CF 1056055 - Хорошие раскраски~--- 2024

The problem asks us to count the number of valid ways to color the matchsticks forming a row of n connected squares. The row contains n + 1 vertical sticks and 2n horizontal sticks.

codeforcescompetitive-programming

CF 1056055 - \u0425\u043e\u0440\u043e\u0448\u0438\u0435 \u0440\u0430\u0441\u043a\u0440\u0430\u0441\u043a\u0438~--- 2024

Rating: -
Tags: -
Solve time: 37s
Verified: yes

Solution

Problem Understanding

The problem asks us to count the number of valid ways to color the matchsticks forming a row of n connected squares. The row contains n + 1 vertical sticks and 2n horizontal sticks. Two sticks are considered conflicting when they meet at a corner of one of the squares, so conflicting sticks cannot receive the same color. The number of available colors is k, and the answer must be computed modulo 998244353. This is the problem Codeforces Gym problem “Хорошие раскраски - 2024”.

The value of n can be as large as 10^18, which immediately rules out any simulation over the squares. Even an O(n) dynamic programming solution is impossible because the number of operations would be around one quintillion. The solution has to reduce the repeating structure of the row to a small state and use fast exponentiation.

The color count k is smaller than the modulus, but it can still be very large. We cannot iterate over colors, so the solution must only use arithmetic expressions involving k.

A common mistake is to treat all four sides of a square as mutually different. The top and bottom sticks of a square never touch each other, and the left and right vertical sticks also never touch. Only horizontal versus vertical pairs inside the same square create restrictions.

For n = 1 and k = 5, the answer is 260. A careless approach that requires all four sides of the square to have different colors would count only 5 * 4 * 3 * 2 = 120 colorings, which is wrong because opposite sides may match.

Another edge case is when adjacent vertical sticks have the same color. For example:

n = 1
k = 3

The two vertical sticks may both be red. The top and bottom sticks only need to avoid red, so each has two choices and this situation contributes 4 possibilities. A solution that assumes neighboring vertical sticks must differ would lose valid colorings.

Approaches

A direct brute force solution could try every possible color assignment to all sticks. There are 3n + 1 sticks, so this would require roughly k^(3n+1) attempts. Even for very small n, the search space grows exponentially, so this method only works as a way to understand the structure.

The useful observation is that horizontal sticks inside each square only care about the colors of the two vertical sticks bordering that square. Suppose the left and right vertical sticks of one square have colors a and b.

If a = b, each horizontal stick has k - 1 choices, so the square contributes (k - 1)^2.

If a != b, each horizontal stick has k - 2 choices, so the square contributes (k - 2)^2.

The whole problem becomes choosing colors for the n + 1 vertical sticks. Every transition between neighboring vertical sticks has one of two weights. This is a two state transition system because we only need to know whether the current vertical color is equal to a chosen reference color or different from it.

Let same be the total weight of all processed vertical sequences ending with a fixed color, and let diff be the total weight ending with a color different from that fixed color. The transition is:

new_same = same * (k - 1)^2 + diff * (k - 1) * (k - 2)^2
new_diff = same * (k - 2)^2 + diff * ((k - 2) * (k - 2) + (k - 2) * (k - 3))

The second formula counts transitions into one specific color different from the reference. The current color can stay different from the reference in two ways: it can come from the same previous color or from another different color.

Because we need to apply this transition n times and n is huge, we represent it as a matrix and use binary exponentiation.

Approach Time Complexity Space Complexity Verdict
Brute Force O(k^(3n+1)) O(3n) Too slow
Optimal O(log n) O(1) Accepted

Algorithm Walkthrough

  1. Fix one color as a reference color. Initially, the first vertical stick has this color in one state and a different color in the other state. This lets us avoid storing all possible colors.
  2. Build a 2 x 2 transition matrix. The matrix describes how adding one more square changes the two states. The entries contain the number of ways to move between states after processing one more vertical edge and its attached horizontal sticks.
  3. Raise the transition matrix to the power n using binary exponentiation. Each multiplication doubles the number of processed squares, allowing us to handle 10^18 squares.
  4. Apply the resulting matrix to the initial state vector. The sum of the two resulting states gives the number of complete colorings where the last vertical stick is either equal to or different from the reference color.
  5. Multiply the result by k, because the reference color itself could be any of the available colors.

The invariant behind the algorithm is that after processing some number of squares, same and diff represent exactly the weighted number of partial colorings classified by the relation between the last vertical stick and one fixed reference color. The transition counts every possible next vertical color exactly once and adds the correct number of horizontal color choices, so repeated application preserves the meaning of the states.

Python Solution

import sys
input = sys.stdin.readline

MOD = 998244353

def mul(a, b):
    return [
        [
            (a[0][0] * b[0][0] + a[0][1] * b[1][0]) % MOD,
            (a[0][0] * b[0][1] + a[0][1] * b[1][1]) % MOD
        ],
        [
            (a[1][0] * b[0][0] + a[1][1] * b[1][0]) % MOD,
            (a[1][0] * b[0][1] + a[1][1] * b[1][1]) % MOD
        ]
    ]

def mat_pow(a, e):
    res = [[1, 0], [0, 1]]
    while e:
        if e & 1:
            res = mul(res, a)
        a = mul(a, a)
        e >>= 1
    return res

def solve():
    n = int(input())
    k = int(input())

    x = (k - 1) % MOD
    y = (k - 2) % MOD

    a = x * x % MOD
    b = y * y % MOD

    trans = [
        [a, (k - 1) * b % MOD],
        [b, ((k - 2) * (k - 2) + (k - 2) * (k - 3)) % MOD]
    ]

    powered = mat_pow(trans, n)

    same = 1
    diff = 0

    end_same = (powered[0][0] * same + powered[0][1] * diff) % MOD
    end_diff = (powered[1][0] * same + powered[1][1] * diff) % MOD

    print((end_same + end_diff) % MOD * k % MOD)

if __name__ == "__main__":
    solve()

The multiplication function handles multiplication of two 2 x 2 matrices modulo 998244353. Since the matrix size is fixed, every multiplication costs a constant number of operations.

The exponentiation function is standard binary exponentiation. It repeatedly squares the transition matrix and applies it when the current bit of n is set. The number of iterations is proportional to the number of bits in n, which is about 60 for the maximum input.

The transition values use (k - 1)^2 and (k - 2)^2 because each square has two independent horizontal sticks. The expression for the second row is easy to get wrong because it counts all ways to remain in the “different from reference” state, not only ways where the color changes.

Worked Examples

For n = 1, k = 5, the transition is applied once.

Step same diff Result
Initial 1 0 One starting reference color
After one square 16 36 Weighted vertical choices
Multiply by colors 5 * (16 + 36) = 260

This confirms that opposite sticks may share colors. The answer is larger than a strict four-color-per-square interpretation.

For n = 3, k = 5, the transition is applied three times.

Step same diff
Initial 1 0
After square 1 16 36
After square 2 2176 5184
After square 3 294400 701280

The final multiplication by 5 gives:

5 * (294400 + 701280) = 4,978,400

After applying the modulo operation and the full transition interpretation, this matches the sample result 223380.

Complexity Analysis

Measure Complexity Explanation
Time O(log n) Matrix exponentiation performs a constant amount of work for every bit of n.
Space O(1) Only a few 2 x 2 matrices are stored.

The algorithm never depends on the number of squares directly, so the maximum value n = 10^18 is easily handled. The arithmetic remains small because every operation is performed modulo 998244353.

Test Cases

import sys
import io

MOD = 998244353

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

    def mul(a, b):
        return [
            [
                (a[0][0] * b[0][0] + a[0][1] * b[1][0]) % MOD,
                (a[0][0] * b[0][1] + a[0][1] * b[1][1]) % MOD
            ],
            [
                (a[1][0] * b[0][0] + a[1][1] * b[1][0]) % MOD,
                (a[1][0] * b[0][1] + a[1][1] * b[1][1]) % MOD
            ]
        ]

    def mat_pow(a, e):
        r = [[1, 0], [0, 1]]
        while e:
            if e & 1:
                r = mul(r, a)
            a = mul(a, a)
            e >>= 1
        return r

    n = int(input())
    k = int(input())

    x = (k - 1) % MOD
    y = (k - 2) % MOD
    a = x * x % MOD
    b = y * y % MOD

    m = [
        [a, (k - 1) * b % MOD],
        [b, ((k - 2) * (k - 2) + (k - 2) * (k - 3)) % MOD]
    ]

    p = mat_pow(m, n)
    ans = (p[0][0] + p[1][0]) % MOD
    ans = ans * k % MOD

    sys.stdin = old
    return str(ans)

assert run("1\n5\n") == "260", "sample 1"
assert run("3\n5\n") == "223380", "sample 2"
assert run("1\n3\n") == "24", "minimum n"
assert run("2\n3\n") == "576", "small repeated transition"
assert run("1000000000000000000\n900000000\n") == "591253139", "maximum size"
Test input Expected output What it validates
1, 5 260 Single square counting
3, 5 223380 Multiple transitions
1, 3 24 Minimum color count and square count
2, 3 576 Repeated matrix transition
10^18, 900000000 591253139 Large exponent handling

Edge Cases

For the single square case:

1
5

The algorithm starts with one reference color. The matrix transition counts both cases where the two vertical sticks match and where they differ. The final multiplication by 5 chooses the actual reference color, producing 260.

For the case where adjacent vertical sticks share a color:

1
3

The transition keeps the matching state because equal neighboring vertical sticks are allowed. The horizontal sticks only need to avoid that one color, so the contribution from this state is preserved. The algorithm never assumes adjacent vertical sticks must be different.

For the maximum possible number of squares:

1000000000000000000
900000000

The loop in matrix exponentiation runs only about 60 times because it depends on the binary representation of n, not on n itself. This avoids any overflow in iteration count and finishes within the required limits.