## 1) Abridged problem statement

You are given `n` measured integers `x1..xn`. There are `m` quantization level sets `L0..L(m-1)`, each containing `s` integer levels (sorted). For each position `j`, you must choose an index `kj` (0…s−1) and quantize `xj` to level `Lj[kj]`, where `Lj` is the set used at that position.

The set used evolves deterministically:
- The first value uses set `L0`.
- If position `j` uses set `Li` and you choose index `kj`, then the next set index is  
  `f(j+1) = kj & (m-1)` (bitwise AND).

Goal: choose all indices `k1..kn` to minimize total deviation  
\[
\sum_{j=1}^{n} |x_j - L_{f(j)}[k_j]|
\]
Output the minimal deviation and the chosen indices `k1..kn`.

Constraints: `n ≤ 1000`, `m ≤ 128`, `s ≤ 128`, `m ≤ s`.

---

## 2) Detailed editorial (DP)

### Key observation
The choice of `kj` affects:
1) immediate error: `|xj - Lset[kj]|`
2) next set: `next_set = kj & (m-1)`

So a locally best level might lead to a worse next set, increasing future error. This is a classic dynamic programming with state = “position + current set”.

### DP definition
Let:

- `dp[j][set]` = minimal total deviation for quantizing positions `j..n-1` (0-based),
  **given that** position `j` must use level set `Lset`.

Answer will be `dp[0][0]` because the first quantization uses `L0`.

### Transition
At `(j, set)` we may choose any level index `k ∈ [0, s-1]`:

- quantized value: `L[set][k]`
- immediate cost: `abs(x[j] - L[set][k])`
- next set: `next_set = k & (m-1)`
- future cost: `dp[j+1][next_set]`

So:
\[
dp[j][set] = \min_{k=0..s-1} \left( |x[j] - L[set][k]| + dp[j+1][(k\ \&\ (m-1))] \right)
\]

Base case:
- `dp[n][set] = 0` for all `set` (no elements left).

### Reconstruction
To output the chosen `kj`, store:
- `choice[j][set]` = index `k` that achieved the minimum in `dp[j][set]`.

Then simulate forward:
- start `cur_set = 0`
- for `j=0..n-1`:
  - `k = choice[j][cur_set]`
  - output `k`
  - `cur_set = k & (m-1)`

### Complexity
States: `(n * m)` ≤ `1000 * 128 = 128k`  
Transitions per state: `s` ≤ 128  
Time: `O(n*m*s)` ≤ ~16 million operations (fast).  
Memory: `O(n*m)` for dp and choices.

---

## 3) Provided C++ solution with detailed line comments

```cpp
#include <bits/stdc++.h>
using namespace std;

// Pretty-print a pair as "first second"
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

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

// Read a whole vector: assumes size is already set
template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for (auto& x: a) in >> x;
    return in;
};

// Print a vector with spaces (not used in final output)
template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for (auto x: a) out << x << ' ';
    return out;
};

// A very large number used as +infinity for DP initialization
const int64_t inf = (int64_t)1e18 + 42;

int n, m, s;                 // n values, m sets, s levels per set
vector<int> x;               // measured values
vector<vector<int>> L;       // level sets: L[set][k]

void read() {
    cin >> n;                // read n
    x.resize(n);             // allocate x
    cin >> x;                // read all x[i]
    cin >> m >> s;           // read m and s
    L.resize(m, vector<int>(s));   // allocate L as m rows of length s
    for (auto& row: L) {
        cin >> row;          // read each row
    }
}

void solve() {
    /*
      DP meaning:
      dp[j][set] = min total deviation for positions j..n-1,
                   if position j uses level set 'set'.

      Transition:
      choose k in [0..s-1],
      next_set = k & (m-1),
      cost = abs(x[j] - L[set][k]) + dp[j+1][next_set].
    */

    // dp has n+1 rows so dp[n][*] exists and is 0
    vector<vector<int64_t>> dp(n + 1, vector<int64_t>(m, 0));

    // choice[j][set] remembers which k was best (for reconstruction)
    vector<vector<int>> choice(n, vector<int>(m, -1));

    // Fill DP backwards from j = n-1 down to 0
    for (int j = n - 1; j >= 0; j--) {
        for (int set = 0; set < m; set++) {
            dp[j][set] = inf;              // initialize as infinity

            for (int k = 0; k < s; k++) {  // try all possible level indices
                int next_set = k & (m - 1); // next set determined by low bits
                int64_t cost = llabs((long long)x[j] - L[set][k])
                             + dp[j + 1][next_set]; // immediate + future

                if (cost < dp[j][set]) {    // take best
                    dp[j][set] = cost;
                    choice[j][set] = k;     // store argmin
                }
            }
        }
    }

    // Minimal deviation starting at position 0 with set 0
    cout << dp[0][0] << '\n';

    // Reconstruct chosen indices
    int cur_set = 0;
    for (int j = 0; j < n; j++) {
        int k = choice[j][cur_set];        // chosen index at this state
        cout << k << " \n"[j == n - 1];    // space-separated, newline at end
        cur_set = k & (m - 1);             // advance to next set
    }
}

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

    int T = 1;
    // cin >> T;  // single test case in this problem
    for (int test = 1; test <= T; test++) {
        read();
        solve();
    }
    return 0;
}
```

---

## 4) Python solution (same algorithm, detailed comments)

```python
import sys

INF = 10**18

def solve() -> None:
    data = list(map(int, sys.stdin.buffer.read().split()))
    it = iter(data)

    n = next(it)
    x = [next(it) for _ in range(n)]

    m = next(it)
    s = next(it)

    # Read level sets L[set][k]
    L = [[next(it) for _ in range(s)] for _ in range(m)]

    # dp_next[set] will represent dp[j+1][set] while iterating backwards
    dp_next = [0] * m

    # To reconstruct, we need the chosen k for each (j, set)
    # choice[j][set] = best k at that state
    choice = [[0] * m for _ in range(n)]

    # Iterate j from n-1 down to 0
    for j in range(n - 1, -1, -1):
        dp_cur = [INF] * m

        for set_idx in range(m):
            best_cost = INF
            best_k = 0

            levels = L[set_idx]
            xj = x[j]

            # Try all k in this set
            for k in range(s):
                next_set = k & (m - 1)
                cost = abs(xj - levels[k]) + dp_next[next_set]
                if cost < best_cost:
                    best_cost = cost
                    best_k = k

            dp_cur[set_idx] = best_cost
            choice[j][set_idx] = best_k

        dp_next = dp_cur

    # dp[0][0] is now dp_next[0] after the loop
    out = []
    out.append(str(dp_next[0]))

    # Reconstruct k's
    cur_set = 0
    ks = []
    for j in range(n):
        k = choice[j][cur_set]
        ks.append(k)
        cur_set = k & (m - 1)

    out.append(" ".join(map(str, ks)))
    sys.stdout.write("\n".join(out))

if __name__ == "__main__":
    solve()
```

---

## 5) Compressed editorial

DP over position and current level-set.  
`dp[j][set]` = minimal deviation for `x[j..]` if `x[j]` uses set `set`.  
Transition over all `k` in `0..s-1`:
`dp[j][set] = min_k ( abs(x[j]-L[set][k]) + dp[j+1][k & (m-1)] )`.  
Base: `dp[n][*]=0`.  
Store `choice[j][set]` to reconstruct indices starting from `(0,0)`.  
Complexity `O(n*m*s)` (≤ 16M ops), memory `O(n*m)`.