# Restocking the Holy Palanquin Ship's Confectionery: Alternative Solution

Problem setter’s solution: enumerate plans by nondecreasing prices. Codex (GPT-6) organizes the recursion, modular arithmetic, and output. The preferred method is the [pentagonal recurrence](solution.en.md).

First add the leftover money of the empty plan. A search state specifies the lowest current price and remaining money. Either skip this price or buy one item of it. After a purchase, add the new plan's leftover money, then allow another item of the same price. Expand branches only when they can produce a positive remainder; zero-remainder plans contribute nothing and may be omitted.

Every nonempty plan has a unique nondecreasing price sequence. Skipping prices and buying the current price uniquely constructs that sequence, so no plan is repeated. Every positive-remainder plan reaches its corresponding node by this sequence. Counting the empty plan separately and omitting zero contributions therefore gives the exact sum. For zero budget, do not recurse.

Let $K=\sum_{s=0}^{X}P_s$ be the number of partitions of costs up to the budget. Every purchase prefix is itself such a plan, and a chain of skipped prices has length at most $X$ . A conservative time bound is $O((X+1)K)$ , with $O(X+1)$ recursion space. Runtime grows with the number of plans, making this useful for small-case understanding and independent checks. The formula is correct throughout the range; whether this implementation meets the reference time limit is recorded in the [benchmark](../../../docs/benchmark-R20.json). It is not recommended in place of the low-polynomial methods.

Program: [alternative-search.cpp](alternative-search.cpp).

On the reference machine, the largest five-run median is about 0.410 seconds against a 0.5-second limit. This is only about 1.22 times headroom, below the recommended factor of two. Keep this as an enumeration and checking alternative; full-range acceptance is not promised on deployment. Prefer the pentagonal or knapsack implementation.