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VOSFENCE

Why Practice This

This is a nice example of splitting one messy counting problem into two much cleaner layers:

  • count the red/blue skeleton by how many color changes it has
  • count how to insert white bars into the gaps under upper-bound constraints

The final answer is the product of those two counts, summed over all possible skeleton transition counts.

Recognition Cue

Reach for a decomposition like this when:

  • one color or symbol behaves like separators or gaps
  • adjacency constraints can be described on the non-separator skeleton first
  • after fixing that skeleton, the leftover placements become bounded compositions

The strongest smell is:

  • "count strings/colorings under both adjacency rules and maximum-run rules"

That often means split the structure into:

  • a core sequence
  • and bounded fillers between its parts

Problem-Specific Transformation

The raw fence-coloring statement is messy because all three colors interact.

The reusable rewrite is:

  • first ignore white and count only the B/R skeleton
  • then reinsert white runs into the gaps

That turns one tangled counting problem into two smaller ones:

  • skeleton counting by run structure
  • bounded gap counting by DP / stars-and-bars-with-caps

Core Idea

Let the non-white sequence be the B/R skeleton first.

If the skeleton has t adjacent BR or RB transitions, then after inserting white bars:

  • exactly M of those t transition positions must stay adjacent
  • the other t - M transition positions must get at least one white bar inserted between them

Now look at all b + r + 1 white gaps around the skeleton:

  • t - M special gaps must contain at least 1 white bar
  • the remaining b + r + 1 - t free gaps may contain 0 or more
  • every gap can contain at most K - 1 white bars, because K consecutive whites are forbidden

So for a fixed skeleton transition count t, the white-placement problem becomes a bounded composition problem.

Counting The Skeleton

For a B/R sequence with:

  • b blue bars
  • r red bars
  • exactly t color changes

we count it by runs.

Examples:

  • if t is odd, the sequence starts and ends with different colors
  • if t is even, it starts and ends with the same color

That gives a direct composition formula with binomial coefficients.

Counting The White Gaps

For a fixed t:

  • choose which M of the t transition gaps stay empty: C(t, M)
  • the other t - M transition gaps must be positive
  • the remaining b + r + 1 - t gaps are nonnegative

After subtracting 1 from each forced-positive gap, we need to count:

  • t - M variables in [0, K - 2]
  • b + r + 1 - t variables in [0, K - 1]
  • total sum W - (t - M)

The solution precomputes these bounded-composition counts with small DP because W, B, R <= 100.

Complexity

  • run-count layer: O(B + R)
  • bounded-composition DP: O((B + R) * W * K)
  • final summation over t: O((B + R) * W)

This is easily fast enough for the official limits.

Pitfalls / Judge Notes

  • The “every segment of length K contains a blue or red bar” rule is exactly the same as saying there is no run of K consecutive white bars.
  • The official mirror shows 1 <= W, B, R <= 100, but also notes a subtask with W = 0, so the accepted solution should handle W = 0 safely.
  • If K = 1, then no white bar can appear at all. The implementation handles this as a special edge case automatically.
  • When B = 0 or R = 0, the skeleton has only one color, so the number of BR/RB adjacencies is always 0.

Reusable Pattern

Solutions