Skip to content

VMSCALE

Why Practice This

This is a very good “DP hidden inside a story problem” exercise.

At first glance it looks like a weighing puzzle or information-theory problem. The real structure is a minimax decision tree where every comparison pivot has a non-uniform price.

Recognition Cue

Reach for minimax DP on a decision tree when:

  • each question partitions the remaining answer range into branches
  • the objective is worst-case total cost, not expected cost
  • different questions have different prices
  • the statement looks like a guessing game or measuring puzzle

The strongest smell is:

  • "minimum worst-case total cost to identify one hidden value"

That usually means states are intervals and transitions are chosen questions.

Problem-Specific Transformation

The weighing story is rewritten as:

  • each chosen target T is one ternary comparison question
  • asking that question costs cost(T)
  • after the answer, the remaining candidate set is one sub-interval

So the whole problem becomes:

  • choose pivots
  • pay their balanced-ternary cost
  • minimize the worst-case future budget

That is why the right state is not one current weight configuration, but an interval budget DP.

Core Idea

One weighing can place powers of 3 on both sides, so any integer pivot T can be written in balanced ternary:

T = sum d_i * 3^i, where d_i in {-1, 0, 1}

The weighing result tells us whether:

x < T, x = T, or x > T

and the money spent for that weighing is exactly:

cost(T) = number of nonzero digits d_i

So the original problem becomes:

  • unknown integer x lies in [L, R]
  • one question chooses a pivot T
  • asking that question costs cost(T)
  • we want the minimum worst-case total cost

Define:

far[B][L] = farthest R such that every x in [L, R]
            can be identified with worst-case total cost <= B

If the first pivot is T with cost c, then:

  • the left branch [L, T - 1] must fit in budget B - c
  • the right branch [T + 1, R] must fit in budget B - c

So T is usable exactly when:

T - 1 <= far[B - c][L]

and then the best right endpoint we can guarantee is:

far[B - c][T + 1]

This gives a budget DP from small budgets upward.

Complexity

Let W = 10000.

  • balanced ternary cost preprocessing: O(W log W)
  • DP: O(Bmax * W * C)
  • here C <= 10 is the maximum balanced-ternary weight for numbers up to 10000
  • in practice Bmax = 61
  • each query is answered with binary search on the budget: O(log Bmax)

That is fast enough even for Q <= 10^6.

Pitfalls / Judge Notes

  • The objective is worst-case total money, not the number of weighings.
  • A pivot with many nonzero balanced-ternary digits may still be worth it if it splits the interval well.
  • The cost of T is not the number of ternary digits, but the number of nonzero digits in balanced ternary.
  • Because Q is huge, slow per-query logic will time out even if the recurrence is correct.
  • The hidden trick is to precompute the whole answer table by budget, then answer each query with only a binary search.

Reusable Pattern

  • Topic page: Foundations
  • Practice ladder: Foundations ladder
  • Starter template: contest-main.cpp
  • Notebook refresher: DP cheatsheet
  • Carry forward: write the state meaning in one sentence before you derive any recurrence.
  • This note adds: the recurrence shape and optimization that belong to this specific problem.

Solutions