## 1) Concise, abridged problem statement

A ramen shop has **N counters**; counter *c* has **C[c] seats in a line**. There are **M customer groups** arriving in increasing time order. Group *i* arrives at time **Ti**, has **Pi** people (must sit in **Pi consecutive seats within a single counter**), will wait up to **Wi** time units (inclusive) for seats, and then eats for **Ei** time units once seated. The shop closes at time **T**: groups can only start eating **strictly before T**, but if they start, they may finish after T.

Groups form a **FIFO queue**: if earlier groups are waiting, a newly arrived group cannot be seated before them (even if seats are available), until earlier groups either get seated or give up.

When a group is to be seated, it chooses among all possible contiguous empty blocks of length Pi using these tie-breakers:

1. Maximize `min(SL, SR)`  
2. Then maximize `max(SL, SR)`  
3. Then smallest counter index  
4. Then leftmost position  

Here `SL`/`SR` are the numbers of consecutive empty seats directly to the left/right *after seating* within that counter; if there is no occupied seat on that side, it is treated as infinity.

Satisfaction per customer in group *i*:
- If the group leaves without eating: **-1**
- Else: `(Wi - wait_time) / Wi` (in [0,1])

Output the **average satisfaction over all customers** (weighted by group sizes).

---

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

### Key observations

- Constraints are small for the seating layout:  
  `N ≤ 100`, each `Ci ≤ 100` ⇒ at most **10,000 seats total**.
- But groups can be many: `M ≤ 10,000`.
- We must simulate time with events (arrivals, departures, and also “deadline checks” due to FIFO).

The tricky parts are:
1. **FIFO waiting rule**: only the front of the waiting queue is considered for seating; others behind cannot skip ahead.
2. **Seat choice rule**: choose the “most isolated” block by the `min/max(SL,SR)` criteria.

Given small seat counts, the C++ solution uses a straightforward scan for each seating attempt rather than complex interval trees.

---

### Data structures in the solution

- `cc[c][s]` (0/1): occupancy of seat `s` in counter `c`.
- `waiting`: a FIFO queue of group indices that have arrived and are not yet seated/gone.
- `ev`: a min-heap of events `(time, id)` where:
  - `id >= 0 and id < m`: an **arrival** of group `id`
  - `id < 0`: a **departure** event for group `(-id - 1)` (they finish eating and free seats)
  - `id == m`: a **dummy event** at `Ti + Wi` (deadline time) used to “wake up” the simulator even if nothing else happens; it does not directly do anything, but ensures the loop revisits the queue at that time.

Also stored:
- `opts[i]`: where group `i` was seated (counter index and starting position), so we can remove them on departure.
- `satisfaction[i]`: satisfaction value for group i (per customer).

---

### Event processing order

The simulator repeatedly takes the earliest event time `cur_time` from `ev`. If `cur_time >= T`, stop (no one can start after closing).

At each `cur_time`:
1. Pop **all events with that same time**.
2. Apply them:
   - Departures first (in practice: the code checks `id < 0` and removes seats)
   - Arrivals: push group into `waiting`
   - Dummy deadline events: ignored besides being popped
3. After all departures/arrivals at this time, try to seat groups from the **front** of `waiting`:
   - Let `g = waiting.front()`.
   - If `cur_time <= Ti + Wi`, attempt to seat them now.
     - If can seat:
       - schedule departure at `cur_time + Ei`
       - compute satisfaction: `(Ti + Wi - cur_time) / Wi`
       - pop from waiting and continue with next front
     - If cannot seat: stop trying now (because FIFO blocks later groups).
   - If `cur_time >= Ti + Wi`, the group’s waiting time expired; they leave with satisfaction -1, pop and continue.

The dummy event at `Ti+Wi` ensures that even if nothing changes (no arrivals/departures), the simulator will still wake up exactly when a waiting group expires and can be popped.

---

### How seat selection is computed (`can_seat(g)`)

For a group `g` of size `P`:

For each counter `c`:
- For each possible block start position `i` such that `i+P <= C[c]`:
  - Check whether seats `[i, i+P-1]` are all empty.
  - If yes, compute:
    - `left_span`: distance (in empty seats) from `i-1` going left until the nearest occupied seat; if none exists, **INF**.
    - `right_span`: distance from `i+P` going right until the nearest occupied seat; if none exists, **INF**.
  - Define:
    - `min_dist = min(left_span, right_span)`
    - `max_dist = max(left_span, right_span)`
  - Choose the best option by:
    1) larger `min_dist`
    2) larger `max_dist`
    3) smaller counter index
    4) smaller position

Implementation detail: the C++ code maintains a sliding window sum to test emptiness quickly, and tracks nearest occupied on both sides as it moves the window.

Once best option found:
- Mark those seats occupied in `cc`.
- Store option in `opts[g]`, set `seated[g]=true`.

Complexity:
- Worst-case per seating attempt: `O(total seats)` ≈ `O(10,000)` which is fine here.
- Each group is seated at most once; plus they may be checked a few times when they are at queue front.
- With small constants and tight time limit, the naive approach still passes given constraints (as intended by the author comment).

---

### Final answer computation

Total satisfaction over customers:
\[
\frac{\sum_i satisfaction[i] \cdot P_i}{\sum_i P_i}
\]
Groups that never eat keep satisfaction `-1`.

---

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

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

// Pretty-print a pair (not required for solution, just utility).
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

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

// Read a vector: reads exactly a.size() items into the vector.
template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
}

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

// A large integer used as "infinity" for spans when no neighbor exists.
const int inf = (int)2e9 + 42;

// Input variables:
// n = number of counters
// m = number of groups
// t = closing time
int n, m, t;

// Per-counter capacity.
vector<int> cap;

// Per-group parameters.
vector<int> start, group_size, deadline, eat_time;

// Read all input.
void read() {
    cin >> n >> m >> t;

    // Read capacities C1..Cn
    cap.resize(n);
    for(int i = 0; i < n; i++) {
        cin >> cap[i];
    }

    // Read group data (Ti, Pi, Wi, Ei) for i=0..m-1
    start.resize(m);
    group_size.resize(m);
    deadline.resize(m);
    eat_time.resize(m);
    for(int i = 0; i < m; i++) {
        cin >> start[i] >> group_size[i] >> deadline[i] >> eat_time[i];
    }
}

// cc[c][s] = 1 if seat s in counter c is occupied, else 0.
vector<vector<int>> cc;

// Satisfaction per group (per customer).
vector<double> satisfaction;

// Whether a group got seated (mostly bookkeeping).
vector<bool> seated;

// A seating option for a group:
struct Option {
    int min_dist; // min(SL, SR) to maximize first
    int max_dist; // max(SL, SR) to maximize second
    int idx;      // counter index
    int pos;      // starting seat position within counter
};

// Stores chosen option for each group (needed for removing later).
vector<Option> opts;

// Event priority queue: (time, id), ordered by smallest time.
// We use greater<> to make it a min-heap.
priority_queue<
    pair<int64_t, int>, vector<pair<int64_t, int>>, greater<pair<int64_t, int>>>
    ev;

// FIFO queue of waiting groups (indices).
queue<int> waiting;

// Comparator implementing the problem's preference ordering.
// Returns true if a is better than b.
bool is_better(const Option& a, const Option& b) {
    // Prefer larger min_dist
    if(a.min_dist != b.min_dist) {
        return a.min_dist > b.min_dist;
    }
    // Then prefer larger max_dist
    if(a.max_dist != b.max_dist) {
        return a.max_dist > b.max_dist;
    }
    // Then smaller counter index
    if(a.idx != b.idx) {
        return a.idx < b.idx;
    }
    // Then leftmost position
    return a.pos < b.pos;
}

// Try to seat group g at current time, following all seat preference rules.
// If successful, occupy seats and store the chosen option in opts[g].
bool can_seat(int g) {
    bool found = false;
    Option best_opt; // will hold best option found so far

    // Try every counter independently (group cannot span counters).
    for(int c = 0; c < n; c++) {
        int sum = 0;         // number of occupied seats in current sliding window
        int last_left, last_right; // nearest occupied seat info around the window

        // Try all starting positions i for a block of length group_size[g].
        for(int i = 0; i + group_size[g] <= cap[c]; i++) {

            // Maintain sliding window occupancy sum over seats [i, i+P-1].
            if(i == 0) {
                // Initialize sum for the first window.
                for(int j = 0; j < group_size[g]; j++) {
                    sum += cc[c][j];
                }
                // last_left: index of nearest occupied seat to the left of window.
                // For i==0 there is nothing to the left.
                last_left = -1;

                // last_right will be advanced to the first occupied seat on the right.
                last_right = group_size[g];
            } else {
                // Slide window by 1:
                // add new rightmost seat and remove old leftmost seat.
                sum += cc[c][i + group_size[g] - 1] - cc[c][i - 1];

                // If the seat that just moved out on the left was occupied,
                // it becomes the nearest occupied seat on the left side.
                if(cc[c][i - 1]) {
                    last_left = i - 1;
                }

                // Ensure last_right is at least the end of window (monotone).
                last_right = max(last_right, i + (int)group_size[g]);
            }

            // Move last_right rightwards until it hits an occupied seat or end.
            while(last_right < cap[c] && !cc[c][last_right]) {
                last_right++;
            }

            // If sum==0, window is fully empty: candidate placement.
            if(!sum) {
                // Compute SL (empty seats to left until occupied):
                // If no occupied on the left, treat as infinity.
                int left_span = (last_left == -1) ? inf : (i - 1 - last_left);

                // Compute SR (empty seats to right until occupied):
                // If no occupied on the right, treat as infinity.
                int right_span = (last_right == cap[c])
                                     ? inf
                                     : (last_right - i - group_size[g]);

                // Build the option according to preference keys.
                Option cur_opt = {
                    min(left_span, right_span), // min(SL,SR)
                    max(left_span, right_span), // max(SL,SR)
                    c,                          // counter
                    i                           // start position
                };

                // Keep the best option across all counters/positions.
                if(!found || is_better(cur_opt, best_opt)) {
                    found = true;
                    best_opt = cur_opt;
                }
            }
        }
    }

    // If found a seat block, commit it: mark seats occupied.
    if(found) {
        opts[g] = best_opt;
        seated[g] = true;
        for(int i = 0; i < group_size[g]; i++) {
            cc[best_opt.idx][best_opt.pos + i] = 1;
        }
    }

    return found;
}

// Remove (free) seats occupied by group g when they finish eating.
void remove_group(int g) {
    Option& opt = opts[g];
    for(int i = 0; i < group_size[g]; i++) {
        cc[opt.idx][opt.pos + i] = 0;
    }
}

void solve() {
    // Seat occupancy arrays per counter: initially all empty.
    cc.resize(n);
    for(int i = 0; i < n; i++) {
        cc[i].assign(cap[i], 0);
    }

    satisfaction.resize(m);
    opts.resize(m);
    seated.assign(m, false);

    // Initialize satisfaction to -1 (meaning "never ate" unless updated).
    // Also push events: arrivals at start[i], plus a dummy at start[i]+deadline[i].
    for(int i = 0; i < m; i++) {
        satisfaction[i] = -1;
        ev.emplace(start[i], i);                  // arrival event for group i
        ev.emplace(start[i] + deadline[i], m);    // dummy event at deadline time
    }

    // Process events in chronological order.
    while(!ev.empty()) {
        int64_t cur_time = ev.top().first;

        // No new seating can happen at/after closing time T.
        if(cur_time >= t) {
            break;
        }

        // Apply all events occurring at cur_time.
        while(!ev.empty() && ev.top().first == cur_time) {
            int id = ev.top().second;
            ev.pop();

            if(id < 0) {
                // Departure event: id encodes group index as -(g+1).
                remove_group(-id - 1);
            }
            if(id >= 0 && id < m) {
                // Arrival event: push group into FIFO waiting queue.
                waiting.push(id);
            }
            // id == m is dummy; nothing to do.
        }

        // After processing all departures/arrivals at this time,
        // repeatedly try to seat the front waiting group (FIFO rule).
        while(!waiting.empty()) {
            int g = waiting.front();

            // If still within allowed waiting time (inclusive at the boundary),
            // try to seat now.
            if(cur_time <= start[g] + deadline[g]) {
                if(can_seat(g)) {
                    // Schedule departure (free seats) after eating time.
                    ev.emplace(cur_time + eat_time[g], -(g + 1));

                    // Satisfaction = (remaining patience) / Wi
                    satisfaction[g] =
                        (double)(start[g] + deadline[g] - cur_time) /
                        deadline[g];

                    // Remove group from waiting; proceed to next.
                    waiting.pop();
                    continue;
                }
            }

            // If patience expired (or exactly at end but couldn't seat),
            // group leaves now with satisfaction -1 (already default).
            if(cur_time >= start[g] + deadline[g]) {
                waiting.pop();
                continue;
            }

            // Otherwise, cannot seat now and hasn't expired => stop.
            // FIFO means later groups cannot be tried.
            break;
        }
    }

    // Compute average satisfaction weighted by customer counts.
    int64_t tot_sz = 0;
    double tot = 0.0;
    for(int i = 0; i < m; i++) {
        tot += satisfaction[i] * group_size[i];
        tot_sz += group_size[i];
    }

    cout << fixed << setprecision(16) << tot / tot_sz << "\n";
}

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

    int T = 1;
    // cin >> T; // single test in this problem

    for(int test = 1; test <= T; test++) {
        read();
        solve();
    }
    return 0;
}
```

---

## 4) Python solution (same approach) with detailed comments

```python
import sys
import heapq
from collections import deque

INF = 2_000_000_042  # "infinity" span used in comparisons


def solve() -> None:
    data = sys.stdin.read().strip().split()
    it = iter(data)

    n = int(next(it))  # number of counters
    m = int(next(it))  # number of groups
    Tclose = int(next(it))  # closing time

    cap = [int(next(it)) for _ in range(n)]  # seats per counter

    start = [0] * m
    group_size = [0] * m
    deadline = [0] * m
    eat_time = [0] * m
    for i in range(m):
        start[i] = int(next(it))
        group_size[i] = int(next(it))
        deadline[i] = int(next(it))
        eat_time[i] = int(next(it))

    # cc[c][s] = 1 if occupied, else 0
    cc = [[0] * cap[c] for c in range(n)]

    # satisfaction per group, default -1 (means "gave up")
    satisfaction = [-1.0] * m

    # opts[i] stores (counter_idx, pos, group_len) to remove later
    # We'll also store spans if we want, but only counter and pos are necessary.
    opts = [None] * m

    # Event heap: (time, id)
    # id meanings:
    #   0..m-1 : arrival of group id
    #   <0     : departure of group (-id-1)
    #   == m   : dummy wake-up at start[i] + deadline[i]
    ev = []
    for i in range(m):
        heapq.heappush(ev, (start[i], i))
        heapq.heappush(ev, (start[i] + deadline[i], m))  # dummy

    waiting = deque()

    def better(a, b):
        """Return True if option a is better than b.
        option = (min_dist, max_dist, counter_idx, pos)
        """
        if a[0] != b[0]:
            return a[0] > b[0]
        if a[1] != b[1]:
            return a[1] > b[1]
        if a[2] != b[2]:
            return a[2] < b[2]
        return a[3] < b[3]

    def can_seat(g):
        """Try to seat group g now. If success, occupy seats and save opts[g]."""
        P = group_size[g]
        found = False
        best = None  # (min_dist, max_dist, c, pos)

        for c in range(n):
            C = cap[c]
            if P > C:
                continue

            # Sliding window sum over occupancy
            window_sum = 0
            last_left = -1
            last_right = P  # will move to first occupied seat to the right

            # init window [0, P-1]
            for j in range(P):
                window_sum += cc[c][j]

            # For each possible start i
            for i0 in range(0, C - P + 1):
                if i0 > 0:
                    # slide window by 1:
                    window_sum += cc[c][i0 + P - 1] - cc[c][i0 - 1]

                    # if seat leaving on the left was occupied, update last_left
                    if cc[c][i0 - 1]:
                        last_left = i0 - 1

                    # last_right must be at least end of window
                    if last_right < i0 + P:
                        last_right = i0 + P

                # move last_right to the next occupied seat or end
                while last_right < C and cc[c][last_right] == 0:
                    last_right += 1

                # If window is empty, compute option score
                if window_sum == 0:
                    left_span = INF if last_left == -1 else (i0 - 1 - last_left)
                    right_span = INF if last_right == C else (last_right - i0 - P)

                    mn = left_span if left_span < right_span else right_span
                    mx = right_span if left_span < right_span else left_span
                    cur = (mn, mx, c, i0)

                    if (not found) or better(cur, best):
                        found = True
                        best = cur

        if not found:
            return False

        # Occupy seats according to best option
        _, _, c, pos = best
        for k in range(group_size[g]):
            cc[c][pos + k] = 1

        opts[g] = (c, pos, group_size[g])
        return True

    def remove_group(g):
        """Free seats occupied by group g."""
        c, pos, P = opts[g]
        for k in range(P):
            cc[c][pos + k] = 0

    # Main simulation loop
    while ev:
        cur_time, eid = ev[0]

        # cannot start seating at/after closing time
        if cur_time >= Tclose:
            break

        # pop all events at cur_time
        while ev and ev[0][0] == cur_time:
            _, eid = heapq.heappop(ev)

            if eid < 0:
                g = -eid - 1
                remove_group(g)
            elif 0 <= eid < m:
                waiting.append(eid)
            # eid == m: dummy, do nothing

        # after applying all changes at this time, try to seat from the front
        while waiting:
            g = waiting[0]
            # If still within patience, try to seat now
            if cur_time <= start[g] + deadline[g]:
                if can_seat(g):
                    # schedule departure
                    heapq.heappush(ev, (cur_time + eat_time[g], -(g + 1)))

                    # compute satisfaction (per customer)
                    satisfaction[g] = (start[g] + deadline[g] - cur_time) / deadline[g]

                    waiting.popleft()
                    continue

            # If patience expired, group leaves now
            if cur_time >= start[g] + deadline[g]:
                waiting.popleft()
                continue

            # Otherwise cannot seat now and still waiting => stop (FIFO blocks)
            break

    # Weighted average satisfaction
    tot = 0.0
    tot_sz = 0
    for i in range(m):
        tot += satisfaction[i] * group_size[i]
        tot_sz += group_size[i]

    ans = tot / tot_sz
    # print with enough precision
    sys.stdout.write(f"{ans:.16f}\n")


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

---

## 5) Compressed editorial

Simulate the shop with a min-heap of time events and a FIFO waiting queue. Push an arrival event at `Ti` for each group, and also a dummy event at `Ti+Wi` to ensure the simulation wakes up when a front group’s patience expires. When processing time `cur`, first apply all departures (free seats) and arrivals (enqueue). Then repeatedly process only the front waiting group: if `cur > Ti+Wi` it leaves; otherwise attempt to seat it. Seating is done by scanning all counters and all contiguous blocks of length `Pi`, selecting the block maximizing `min(SL,SR)`, then `max(SL,SR)`, then smallest counter, then leftmost. Mark seats occupied, schedule a departure at `cur+Ei`, and set satisfaction `(Ti+Wi-cur)/Wi`. At the end output the customer-weighted average satisfaction, with groups that never ate contributing `-1`.