1. Abridged Problem Statement
Given integers N (1 ≤ N ≤ 10^7) and K (1 ≤ K ≤ 5000), and a list of K positions s₁,…,sₖ, compute all "self-numbers" in the interval [1..N]. A self-number is an integer that cannot be written in the form m + sum_of_digits(m) for any positive m. Let a[i] be the i-th self-number in ascending order. Output:
• First line: the total count of self-numbers in [1..N].
• Second line: K numbers a[s₁], a[s₂], …, a[sₖ]. It is guaranteed that all these requested a[sᵢ] lie within [1..N].

2. Detailed Editorial
Definition and goal
- Define d(m) = m + (sum of digits of m). A number x is called a generator of y if d(x) = y.
- A "self-number" is one that has no generator.
- We list all self-numbers a[1], a[2], … up to N, count how many there are, and answer K queries a[sᵢ].

Naïve vs. efficient approach
- Naïvely checking for each y whether any x < y satisfies d(x)=y is O(N²) in the worst case, too slow for N up to 10^7.
- Instead, we run a single pass for x = 1..N, compute y = d(x), and if y ≤ N, mark y as "has a generator." Then unmarked numbers are self-numbers. This is O(N · cost(sum_of_digits)) = O(N · log₁₀N), which for N=10^7 is fine in optimized C++.

Implementation details
1. Use a bitset `dp` over [0..len) where len is the smallest power of two exceeding N (chosen at compile time via template recursion `solve_fixed_len`), so the bitset is sized statically. `dp[i]` is set if i has a generator.
2. For x in 1..N:
   - Compute y = d(x) = x + sum_of_digits(x).
   - If y < len, set dp[y] = true.
3. Mark `useful_indices[s]` for every requested rank s.
4. Traverse i = 1..N in order; whenever dp[i] is false, i is the next self-number, so increment cnt. If `useful_indices[cnt]` is set, append i to the answer list.
5. Finally, print the total cnt on the first line, then for each query (in the original order) look up its answer through the coordinate-compressed list of distinct requested ranks and print it.

Time complexity
- O(N · digit_count) to mark generated numbers.
- O(N) to scan and count self-numbers.
- O(K log K) for query lookups.
Overall O(N) for N up to 10^7 is feasible in C++.

3. C++ Solution
```cpp
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

int n, k;
vector<int> a;

void read() {
    cin >> n >> k;
    a.resize(k);
    cin >> a;
}

const int MAXLEN = (int)1e7 + 42;

template<int len = 1>
void solve_fixed_len() {
    if(len <= n) {
        solve_fixed_len<min(len * 2, MAXLEN)>();
        return;
    }

    function<int(int)> nxt = [&](int x) {
        int res = x;
        while(x) {
            res += x % 10;
            x /= 10;
        }
        return res;
    };

    bitset<len> dp;
    bitset<len> useful_indices;
    for(int i: a) {
        useful_indices[i] = true;
    }

    for(int i = 1; i <= n; i++) {
        int nxt_i = nxt(i);
        if(nxt_i >= len) {
            continue;
        }
        dp[nxt_i] = true;
    }

    vector<int> ans;
    vector<int> compressed = a;
    sort(compressed.begin(), compressed.end());
    compressed.erase(
        unique(compressed.begin(), compressed.end()), compressed.end()
    );

    int cnt = 0;
    for(int i = 1; i <= n; i++) {
        if(!dp[i]) {
            cnt++;
            if(useful_indices[cnt]) {
                ans.push_back(i);
            }
        }
    }

    cout << cnt << '\n';
    for(int i: a) {
        int real_i = lower_bound(compressed.begin(), compressed.end(), i) -
                     compressed.begin();
        cout << ans[real_i] << ' ';
    }
    cout << '\n';
}

void solve() {
    // A number m is a self-number if no smaller k satisfies k + digitsum(k) ==
    // m. Mark every value d(i) = i + digitsum(i) as "has a generator", then the
    // unmarked values in [1, n] are exactly the self-numbers, in order.
    //
    // dp is a bitset over [0, len) flagging generated numbers; len is chosen at
    // compile time as the smallest power of two exceeding n (via the
    // solve_fixed_len template recursion) so the bitset is sized statically.
    // We sweep 1..n, count self-numbers, and for the queried 1-based indices
    // s_i collect the corresponding self-numbers. useful_indices marks which
    // ranks were requested; the queries are then answered through the
    // coordinate-compressed list of distinct requested ranks.

    solve_fixed_len();
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}
```

4. Python Solution with Detailed Comments
```python
import sys
data = sys.stdin.read().split()
# Parse inputs
n, k = map(int, data[:2])
queries = list(map(int, data[2:]))

# Step 1: mark which numbers have at least one generator
# We use a bytearray (faster than a list of bools)
has_generator = bytearray(n+1)

def d(x):
    """Compute d(x) = x + sum of digits of x."""
    s = x
    while x:
        s += x % 10
        x //= 10
    return s

# Mark all generated numbers up to n
for x in range(1, n+1):
    y = d(x)
    if y <= n:
        has_generator[y] = 1

# Step 2: collect self-numbers and answer queries
# We only need to store answers for requested positions s_i
needed = set(queries)
answers_for_pos = {}
count = 0

# Go through each i, if has_generator[i]==0 it's a self-number
for i in range(1, n+1):
    if not has_generator[i]:
        count += 1
        # if this rank is requested, record it
        if count in needed:
            answers_for_pos[count] = i

# Output total count
out = [str(count)]
# Output a[s1], a[s2], ..., a[sK] in original order
out.append(" ".join(str(answers_for_pos[s]) for s in queries))

sys.stdout.write("\n".join(out))
```

5. Compressed Editorial
Compute self-numbers by a single sweep:
1. Create a bitset `dp[0..len)` with len just above N.
2. For x from 1 to N, compute y = x + sum_of_digits(x); if y < len, set `dp[y]=true`.
3. Traverse i=1..N: if `dp[i]` is false, increment a counter `cnt` and, if `cnt` matches any query, store `i` as the answer for that query.
4. Print total `cnt` and the stored answers in query order.
