## 1) Abridged problem statement (concise)

You are given a description of a synchronous digital circuit with named junctions (wires). Some junctions are **primary inputs** (`INPUT(name)`), some are **requested outputs** (`OUTPUT(name)`), and the rest are defined by logic gates:

- `NOT(x)`
- `AND(x1, x2, ...)`, `NAND(...)`
- `OR(x1, x2, ...)`, `NOR(...)`
- `DFF(x)` — a D-type flip-flop: its output equals its input **delayed by one tick**; before the first tick all DFF outputs are `0`.

After the line `INPUT VALUES`, each following line gives a bitstring (in the same order as `INPUT(...)` lines) for one clock tick. For each tick, output a bitstring of the requested `OUTPUT(...)` junction values (in the same order as outputs were listed).
Constraints: up to 5000 junctions, up to 500 ticks.

---

## 2) Detailed Editorial

### Key observation
The circuit forms a directed dependency graph: each junction's value is a function of other junction values. The only sequential element is `DFF`, which outputs a stored state from the previous tick.

So at each tick we must:
1. Assign the primary input values for this tick.
2. Evaluate the requested output junctions by recursively evaluating dependencies.
3. Compute the **next** DFF states by evaluating each DFF's input (still within the current tick).
4. Update DFF states for the next tick.

The input guarantees the circuit "operates correctly" (i.e., no invalid combinational cycles). DFF breaks cycles by using previous state.

### Parsing
We read the file line-by-line, ignoring:
- comment lines starting with `#`
- empty lines

Before `INPUT VALUES`:
- `INPUT(name)` → create/get an id for `name`, mark it as an input node, and remember input order.
- `OUTPUT(name)` → remember its id in output order.
- `lhs = OP(arg1, arg2, ...)` → create/get id for `lhs`, set its gate type, store ids of its input junctions. If `OP` is `DFF`, also remember this node in a list of DFF nodes.

We map junction name → integer id (`unordered_map`) and store nodes in a vector.

### Evaluation per tick: DFS + memoization
We compute values on demand using a recursive function `eval(u)`.

To avoid recomputing the same node multiple times in one tick, we memoize:
- `val[u]` = computed value for current tick
- `seen[u]` = stamp of last tick in which `val[u]` was computed

We increment a global `stamp` each tick. Then "computed in this tick" is `seen[u] == stamp`.

Gate semantics:
- `INPUT`: value is directly set from the tick's input bitstring
- `DFF`: value is `dff_out[pos]` (stored previous-tick state)
- `NOT`: `1 - eval(in0)`
- `AND`: bitwise AND across inputs
- `NAND`: `1 - AND(...)`
- `OR`: bitwise OR across inputs
- `NOR`: `1 - OR(...)`

### Updating DFFs
Important: all DFF outputs for the tick come from the *old* `dff_out`.
After producing outputs, compute `next_dff[d] = eval(input_of_that_DFF)` for each DFF node.
Then swap `dff_out = next_dff`.

This implements the "one tick delay" exactly.

### Complexity
Let:
- `N` = number of junctions (≤ 5000)
- `Q` = number of ticks (≤ 500)

Each tick evaluates only the part of the circuit reachable from outputs and DFF inputs, and each node is evaluated at most once per tick due to memoization.
Time: `O(Q * (N + E))` in worst case, easily fits.
Memory: `O(N + E)`.

---

## 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;
};

enum GateType { G_INPUT, G_NOT, G_DFF, G_AND, G_NAND, G_OR, G_NOR };

struct Junction {
    int type = -1;
    vector<int> inputs;
};

vector<Junction> junctions;
unordered_map<string, int> name_id;
vector<int> input_order;
vector<int> output_order;
vector<int> dff_ids;
vector<string> tick_inputs;

static string trim(const string& s) {
    size_t a = s.find_first_not_of(" \t\r\n");
    if(a == string::npos) {
        return "";
    }
    size_t b = s.find_last_not_of(" \t\r\n");
    return s.substr(a, b - a + 1);
}

static int get_id(const string& name) {
    auto it = name_id.find(name);
    if(it != name_id.end()) {
        return it->second;
    }
    int id = (int)junctions.size();
    name_id[name] = id;
    junctions.emplace_back();
    return id;
}

static int op_type(const string& op) {
    if(op == "NOT") {
        return G_NOT;
    }
    if(op == "DFF") {
        return G_DFF;
    }
    if(op == "AND") {
        return G_AND;
    }
    if(op == "NAND") {
        return G_NAND;
    }
    if(op == "OR") {
        return G_OR;
    }
    if(op == "NOR") {
        return G_NOR;
    }
    return -1;
}

void read() {
    string line;
    bool reading_values = false;
    while(getline(cin, line)) {
        if(!line.empty() && line[0] == '#') {
            continue;
        }
        string t = trim(line);
        if(t.empty()) {
            continue;
        }
        if(!reading_values) {
            if(t == "INPUT VALUES") {
                reading_values = true;
                continue;
            }
            if(t.rfind("INPUT(", 0) == 0) {
                size_t l = t.find('('), r = t.find(')');
                string name = trim(t.substr(l + 1, r - l - 1));
                int id = get_id(name);
                junctions[id].type = G_INPUT;
                input_order.push_back(id);
            } else if(t.rfind("OUTPUT(", 0) == 0) {
                size_t l = t.find('('), r = t.find(')');
                string name = trim(t.substr(l + 1, r - l - 1));
                int id = get_id(name);
                output_order.push_back(id);
            } else {
                size_t eq = t.find('=');
                string lhs = trim(t.substr(0, eq));
                string rhs = trim(t.substr(eq + 1));
                size_t l = rhs.find('('), r = rhs.find(')');
                string op = trim(rhs.substr(0, l));
                string args = rhs.substr(l + 1, r - l - 1);
                int id = get_id(lhs);
                junctions[id].type = op_type(op);
                if(junctions[id].type == G_DFF) {
                    dff_ids.push_back(id);
                }
                stringstream ss(args);
                string tok;
                while(getline(ss, tok, ',')) {
                    tok = trim(tok);
                    if(!tok.empty()) {
                        int arg_id = get_id(tok);
                        junctions[id].inputs.push_back(arg_id);
                    }
                }
            }
        } else {
            tick_inputs.push_back(t);
        }
    }
}

void solve() {
    // This is again a problem where arguably parsing is the hardest part. We
    // can an AST for the expression and then process through it at each query:
    // there are of the order of 5000 nodes, and of the order of 500 queries, so
    // a naive implementation should work.

    int N = (int)junctions.size();
    int D = (int)dff_ids.size();
    vector<int> dff_pos(N, -1);
    for(int i = 0; i < D; i++) {
        dff_pos[dff_ids[i]] = i;
    }
    vector<int> dff_out(D, 0);
    vector<int> val(N, 0), seen(N, 0);
    int stamp = 0;

    function<int(int)> eval = [&](int u) -> int {
        if(seen[u] == stamp) {
            return val[u];
        }
        seen[u] = stamp;
        const Junction& j = junctions[u];
        switch(j.type) {
            case G_INPUT:
                break;
            case G_DFF:
                val[u] = dff_out[dff_pos[u]];
                break;
            case G_NOT:
                val[u] = 1 - eval(j.inputs[0]);
                break;
            case G_AND: {
                int r = 1;
                for(int v: j.inputs) {
                    r &= eval(v);
                }
                val[u] = r;
                break;
            }
            case G_NAND: {
                int r = 1;
                for(int v: j.inputs) {
                    r &= eval(v);
                }
                val[u] = 1 - r;
                break;
            }
            case G_OR: {
                int r = 0;
                for(int v: j.inputs) {
                    r |= eval(v);
                }
                val[u] = r;
                break;
            }
            case G_NOR: {
                int r = 0;
                for(int v: j.inputs) {
                    r |= eval(v);
                }
                val[u] = 1 - r;
                break;
            }
        }
        return val[u];
    };

    string out;
    for(const string& tick: tick_inputs) {
        stamp++;
        for(size_t i = 0; i < input_order.size(); i++) {
            int u = input_order[i];
            val[u] = tick[i] - '0';
            seen[u] = stamp;
        }
        out.assign(output_order.size(), '0');
        for(size_t i = 0; i < output_order.size(); i++) {
            out[i] = char('0' + eval(output_order[i]));
        }
        cout << out << '\n';
        vector<int> next_dff(D);
        for(int d = 0; d < D; d++) {
            next_dff[d] = eval(junctions[dff_ids[d]].inputs[0]);
        }
        dff_out.swap(next_dff);
    }
}

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

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        solve();
    }

    return 0;
}
```

---

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

```python
import sys

# Gate type constants
G_INPUT, G_NOT, G_DFF, G_AND, G_NAND, G_OR, G_NOR = range(7)

class Junction:
    __slots__ = ("type", "inputs")
    def __init__(self):
        self.type = -1       # will be set once we parse its definition
        self.inputs = []     # list of upstream junction indices

def trim(s: str) -> str:
    return s.strip()

def op_type(op: str) -> int:
    if op == "NOT":  return G_NOT
    if op == "DFF":  return G_DFF
    if op == "AND":  return G_AND
    if op == "NAND": return G_NAND
    if op == "OR":   return G_OR
    if op == "NOR":  return G_NOR
    return -1

def solve() -> None:
    # Circuit storage
    junctions = []            # id -> Junction
    name_id = {}              # name -> id
    input_order = []          # ids in INPUT(...) order
    output_order = []         # ids in OUTPUT(...) order
    dff_ids = []              # ids of DFF junctions
    tick_inputs = []          # list of tick bitstrings

    def get_id(name: str) -> int:
        """Return existing integer id for junction name, or create a new one."""
        if name in name_id:
            return name_id[name]
        idx = len(junctions)
        name_id[name] = idx
        junctions.append(Junction())
        return idx

    # -------- Parse input --------
    reading_values = False
    for raw in sys.stdin:
        if raw.startswith("#"):
            continue
        t = trim(raw)
        if not t:
            continue

        if not reading_values:
            if t == "INPUT VALUES":
                reading_values = True
                continue

            if t.startswith("INPUT("):
                l = t.find("(")
                r = t.find(")")
                name = trim(t[l+1:r])
                idx = get_id(name)
                junctions[idx].type = G_INPUT
                input_order.append(idx)

            elif t.startswith("OUTPUT("):
                l = t.find("(")
                r = t.find(")")
                name = trim(t[l+1:r])
                idx = get_id(name)
                output_order.append(idx)

            else:
                # Parse: lhs = OP(arg1, arg2, ...)
                eq = t.find("=")
                lhs = trim(t[:eq])
                rhs = trim(t[eq+1:])

                l = rhs.find("(")
                r = rhs.find(")")
                op = trim(rhs[:l])
                args = rhs[l+1:r]

                u = get_id(lhs)
                junctions[u].type = op_type(op)
                if junctions[u].type == G_DFF:
                    dff_ids.append(u)

                # Split args by comma and record their ids
                for part in args.split(","):
                    tok = trim(part)
                    if tok:
                        junctions[u].inputs.append(get_id(tok))
        else:
            tick_inputs.append(t)

    # -------- Simulation --------
    n = len(junctions)
    d = len(dff_ids)

    # Map junction id -> position in dff_out array (or -1 if not a DFF)
    dff_pos = [-1] * n
    for i, u in enumerate(dff_ids):
        dff_pos[u] = i

    # DFF outputs (state) before first tick are all 0
    dff_out = [0] * d

    # Memoization for each tick using stamp technique
    val = [0] * n
    seen = [0] * n
    stamp = 0

    sys.setrecursionlimit(1000000)

    def eval_node(u: int) -> int:
        """Compute junction u's value for current tick, memoized by stamp."""
        nonlocal stamp
        if seen[u] == stamp:
            return val[u]
        seen[u] = stamp

        j = junctions[u]
        t = j.type

        if t == G_INPUT:
            # value already set at tick start
            return val[u]

        if t == G_DFF:
            val[u] = dff_out[dff_pos[u]]
            return val[u]

        if t == G_NOT:
            val[u] = 1 - eval_node(j.inputs[0])
            return val[u]

        if t == G_AND:
            r = 1
            for v in j.inputs:
                r &= eval_node(v)
            val[u] = r
            return val[u]

        if t == G_NAND:
            r = 1
            for v in j.inputs:
                r &= eval_node(v)
            val[u] = 1 - r
            return val[u]

        if t == G_OR:
            r = 0
            for v in j.inputs:
                r |= eval_node(v)
            val[u] = r
            return val[u]

        if t == G_NOR:
            r = 0
            for v in j.inputs:
                r |= eval_node(v)
            val[u] = 1 - r
            return val[u]

        # Should not happen with correct input
        return 0

    out_lines = []
    for tick in tick_inputs:
        stamp += 1

        # Load primary inputs for this tick
        for i, u in enumerate(input_order):
            val[u] = ord(tick[i]) - ord("0")
            seen[u] = stamp

        # Produce outputs
        bits = []
        for u in output_order:
            bits.append("1" if eval_node(u) else "0")
        out_lines.append("".join(bits))

        # Compute and latch next DFF outputs
        next_dff = [0] * d
        for i, u in enumerate(dff_ids):
            # DFF has exactly one input
            next_dff[i] = eval_node(junctions[u].inputs[0])
        dff_out = next_dff

    sys.stdout.write("\n".join(out_lines))

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

---

## 5) Compressed editorial

Parse the circuit into nodes (junctions) with types and input edges; map names to ids. Keep arrays of input ids (order), output ids (order), and DFF ids. Maintain `dff_out[]` state initialized to 0.

For each tick bitstring:
1) set `val[input_id]` accordingly;
2) evaluate each output via DFS recursion with memoization (stamp trick) implementing gate logic; DFF returns `dff_out[pos]`;
3) compute `next_dff[d] = eval(DFF_input)` for all DFFs;
4) assign `dff_out = next_dff`; print output bitstring.

Each node is evaluated at most once per tick ⇒ fast for N ≤ 5000, ticks ≤ 500.
