(i)] $
Let $ T = \begin{cases}
\{1\} & \text{if } k = 0 \\
\{ m \in [1, 1000] \mid G[m] = k - 1 \} & \text{if } k \geq 1
\end{cases} $
Define $ \text{dp}[i][tight][ones] $: number of ways to fill the first $ i $ bits of $ n $, with tight constraint, and having accumulated $ ones $ 1-bits.
Then compute:
$$
\text{ans} = \sum_{m \in T} \text{dp}[L][0][m] + \text{dp}[L][1][m]
$$
But note: we must exclude $ x = 0 $. Since we start from the first bit and require at least one 1, and $ n \geq 1 $, we are counting only $ x \in [1, n] $.
Actually, standard digit DP for numbers from 1 to $ n $ can be implemented by counting numbers from 0 to $ n $, then subtracting the count for 0 (which is 1 if 0 is included). But since we are counting numbers with exactly $ m \geq 1 $ bits set, 0 has 0 bits and won't be counted. So we can do DP from 0 to $ n $, and it's fine.
But to be safe: we can design DP to count numbers in $ [0, n] $ with popcount in $ T $, then subtract 1 if $ 0 \in T $ — but $ T \subseteq [1,1000] $, so 0 is never in $ T $. So we are safe.
Thus:
> Compute $ C = \sum_{m \in T} \text{count}_{\leq n}(m) $, where $ \text{count}_{\leq n}(m) $ = number of integers $ x \in [0, n] $ with exactly $ m $ bits set.
Then output $ C \mod (10^9 + 7) $
But note: if $ k = 0 $, then $ T = \{1\} $, but we want $ x = 1 $, which has popcount 1. So this matches.
Wait — if $ k = 0 $, then $ g(x) = 0 \iff x = 1 $, so we want numbers with $ g(x) = 0 $, which is only $ x = 1 $, which has popcount 1.
So yes, $ T = \{ m \mid G(m) = -1 \} $? No.
Wait — correction:
We defined:
- $ g(x) = k \iff G[\text{popcount}(x)] = k - 1 $
So for $ k = 0 $: $ G[\text{popcount}(x)] = -1 $ → impossible.
Mistake here.
Let’s re-derive carefully.
Define:
- $ g(1) = 0 $
- $ g(x) = 1 + g(f(x)) $ for $ x > 1 $
So:
- $ g(x) = 0 \iff x = 1 $
- $ g(x) = 1 \iff f(x) = 1 \iff \text{popcount}(x) = 1 $
- $ g(x) = 2 \iff f(x) \in \{2,3\} $ because $ g(2) = g(3) = 1 $, since $ f(2) = f(3) = 1 \Rightarrow g(2) = g(3) = 1 $
So:
- $ g(x) = k \iff g(f(x)) = k - 1 $
So for $ k \geq 1 $: $ g(x) = k \iff G[\text{popcount}(x)] = k - 1 $
But for $ k = 0 $: $ g(x) = 0 \iff x = 1 $
So we must handle $ k = 0 $ separately.
Thus:
- If $ k = 0 $: answer = 1 if $ n \geq 1 $, else 0 → always 1 (since $ n \geq 1 $)
- If $ k \geq 1 $: answer = number of $ x \in [1, n] $ such that $ \text{popcount}(x) \in T $, where $ T = \{ m \in [1, 1000] \mid G(m) = k - 1 \} $
So final formulation:
---
Let $ G[1] = 0 $
For $ m = 2 $ to $ 1000 $:
$ G[m] = 1 + G[\text{popcount}(m)] $
Let $ T = \begin{cases}
\emptyset & \text{if } k = 0 \text{ (handled separately)} \\
\{ m \in [1, 1000] \mid G[m] = k - 1 \} & \text{if } k \geq 1
\end{cases} $
Then:
$$
\text{Answer} =
\begin{cases}
1 & \text{if } k = 0 \\
\sum_{m \in T} \left( \text{number of integers } x \in [1, n] \text{ with } \text{popcount}(x) = m \right) & \text{if } k \geq 1
\end{cases}
\mod (10^9 + 7)
$$
And the count for each $ m $ is computed via digit DP on binary string $ n $.
---
**Note**: Since $ n < 2^{1000} $, its binary representation has at most 1000 bits. We can do DP with state:
$ dp[pos][tight][ones] $, where
- $ pos \in [0, L] $
- $ tight \in \{0,1\} $
- $ ones \in [0, 1000] $
Total states: $ 1000 \times 2 \times 1001 \approx 2 \times 10^6 $, feasible.
We precompute $ G[1..1000] $ in $ O(1000 \cdot \log 1000) $, trivial.
We then compute the digit DP.
Final output: answer mod $ 10^9 + 7 $
---
### Final Answer (Formal Mathematical Statement):
Let $ n \in \mathbb{N} $ be given in binary representation as a string of length $ L \leq 1000 $, with $ n \geq 1 $.
Define $ f(x) = \text{popcount}(x) $, and recursively define $ g: \mathbb{N}^+ \to \mathbb{N} $ by:
$$
g(x) =
\begin{cases}
0 & \text{if } x = 1 \\
1 + g(f(x)) & \text{if } x > 1
\end{cases}
$$
Let $ k \in \mathbb{N} $, $ 0 \leq k \leq 1000 $.
Define the set:
$$
S_k = \{ x \in \mathbb{N}^+ \mid x \leq n \text{ and } g(x) = k \}
$$
Compute $ |S_k| \mod (10^9 + 7) $.
---
**Note**: The problem reduces to counting numbers $ \leq n $ with popcount in a precomputed set $ T $, with $ T $ derived from $ G = g|_{[1,1000]} $, and handling $ k = 0 $ as a special case.