## 1) Abridged problem statement

A platform has length \(L\) (0…\(L\)). There are \(M\) passengers at sorted positions \(P_1 \le \dots \le P_M\). A train has \(N\) doors: door 1 is at offset \(0\), and the remaining door offsets \(D_2,\dots,D_N\) are given, strictly increasing (so \(D_1=0\)).

If the train stops with its first door at position \(S\) on the platform, then door \(j\) is at position \(S + D_j\). The train must satisfy \(0 \le S\) and \(S + D_N \le L\).

For each passenger \(i\), define their distance to the closest door as
\[
\min_j |(S + D_j) - P_i|.
\]
Find a stop position \(S\) that **maximizes the sum** of these minimal distances over all passengers, and output \(S\) and the maximum sum. If multiple \(S\) work, output any.

Constraints: \(L \le 5000\), \(M,N \le 300\).

---

## 2) Detailed editorial (explaining the provided solution)

### Key observation: optimal \(S\) lies on integer or half-integer grid
For a fixed passenger at \(P\), the distance to the nearest door among positions \(\{S + D_j\}\) is a “V-shaped” piecewise-linear function of \(S\), and the *identity of the nearest door* changes when the passenger is exactly midway between two adjacent doors.

Midpoints between two doors \(S + D_j\) and \(S + D_{j+1}\) occur at:
\[
P = S + \frac{D_j + D_{j+1}}{2}
\quad\Rightarrow\quad
S = P - \frac{D_j + D_{j+1}}{2}.
\]
All \(P\) and \(D\) are integers, so \((D_j + D_{j+1})/2\) is integer or half-integer; therefore these “change points” for \(S\) are integer or half-integer. The sum over passengers is still piecewise-linear, with slope changes only at such points. Thus a maximum can be attained at some boundary point, i.e., on the grid of step \(0.5\).

So we can brute-force all feasible \(S \in \{0, 0.5, 1.0, \dots\}\) within \([0, L - D_N]\).

### Avoid floating point: scale all coordinates by 2
Multiply everything by 2:
- \(L_2 = 2L\)
- \(P_{i,2} = 2P_i\)
- \(D_{j,2} = 2D_j\)
- \(S_2 = 2S\) is now an integer iterating by 1.

Feasible range:
\[
0 \le S_2 \le L_2 - D_{N,2}.
\]

### Evaluating the score for a fixed \(S\)
For a passenger at \(P_{i,2}\), define the passenger coordinate in the “door-offset frame”:
\[
T = P_{i,2} - S_2.
\]
Now we need the nearest door offset \(D_{j,2}\) to \(T\) (minimizing \(|T - D_{j,2}|\)).

Because door offsets are sorted, the nearest door index \(j\) moves monotonically as \(T\) increases. Passengers \(P_i\) are given sorted, hence \(T\) is also sorted as we scan passengers for a fixed \(S\). Therefore we can use a single pointer \(j\) and advance it when the midpoint condition says “next door is closer”.

Nearest-door switch condition between \(D_j\) and \(D_{j+1}\) happens when \(T\) crosses their midpoint:
\[
T > \frac{D_j + D_{j+1}}{2}
\quad\Leftrightarrow\quad
2T > D_j + D_{j+1}.
\]
All in doubled integers, so we can test:
```cpp
while (j+1 < n && 2*T > D2[j] + D2[j+1]) j++;
```
Then add \(|T - D2[j]|\) to the sum.

This computes the total in \(O(M+N)\) per \(S\) (because \(j\) only moves forward), and we try \(O(L)\) half-steps, so:
\[
O((2L)\cdot(M+N)) \le 10000 \cdot 600 = 6\cdot 10^6
\]
which is fine.

### Track best and print in halves
Keep the best (maximum) doubled sum and doubled \(S\). To print:
- if value is even → print integer `/2`
- else → print integer part with `.5`

---

## 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 l, m, n;
vector<int> p;
vector<int> d;

void read() {
    cin >> l >> m;
    p.resize(m);
    cin >> p;
    cin >> n;
    d.resize(n);
    d[0] = 0;
    for(int i = 1; i < n; i++) {
        cin >> d[i];
    }
}

void solve() {
    // This is more of an implementation problem. As the L is fairly small, we
    // can try all possible starts S on either the integer, or .5 grid, and
    // simply evaluate linearly how much the utility would be.

    int l2 = l * 2;
    vector<int> p2(m), d2(n);
    for(int i = 0; i < m; i++) {
        p2[i] = p[i] * 2;
    }
    for(int i = 0; i < n; i++) {
        d2[i] = d[i] * 2;
    }

    int max_s = l2 - d2[n - 1];

    int64_t best_sum = -1;
    int best_s = 0;

    for(int s = 0; s <= max_s; s++) {
        int64_t sum = 0;
        int j = 0;
        for(int i = 0; i < m; i++) {
            int target = p2[i] - s;
            while(j + 1 < n && 2 * target > d2[j] + d2[j + 1]) {
                j++;
            }
            sum += abs(target - d2[j]);
        }
        if(sum > best_sum) {
            best_sum = sum;
            best_s = s;
        }
    }

    auto print_half = [](int64_t x) {
        if(x % 2 == 0) {
            cout << x / 2;
        } else {
            cout << x / 2 << ".5";
        }
    };

    print_half(best_s);
    cout << " ";
    print_half(best_sum);
    cout << "\n";
}

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 (same approach, with detailed comments)

```python
import sys

def print_half(x2: int) -> str:
    """Convert doubled integer to string with .5 if needed."""
    if x2 % 2 == 0:
        return str(x2 // 2)
    else:
        return f"{x2//2}.5"

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

    # Read platform
    L = next(it)               # platform length
    M = next(it)               # number of passengers
    P = [next(it) for _ in range(M)]  # passenger positions (sorted)

    # Read train
    N = next(it)               # number of doors
    D = [0] * N                # door offsets, with D1 = 0
    for i in range(1, N):
        D[i] = next(it)        # read D2..DN

    # Work in doubled coordinates to avoid floating point and to enumerate half-steps.
    L2 = 2 * L
    P2 = [2 * x for x in P]
    D2 = [2 * x for x in D]

    # Train must fit: 0 <= S <= L - D_N  =>  0 <= S2 <= 2L - 2D_N
    max_s2 = L2 - D2[-1]

    best_sum2 = -1
    best_s2 = 0

    # Enumerate all possible S2 (integer), corresponding to S in steps of 0.5.
    for s2 in range(0, max_s2 + 1):
        total2 = 0
        j = 0  # index of currently closest door for increasing targets

        # Scan passengers in sorted order; target = P2[i] - s2 is also sorted.
        for p2 in P2:
            target = p2 - s2

            # Advance to the next door while it becomes closer than current door.
            # Condition: target is beyond midpoint of D2[j] and D2[j+1]
            # midpoint test: target > (D_j + D_{j+1}) / 2
            # doubled: 2*target > D2[j] + D2[j+1]
            while j + 1 < N and 2 * target > D2[j] + D2[j + 1]:
                j += 1

            # Add distance to nearest door
            total2 += abs(target - D2[j])

        # Maximize the total distance
        if total2 > best_sum2:
            best_sum2 = total2
            best_s2 = s2

    # Output S and sum, both in original units (possibly .5)
    sys.stdout.write(f"{print_half(best_s2)} {print_half(best_sum2)}\n")

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

---

## 5) Compressed editorial

- Optimal nearest-door assignments change only when a passenger is at a midpoint between two adjacent doors; these midpoints imply \(S\) is integer or half-integer.
- Multiply coordinates by 2, enumerate all integer \(S_2 \in [0, 2L-2D_N]\).
- For fixed \(S_2\), for each passenger compute \(T = P_2 - S_2\) and find nearest door offset among sorted \(D_2\).
- Because passengers are sorted, \(T\) increases; maintain a pointer \(j\) to nearest door and advance while \(2T > D_2[j]+D_2[j+1]\).
- Sum \(|T - D_2[j]|\); keep maximal sum and corresponding \(S_2\).
- Print \(S\) and sum by converting doubled integers to integer/“.5”.