## 1. Abridged problem statement

Starting with one point \(A_0=(0,0)\), add \(N\) points one by one. Each new point must be connected to exactly one previous point by a segment of length at most \(1\), without intersecting or covering existing segments/points except at the chosen endpoint.

After each added point, compute:

1. the maximum number of segments incident to a single point, i.e. the maximum degree of the drawn tree;
2. the largest Euclidean distance between any two drawn points, i.e. the geometric diameter.

The score for that step is the product of these two values. Find the maximum possible total score after all \(N\) points are added.

Constraints: \(0 \le N \le 1000\). Output the answer with accuracy \(10^{-3}\).

---

## 2. Detailed editorial

### Key interpretation

Because every new point is connected to exactly one previous point and intersections are forbidden, the drawn figure is always a tree embedded in the plane.

At any moment:

- the first value is the maximum degree of this tree;
- the second value is the largest Euclidean distance between two vertices.

Since every edge has length at most \(1\), the Euclidean distance between two vertices is at most the number of edges on the tree path between them.

---

### Best shape for fixed number of points and maximum degree

Suppose after adding \(k\) points, we have \(k\) edges total.

Let the maximum degree be \(d\).

If a vertex has degree \(d\), then any simple path in the tree can use at most two of the \(d\) branches incident to that vertex. The other \(d-2\) branches must contain at least one edge each and cannot contribute to the diameter path.

Therefore, for \(d \ge 2\),

\[
\text{diameter} \le k - (d-2) = k-d+2.
\]

For \(d=1\), this only really happens at \(k=1\), but the formula used in the code also handles it safely.

This upper bound is achievable:

- choose one central point;
- grow two long arms in opposite directions;
- put all remaining \(d-2\) arms as single-edge leaves from the center.

Then:

- the maximum degree is \(d\);
- the largest distance is between the endpoints of the two long opposite arms;
- that distance is exactly

\[
k-d+2.
\]

So for fixed \(k,d\), the best possible one-step score is

\[
d \cdot (k-d+2).
\]

The code writes this as `dist = k - max(d - 2, 0)`, which is equivalent for \(d \ge 2\), and also gives the correct value for the base case \(d=1,k=1\).

---

### Dynamic programming idea

The maximum degree can only:

- stay the same when we extend an existing arm;
- increase by \(1\) when we attach a new point to the current central point.

It cannot decrease, and it cannot increase by more than \(1\) in a single step.

Define:

\[
dp[k][d]
\]

as the maximum total score after adding exactly \(k\) points, ending with maximum degree \(d\).

For each state \((k,d)\), the score earned at step \(k\) is:

\[
p = d \cdot \left(k - \max(d-2,0)\right).
\]

There are two possible transitions into \((k,d)\):

1. We already had degree \(d\), and extended one of the long arms:

\[
dp[k][d] = \max(dp[k][d], dp[k-1][d] + p)
\]

2. We had degree \(d-1\), and created a new arm from the center:

\[
dp[k][d] = \max(dp[k][d], dp[k-1][d-1] + p)
\]

Base case:

\[
dp[1][1] = 1
\]

because after drawing one point, there is one segment, maximum degree \(1\), diameter \(1\), score \(1\).

If \(N=0\), the answer is \(0\).

The final answer is:

\[
\max_{d} dp[N][d].
\]

---

### Complexity

There are \(O(N^2)\) states and \(O(1)\) transitions per state.

For \(N \le 1000\), this easily fits.

Time complexity:

\[
O(N^2)
\]

Memory complexity:

\[
O(N^2)
\]

---

## 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;
};

using coord3_t = double;

const coord3_t PI = acos((coord3_t)-1.0);

int n;

void read() { cin >> n; }

void solve() {
    // We solve Game with points by dynamic programming on the optimal way
    // to grow a star-of-arms tree rooted at A0. This structure is optimal
    // because it simultaneously maximizes the current max_degree (number of
    // arms d) and the current max_distance (diameter of the two longest arms)
    // for any number of added points k.
    //
    // At step k (after adding the k-th point) with d arms:
    //   - max_degree = d
    //   - max_distance = k - max(d - 2, 0)   (achieved by placing the two
    //     longest arms opposite each other in straight lines of length 1;
    //     the remaining d-2 arms are single-point leaves)
    //
    // The contribution at step k is d * (k - max(d-2,0)).
    //
    // Transitions between steps respect the geometry and rules:
    //   - "extend" (attach new point to the end of one of the two longest
    //   arms):
    //     d stays the same, diameter increases by 1.
    //   - "new arm" (attach new point directly to A0):
    //     d increases by 1, diameter stays the same (new arm length 1).
    //
    // dp[k][d] stores the maximum total score after exactly k additions ending
    // with exactly d arms. The DP runs in O(N^2) time (N <= 1000) and selects
    // the globally optimal sequence of actions. For N=0 the score is 0.
    //
    // Why double? The problem requires the answer accurate to 1e-3, and output
    // must be printed with exactly three decimal places (fixed
    // setprecision(3)). Although the optimal score under this construction is
    // always an integer, we use double throughout for safety, precision
    // handling, and to match the required output format.
    //
    // Corner case / non-straight lines: bending an arm (non-straight) can never
    // improve the score. It either keeps diameter the same or decreases it
    // (by triangle inequality) while the max degree cannot exceed what the star
    // already achieves. Straight opposite arms strictly maximize diameter for
    // any given d and k, so any deviation is suboptimal or equal at best.
    // The construction also satisfies all geometric constraints (distinct
    // directions, no intersections, lengths <= 1, no point on segment).

    if(n == 0) {
        cout << fixed << setprecision(3) << 0.0 << '\n';
        return;
    }

    const double INF = -1e18;
    vector<vector<double>> dp(n + 1, vector<double>(n + 2, INF));
    dp[1][1] = 1.0;

    for(int k = 2; k <= n; ++k) {
        for(int d = 1; d <= k; ++d) {
            double dist = k - max(d - 2, 0);
            double p = d * dist;

            // came from extend (same d)
            if(d <= k - 1) {
                dp[k][d] = max(dp[k][d], dp[k - 1][d] + p);
            }
            // came from new arm (previous d-1)
            if(d > 1 && d - 1 <= k - 1) {
                dp[k][d] = max(dp[k][d], dp[k - 1][d - 1] + p);
            }
        }
    }

    double ans = INF;
    for(int d = 1; d <= n; ++d) {
        ans = max(ans, dp[n][d]);
    }
    cout << fixed << setprecision(3) << ans << '\n';
}

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

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

    return 0;
}
```

---

## 4. Python solution with detailed comments

```python
import sys

def solve():
    # Read the whole input.
    data = sys.stdin.read().strip()

    # If input is empty, do nothing.
    if not data:
        return

    # The input contains a single integer N.
    n = int(data)

    # If no points are added, the total score is zero.
    if n == 0:
        print("0.000")
        return

    # Negative infinity for impossible DP states.
    NEG_INF = -10**100

    # dp[k][d] = maximum total score after adding k points
    # and ending with maximum degree d.
    #
    # We allocate n + 1 rows for k = 0..n
    # and n + 2 columns for degrees.
    dp = [[NEG_INF] * (n + 2) for _ in range(n + 1)]

    # Base case:
    # After adding the first point:
    # - one segment exists,
    # - maximum degree is 1,
    # - diameter is 1,
    # - score is 1.
    dp[1][1] = 1.0

    # Fill DP for k = 2..n.
    for k in range(2, n + 1):

        # Maximum degree cannot exceed the number of edges k.
        for d in range(1, k + 1):

            # Maximum possible diameter for k edges and maximum degree d.
            #
            # For d >= 2:
            # diameter <= k - (d - 2)
            #
            # The expression below also handles d = 1 safely.
            dist = k - max(d - 2, 0)

            # Score gained after adding the k-th point.
            score = d * dist

            # Transition 1:
            # Previous maximum degree was also d.
            # We extend an existing arm.
            if d <= k - 1:
                dp[k][d] = max(dp[k][d], dp[k - 1][d] + score)

            # Transition 2:
            # Previous maximum degree was d - 1.
            # We add a new arm from the central point.
            if d > 1 and d - 1 <= k - 1:
                dp[k][d] = max(dp[k][d], dp[k - 1][d - 1] + score)

    # The final maximum degree can be anything from 1 to n.
    ans = max(dp[n][1:n + 1])

    # Required output accuracy is 1e-3.
    print(f"{ans:.3f}")


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

---

## 5. Compressed editorial

The drawing is always a tree. After \(k\) added points there are \(k\) edges. If the maximum degree is \(d\), a diameter path can use at most two branches of a degree-\(d\) vertex, so at least \(d-2\) edges are outside the diameter path. Hence the maximum possible diameter is:

\[
k-\max(d-2,0).
\]

This is achievable by making a central vertex with \(d\) arms: two long opposite arms and all other arms of length \(1\).

Use DP:

\[
dp[k][d] = \text{maximum total score after } k \text{ points with max degree } d.
\]

The score at step \(k\) is:

\[
d \cdot (k-\max(d-2,0)).
\]

Transitions:

- extend an existing arm:

\[
dp[k][d] \leftarrow dp[k-1][d] + score
\]

- create a new arm:

\[
dp[k][d] \leftarrow dp[k-1][d-1] + score
\]

Base:

\[
dp[1][1]=1.
\]

If \(N=0\), answer is \(0\). Otherwise output:

\[
\max_d dp[N][d].
\]

Complexity: \(O(N^2)\) time and \(O(N^2)\) memory.
