CF 102788L - Fence

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Solve time: 47s
Verified: yes

Solution

Problem Understanding

The task is not asking us to construct the whole magic square. We only need the sum of the numbers placed in its first row. A normal magic square of size n contains every integer from 1 to n² exactly once, and every row has the same sum. The input gives the side length of the square, and the output must be that common row sum.

The size can be as large as 1000, so simulating the square is unnecessary. Even a straightforward construction would require storing one million cells for the largest case, while the required answer depends only on a mathematical property of magic squares. The intended solution should run in constant time, because any approach that processes all n² cells is doing work that the problem does not require.

The unusual cases are small orders. For n = 1, the only cell contains 1, so the answer is 1. A solution that assumes every square has multiple rows may fail here. For example, input 1 must produce output 1.

The value n = 2 is excluded by the statement because a normal magic square does not exist for that size. A program should not try to build one or rely on a construction formula for this case. The judge will not provide this input.

Approaches

A direct approach would be to generate a magic square and add the elements of its first row. This works because every valid construction would contain the correct first row, and the row sum would match the required answer. However, constructing the square is unnecessary. For n = 1000, there are 1,000,000 cells, so even a simple traversal already performs a million operations and uses significant memory. More complicated constructions add additional overhead without helping us find the requested value.

The key observation is that every number from 1 to n² appears exactly once. The total sum of all cells is the sum of the first n² positive integers:

$$\frac{n^2(n^2+1)}{2}$$

A magic square has n rows with identical sums. Dividing the total sum by the number of rows gives the sum of one row:

$$\frac{n^2(n^2+1)}{2n}$$

which simplifies to:

$$\frac{n(n^2+1)}{2}$$

The first row must have this value because all rows share the same sum. The entire problem is reduced to evaluating this formula.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n²)	O(n²)	Too slow and unnecessary
Optimal	O(1)	O(1)	Accepted

Algorithm Walkthrough

Read the order n of the magic square. The value of n is enough information because the row sum is determined entirely by the square size.
Compute the magic sum using the formula n * (n * n + 1) // 2. The expression comes from dividing the total value of all cells by the number of rows.
Print the computed value.

Why it works:

The numbers inside the square are exactly the integers from 1 to n², so their total sum is fixed. Since every row has the same sum, the total must be split equally among n rows. The formula calculates that single row value directly, which means it must also be the sum of the first row.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    n = int(input())
    print(n * (n * n + 1) // 2)

if __name__ == "__main__":
    solve()

The program reads the only input value and applies the derived formula immediately. The multiplication n * n is safe because the largest possible value is only one million. Python integers also remove any concern about overflow.

The division is performed with integer division because the final result is always an integer. The expression is arranged as n * (n * n + 1) // 2, which keeps the computation simple while preserving exact arithmetic.

Worked Examples

The original statement does not provide usable sample inputs for this task, so consider two direct traces.

For input 3, the algorithm computes:

Step	n	n² + 1	Result
Read input	3	10
Apply formula	3	10	3 × 10 / 2 = 15

The answer is 15. A 3×3 magic square contains numbers from 1 to 9, whose total sum is 45. Since there are three rows, each row must sum to 15.

For input 5, the algorithm computes:

Step	n	n² + 1	Result
Read input	5	26
Apply formula	5	26	5 × 26 / 2 = 65

The answer is 65. The full square contains numbers from 1 to 25, with total sum 325. Splitting that across five equal rows gives 65 per row.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(1)	Only a fixed number of arithmetic operations are performed
Space	O(1)	The program stores only the input value and the result

The constraints allow a constant-time solution easily. The algorithm does not depend on the number of cells in the square, so even the largest allowed order is handled instantly.

Test Cases

import sys
import io

def solve_data(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    input = sys.stdin.readline
    n = int(input())
    return str(n * (n * n + 1) // 2)

# minimum size
assert solve_data("1\n") == "1", "1x1 magic square"

# small non-trivial square
assert solve_data("3\n") == "15", "3x3 magic square"

# another valid odd order
assert solve_data("5\n") == "65", "5x5 magic square"

# maximum allowed order
assert solve_data("1000\n") == "500000500", "largest order"

Test input	Expected output	What it validates
`1`	`1`	Handles the trivial magic square
`3`	`15`	Checks the smallest non-trivial case
`5`	`65`	Checks a larger ordinary case
`1000`	`500000500`	Checks the maximum constraint

Edge Cases

For n = 1, the formula gives:

$$1 \times (1^2 + 1) / 2 = 1$$

The algorithm prints 1, matching the only cell in the square. A construction-based solution that starts with assumptions about multiple rows could incorrectly fail here.

For n = 3, the formula gives:

$$3 \times (9 + 1) / 2 = 15$$

A careless solution might try to derive the answer from one particular magic square layout. That would work for this case, but it would not explain why every valid square has the same first-row sum. The formula uses the invariant shared by all magic squares, so it works regardless of the arrangement.

For n = 1000, the formula gives:

$$1000 \times (1000000 + 1) / 2 = 500000500$$

The algorithm performs the same few operations as it does for n = 1, avoiding the million-cell processing that a simulation approach would require.