## 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:

```cpp
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. Provided C++ solution with detailed comments

```cpp
#include <bits/stdc++.h> // Includes almost all standard C++ library headers.

using namespace std; // Allows using standard library names without std:: prefix.

// Overload output operator for pairs.
// This helper is not actually needed for this problem, but is harmless.
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second; // Print first and second separated by a space.
}

// Overload input operator for pairs.
// Also unused in this solution, but commonly included in templates.
template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second; // Read first and second.
}

// Overload input operator for vectors.
// Reads all elements of a vector.
template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) { // Iterate over every element by reference.
        in >> x;      // Read the element.
    }
    return in;        // Return stream to allow chaining.
};

// Overload output operator for vectors.
// Prints all elements separated by spaces.
template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {      // Iterate over vector elements.
        out << x << ' ';  // Print each element followed by a space.
    }
    return out;           // Return stream to allow chaining.
};

// Type alias for floating-point coordinates.
// Not really used in the dynamic programming solution.
using coord3_t = double;

// Constant pi, also unused here.
const coord3_t PI = acos((coord3_t)-1.0);

// Number of points Petya has to add.
int n;

// Reads input.
void read() {
    cin >> n; // Input contains a single integer N.
}

void solve() {
    // If no points are added, no score is gained.
    if(n == 0) {
        cout << fixed << setprecision(3) << 0.0 << '\n'; // Print 0.000.
        return; // Finish immediately.
    }

    // Very negative value used as negative infinity for impossible DP states.
    const double INF = -1e18;

    // dp[k][d] = maximum total score after adding exactly k points
    // and having maximum degree exactly d.
    //
    // Dimensions:
    // k ranges from 0 to n.
    // d can be at most n, so n+2 is enough.
    vector<vector<double>> dp(n + 1, vector<double>(n + 2, INF));

    // Base case:
    // After adding one point, there is one segment.
    // Maximum degree = 1, diameter = 1, score = 1.
    dp[1][1] = 1.0;

    // Process number of added points from 2 up to n.
    for(int k = 2; k <= n; ++k) {

        // Try every possible final maximum degree d.
        // It cannot exceed k because there are only k edges.
        for(int d = 1; d <= k; ++d) {

            // For fixed k and d, the maximum possible diameter is:
            //
            // k - (d - 2), when d >= 2.
            //
            // Explanation:
            // A diameter path can use at most two branches of a degree-d vertex.
            // The remaining d-2 branches need at least one edge each,
            // so they cannot be part of the diameter path.
            //
            // For d = 1, max(d-2, 0) keeps the expression valid for the base case.
            double dist = k - max(d - 2, 0);

            // Score gained at step k:
            // maximum degree times maximum distance.
            double p = d * dist;

            // Transition 1:
            // We came from a state with the same maximum degree d.
            // Geometrically, this means extending one of the long arms.
            if(d <= k - 1) {
                dp[k][d] = max(dp[k][d], dp[k - 1][d] + p);
            }

            // Transition 2:
            // We came from degree d-1 and created one new arm,
            // increasing the maximum degree by 1.
            if(d > 1 && d - 1 <= k - 1) {
                dp[k][d] = max(dp[k][d], dp[k - 1][d - 1] + p);
            }
        }
    }

    // The answer is the best total score among all possible final degrees.
    double ans = INF;

    // Check every possible final maximum degree.
    for(int d = 1; d <= n; ++d) {
        ans = max(ans, dp[n][d]); // Keep the maximum DP value.
    }

    // Print answer with exactly three digits after the decimal point.
    cout << fixed << setprecision(3) << ans << '\n';
}

int main() {
    // Speeds up C++ input/output.
    ios_base::sync_with_stdio(false);

    // Unties cin from cout for faster input.
    cin.tie(nullptr);

    // There is only one test case.
    int T = 1;

    // Run the single test case.
    for(int test = 1; test <= T; test++) {
        read();  // Read input.
        solve(); // Solve the problem.
    }

    return 0; // Successful program termination.
}
```

---

## 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.