<p>You are given an integer <strong>n</strong>. Your task is to place <strong>n unit squares</strong> (side length 1) inside an <strong>axis-parallel square container</strong> of side length <strong>L</strong> such that:</p> <ul> <li>Every unit square lies entirely inside the container</li> <li>Unit squares have <strong>no common interior points</strong> (touching edges/corners is allowed)</li> <li>Each unit square may be <strong>rotated by an arbitrary angle</strong></li> </ul> <h3>Goal</h3>

<strong>Minimize L</strong> (the container side length).

<h2>Input</h2> <p>A single integer <strong>n</strong> (1 �?n �?100,000).</p> <h2>Output</h2>

<strong>First line:</strong> A real number <strong>L</strong> (the claimed container side length)

<strong>Next n lines:</strong> Three real numbers <code>xi yi ai</code> for i = 1..n:

<ul> <li><code>(xi, yi)</code>: The center of the i-th unit square</li> <li><code>ai</code>: Its rotation angle in degrees counterclockwise (0 �?ai < 180)</li> </ul>

<strong>Precision:</strong> All numbers must be finite reals. At least <strong>6 decimal places</strong> is recommended.

<h3>Sample Output Format</h3> <pre><code class="language-">L x1 y1 a1 x2 y2 a2 ... xn yn an</code></pre> <h2>Validity Constraints</h2> <p>Your output is valid if and only if:</p> <p>1. <strong>Containment:</strong> Every point of every unit square lies inside [0, L] × [0, L] 2. <strong>Non-overlap (interiors):</strong> The interiors of any two unit squares are disjoint - Touching along edges or at corners is <strong>allowed</strong> 3. <strong>Angles:</strong> 0 �?ai < 180 for all i</p> <h3>Judge Verification (epsilon = 1e-7)</h3> <ul> <li><strong>Containment:</strong> A square is accepted if its maximum outward violation beyond the container is �?1e-7</li> <li><strong>Non-overlap:</strong> Two squares are accepted as interior-disjoint if the minimum signed distance between their boundaries is �?�?e-7</li> </ul>

<strong>Invalid submissions score 0 for that test.</strong>

<h2>Baseline</h2> <p>A simple baseline packs all squares <strong>unrotated</strong> (ai = 0) on the unit grid, with centers at (0.5 + x, 0.5 + y), using the smallest integer-sided container that fits all n.</p>

The baseline side length is:

<pre><code class="language-">L₀(n) = ceil(√n)</code></pre>

<strong>Example:</strong> For n = 11, the baseline uses L₀ = 4.000000

<h2>Scoring</h2> <p>Let <strong>L</strong> be your reported container side length (validated by the judge). Define:</p> <ul> <li><strong>LB = √n</strong> (trivial area lower bound; no packing can have L < LB)</li> <li><strong>L₀ = ceil(√n)</strong> (baseline side length)</li> <li><strong>s = s(n)</strong> (reference scale; satisfies LB �?s �?L₀)</li> </ul> <h3>Score Calculation</h3> <table> <thead> <tr> <th>Condition</th> <th>Score</th> </tr> </thead> <tbody> <tr> <td>Invalid submission</td> <td>0 points</td> </tr> <tr> <td>L �?L₀</td> <td>1 point</td> </tr> <tr> <td>L = LB</td> <td>100 points</td> </tr> <tr> <td>LB < L �?s</td> <td>95 + 5 × min(1.0, 1.1 × p�? where p�?= (s �?L) / (s �?LB)</td> </tr> <tr> <td>s < L < L₀</td> <td>94 × min(1.0, 1.1 × p�? + 1 where p�?= (L₀ �?L) / (L₀ �?s)</td> </tr> </tbody> </table> <h3>Scoring Notes</h3> <ul> <li><strong>Baseline (L = L₀):</strong> 1 point</li> <li><strong>Meeting s(n):</strong> At least 95 points</li> <li><strong>Area bound (L = LB):</strong> 100 points</li> <li>Scores vary smoothly between these anchors</li> <li>The +10% amplification is applied within each band and capped at that band's ceiling (95 or 100)</li> </ul> <h3>Reference Scale s(n)</h3> <p>For n �?100: s(n) is the best human score For n > 100: s(n) = 2 × s(ceil(n / 4))</p> <p>This recursive definition allows scoring for large n based on smaller instances.</p> <h2>Example</h2> <h3>Input</h3> <pre><code class="language-">11</code></pre> <h3>Sample Output (Baseline)</h3> <pre><code class="language-">4.000000 0.500000 0.500000 0.000000 1.500000 0.500000 0.000000 2.500000 0.500000 0.000000 3.500000 0.500000 0.000000 0.500000 1.500000 0.000000 1.500000 1.500000 0.000000 2.500000 1.500000 0.000000 3.500000 1.500000 0.000000 0.500000 2.500000 0.000000 1.500000 2.500000 0.000000 2.500000 2.500000 0.000000</code></pre> <p>This is a valid baseline packing for n = 11 with L = 4.000000, achieving the minimum score of 1 point.</p> <h2>Additional Clarifications</h2> <h3>Unit Squares</h3> <ul> <li>Side length <strong>exactly 1</strong></li> <li>Centered at (xi, yi)</li> <li>Rotated by ai degrees around the center</li> </ul> <h3>Ordering</h3> Squares may be listed in <strong>any order</strong>. <h3>Precision</h3> <ul> <li>Inputs are read as doubles</li> <li>Small rounding differences are tolerated per epsilon (1e-7)</li> </ul> <h3>Touching</h3> <ul> <li>Squares <strong>may touch</strong> each other and the container boundary</li> <li>Only <strong>interior overlap is forbidden</strong></li> </ul> <h2>Strategy Tips</h2> <p>1. <strong>Start with baseline:</strong> Implement the simple grid-based baseline first 2. <strong>Local optimization:</strong> Use local search or simulated annealing to improve placements 3. <strong>Smart rotations:</strong> Small rotations can often reduce wasted space 4. <strong>Nonlinear optimization:</strong> Apply optimization techniques with overlap penalties 5. <strong>Known packings:</strong> For small n (�?100), research optimal or near-optimal packings 6. <strong>Recursive approach:</strong> For large n, consider dividing into smaller subproblems</p> <h2>Theoretical Background</h2> <ul> <li><strong>Area lower bound:</strong> Since each unit square has area 1, and n squares must fit in L × L, we have L² �?n, thus L �?√n</li> <li><strong>Compactness:</strong> For fixed n and L, the feasible set is closed and bounded, guaranteeing that a minimal container side length exists</li> <li><strong>Complexity:</strong> Finding the optimal packing is a hard geometric optimization problem; heuristics are essential for large n</li> </ul> <hr> <h2>Problem Metadata</h2> <strong>Time Limit:</strong> 1 second <strong>Memory Limit:</strong> 512 MB <hr> <h2>Model Information</h2> <p><strong>Model:</strong> GPT 5 Thinking</p> <p><strong>Score:</strong> 1.0</p> <h3>Solution Code</h3> <pre><code>#include <bits/stdc++.h> using namespace std;

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

long long n;
if (!(cin >> n)) return 0;

long long k = (long long)ceil(sqrt((long double)n));
double L = (double)k;

cout.setf(std::ios::fixed); 
cout << setprecision(6);
cout << L << "\n";

for (long long i = 0; i < n; ++i) {
    long long x = i % k;
    long long y = i / k;
    double xi = x + 0.5;
    double yi = y + 0.5;
    double ai = 0.0;
    cout << xi << " " << yi << " " << ai << "\n";
}
return 0;

}</code></pre>