CF 1180A - Alex and a Rhombus

Rating: 800
Tags: dp, implementation, math
Solve time: 59s
Verified: yes

Solution

Problem Understanding

We are given a shape that grows on a grid starting from a single cell. At the first stage there is exactly one cell. At every next stage, the shape expands by attaching every grid cell that shares an edge with any cell already in the shape. In other words, each step adds the full “one-cell thick border” around the current figure in the four cardinal directions.

This produces a diamond-like pattern centered at the initial cell. The task is to determine how many unit grid cells are present after the shape has been expanded exactly $n$ times according to this rule.

The input is a single integer $n$, where $1 \le n \le 100$, and the output is the total number of occupied cells in the final shape.

The constraint is extremely small. With $n \le 100$, any method up to roughly $10^8$ simple operations would pass comfortably, but in practice this problem has a closed-form structure that makes simulation unnecessary.

A naive simulation would explicitly maintain the set of cells and expand it layer by layer. This works conceptually but becomes inefficient as the shape grows quadratically. However, even the worst case here is tiny, so simulation would not time out, but it is unnecessary for understanding the structure.

There are no tricky edge cases in terms of input validity. The only boundary condition is $n = 1$, where the answer must be exactly 1 cell.

Approaches

The brute-force interpretation is straightforward. Start with a single cell at the origin. For each step, scan all current cells and add their four neighbors if they are not already present. If we maintain a set, we ensure no duplicates. After $n-1$ expansions, we count the total cells.

This is correct because it directly follows the definition: every step includes all cells that share at least one edge with the current shape.

The issue is that this approach hides the structure. Each layer adds a symmetric ring around the previous shape. If we look at the shape after a few steps, we observe:

At $n = 1$, we have 1 cell.

At $n = 2$, we get a cross of 5 cells.

At $n = 3$, the shape expands to 13 cells.

At $n = 4$, it becomes 25 cells.

The increments are 4, 8, 12, and so on. Each new layer adds $4(n-1)$ cells. This happens because the boundary of the shape forms a diamond whose perimeter grows linearly with $n$.

So instead of simulating geometry, we reduce the problem to summing an arithmetic progression:

$$1 + 4(1 + 2 + \cdots + (n-1))$$

The inner sum is $\frac{(n-1)n}{2}$, giving:

$$1 + 2n(n-1)$$

This closed form removes all simulation and reduces the task to constant time arithmetic.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force Simulation	$O(n^2)$	$O(n^2)$	Accepted but unnecessary
Closed-form formula	$O(1)$	$O(1)$	Accepted

Algorithm Walkthrough

Read integer $n$. This represents how many layers of expansion we apply starting from a single cell.
If $n = 1$, return 1 immediately since no expansion occurs and the shape remains a single unit cell.
Compute the number of cells added after expansions as $2n(n-1)$. This comes from summing the sizes of successive boundary layers.
Add 1 to account for the original central cell.
Output the final result.

Why it works

After the first step, the shape always remains a perfectly symmetric diamond aligned to grid axes. Each new layer contributes a complete border whose length increases linearly with the layer index. Because each layer is fully formed and does not overlap previously added cells, the total number of cells is exactly the sum of independent boundary contributions. This guarantees that the arithmetic series correctly models the growth without missing or double-counting any cell.

Python Solution

import sys
input = sys.stdin.readline

n = int(input().strip())

if n == 1:
    print(1)
else:
    print(1 + 2 * n * (n - 1))

The solution relies entirely on the closed-form expression. The special case $n = 1$ is handled explicitly, although the formula already evaluates correctly for $n = 1$ as well.

The expression $2n(n-1)$ is computed using Python integers, which safely handle values well beyond the constraints. No intermediate overflow concerns exist.

The structure of the code mirrors the mathematical derivation directly, with no simulation or auxiliary storage.

Worked Examples

Example 1: $n = 1$

Step	n	Formula	Result
Input	1	-	-
Base case	1	return 1	1

This confirms that the base configuration contains exactly one cell.

Example 2: $n = 4$

Step	Computation	Value
n(n-1)	4 × 3	12
2n(n-1)	2 × 12	24
+1	24 + 1	25

This matches the intuitive growth pattern where each layer adds 4, 8, and 12 cells respectively, totaling 24 added cells plus the center.

The trace confirms that the formula accumulates exactly the layered structure implied by the geometric construction.

Complexity Analysis

Measure	Complexity	Explanation
Time	$O(1)$	Only a constant number of arithmetic operations are performed regardless of $n$.
Space	$O(1)$	No auxiliary data structures are used, only integer variables.

The constraints allow up to $n = 100$, but the solution does not depend on constraint size at all. Even if $n$ were significantly larger, the same constant-time computation would remain valid.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    return main()

def main():
    n = int(sys.stdin.readline().strip())
    if n == 1:
        return str(1)
    return str(1 + 2 * n * (n - 1))

# provided sample
assert run("1\n") == "1"

# small growth checks
assert run("2\n") == "5", "n=2 cross shape"
assert run("3\n") == "13", "next layer expansion"
assert run("4\n") == "25", "arithmetic growth"

# boundary case
assert run("100\n") == str(1 + 2 * 100 * 99), "maximum n"

Test input	Expected output	What it validates
1	1	base case correctness
2	5	first expansion layer
3	13	second layer accumulation
4	25	consistency of growth pattern
100	19801	upper bound stability

Edge Cases

Case $n = 1$

Input is:

The algorithm immediately returns 1 without applying the formula. This prevents unnecessary arithmetic and aligns with the definition where no expansion has occurred.

Trace:

Step	Action	Value
Input	read n	1
Check	n == 1	true
Output	return	1

The result matches the fact that the initial configuration is a single cell.

Case $n = 100$

Input:

Trace:

Step	Computation	Value
n(n-1)	100 × 99	9900
2n(n-1)	19800	19800
+1	final	19801

Even at maximum input size, the computation remains stable and does not require iteration. The formula handles the full growth implicitly, confirming correctness at scale.