CF 1044392 - Ёлочки

CF 1044392 - \u0401\u043b\u043e\u0447\u043a\u0438

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Solve time: 1m 8s
Verified: yes

Solution

Problem Understanding

The picture contains a single tree drawn on square cells.

The trunk is always one cell wide and runs vertically through the whole tree. The tree has beauty n, which means there are exactly n branches on each side of the trunk. The lowest branch has length n, the next one has length n - 1, and so on until the top branch, whose length is 1.

Between every pair of neighboring branches there is one trunk cell with no branches. There is also one extra trunk cell below the lowest branch and one extra trunk cell above the highest branch.

The task is to compute how many grid cells belong to the entire drawing.

The only input is the beauty n. The output is the total occupied area.

The value of n can be as large as 2 · 10^9, so constructing the picture is impossible. Even an algorithm that loops n times would require billions of iterations. The answer has to be computed directly with a mathematical formula. The statement also warns that the result does not fit into 32 bit integers, so the implementation must rely on 64 bit arithmetic. Python integers already support arbitrary precision.

A common source of mistakes is counting the trunk height incorrectly. For example, when n = 1, the trunk consists of the bottom cell, the branch row, and the top cell, for a total of 3 cells, not 2.

Another easy mistake is forgetting that every branch extends to both sides of the trunk. For example, when n = 2, the branch rows contribute (2 + 2) + (1 + 1) = 6 branch cells, not 3.

The smallest input is also worth checking. For input

there are no branches at all. Only the single trunk cell remains, so the correct answer is

Approaches

A direct simulation would build every row of the drawing, count the occupied cells, and sum them. This approach is correct because it follows the definition exactly. Its running time is O(n) since there are 2n + 1 rows, making it completely impractical when n reaches two billion.

The drawing has a very regular structure, so every contribution can be counted independently.

The trunk occupies one cell in every row. There are n branch rows, n - 1 empty rows between consecutive branches, one row below the lowest branch, and one row above the highest branch. That is

n + (n - 1) + 2 = 2n + 1

trunk cells.

The branches contribute the sum of their lengths on both sides:

2 × (1 + 2 + ... + n).

Using the arithmetic progression formula,

2 × n(n + 1) / 2 = n(n + 1).

Adding both parts gives

(2n + 1) + n(n + 1) = n² + 3n + 1.

The entire computation is reduced to evaluating one polynomial.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n)	O(1)	Too slow
Optimal	O(1)	O(1)	Accepted

Algorithm Walkthrough

Read the value n.
Compute the number of trunk cells as 2 * n + 1. This is the height of the tree because every row contains exactly one trunk cell.
Compute the number of branch cells as n * (n + 1). This equals twice the sum 1 + 2 + ... + n.
Add the two quantities and print the result.

Why it works

Every occupied cell belongs to exactly one of two categories.

The first category is the trunk. Every row contributes exactly one trunk cell, so the trunk contributes 2n + 1 cells.

The second category is the branches. A branch of length k contributes k cells on the left and k on the right. Summing over all branch lengths produces n(n + 1) cells.

The trunk and branches never overlap except at the trunk itself, which is already counted only once because branch lengths do not include the trunk cell. Every occupied cell is counted exactly once, so the total area is n² + 3n + 1.

Python Solution

import sys
input = sys.stdin.readline

n = int(input())
print(n * n + 3 * n + 1)

The implementation is just the closed-form formula derived above. Python automatically handles integers larger than 64 bits, so there is no overflow concern even for the maximum input.

Writing the expression as n * n + 3 * n + 1 avoids floating point arithmetic entirely. Every operation stays in integer arithmetic, eliminating rounding issues.

Worked Examples

Example 1

Input:

Step	Value
`n`	5
Trunk cells	`2 * 5 + 1 = 11`
Branch cells	`5 * 6 = 30`
Total	`11 + 30 = 41`

This confirms the sample answer. The trunk contributes only eleven cells, while the branches dominate the total area.

Example 2

Input:

Step	Value
`n`	0
Trunk cells	`2 * 0 + 1 = 1`
Branch cells	`0 * 1 = 0`
Total	`1 + 0 = 1`

This demonstrates the boundary case where no branches exist. The formula naturally reduces to the single trunk cell.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(1)	Only a constant number of arithmetic operations are performed.
Space	O(1)	No additional data structures are allocated.

The algorithm performs the same amount of work regardless of the input value, making it easily fast enough even when n is two billion.

Test Cases

# helper: run solution on input string, return output string
import sys, io

def solve():
    input = sys.stdin.readline
    n = int(input())
    print(n * n + 3 * n + 1)

def run(inp: str) -> str:
    backup_stdin = sys.stdin
    backup_stdout = sys.stdout
    sys.stdin = io.StringIO(inp)
    out = io.StringIO()
    sys.stdout = out
    solve()
    sys.stdin = backup_stdin
    sys.stdout = backup_stdout
    return out.getvalue()

# provided sample
assert run("5\n") == "41\n", "sample 1"

# custom cases
assert run("0\n") == "1\n", "minimum beauty"
assert run("1\n") == "5\n", "first nontrivial tree"
assert run("2\n") == "11\n", "checks branch summation"
assert run("2000000000\n") == "4000000006000000001\n", "maximum input"

Test input	Expected output	What it validates
`0`	`1`	No branches, only the trunk remains.
`1`	`5`	Correct trunk height and two-sided branch counting.
`2`	`11`	Arithmetic progression is summed correctly.
`2000000000`	`4000000006000000001`	Very large values and integer overflow safety.

Edge Cases

When the input is

the algorithm computes 0² + 3·0 + 1 = 1. No special handling is required because the branch contribution becomes zero automatically.

When the input is

the algorithm computes one branch row, one bottom trunk cell, and one top trunk cell, giving 3 trunk cells. The branches contribute 2 more cells, producing 5. This catches the common mistake of counting only two trunk rows.

When the input is

the branch contribution is 2 × (1 + 2) = 6, which becomes 2 × 3 = 6 through the formula n(n + 1). The total is 5 + 6 = 11, confirming that branches are counted on both sides of the trunk.