ICPC 2026 Domestic Qualifiers Participation Report

2026/07/04

Introduction

It’s me.

On July 3rd, I participated in the ICPC Domestic Qualifiers as part of team WaSanBon (Twil3akine, Caffeineholic, Akagi_Shrine).

Standings

LiveSite

Problem List

https://storage.googleapis.com/files.icpc.jp/domestic2026/problems_ja.pdf

To get straight to the point, we solved only two problems, A and B. I have absolutely no intention of insulting my teammates (in fact, I am full of gratitude toward them), but personally, this was truly a disgrace.

I feel so pathetic that I’m at a total loss for words. I feel like I could spend a whole day talking about “what if I had done this” or “if only I had done that.”

Results of team WaSanBon in the ICPC 2026 Domestic Qualifiers

I acted as the general consultant for the team and was responsible for analyzing problems C and E.

If I were better at implementation, I would have wanted to implement everything myself… I’m truly frustrated.

Problem A: “Find the Strongest Card”

The task is to output the strongest card for a game of Daifugo from a given set of cards.

Letting the number on the card be $x$ , I defined the value $f(x)$ used for comparing strength as follows:

f(x) = \begin{cases} x + 13 & (x \in \{1, 2\}) \\ x & (3 \le x \le 13) \end{cases}

I selected the card that maximizes $f(x)$ and output its original number. In the implementation, if the maximum value was $14$ or higher, I subtracted $13$ ; otherwise, I output it as is.

After the contest, I heard a solution that suggested “if 1 or 2 is present, output the stronger of the two; otherwise, output the maximum value,” and I thought that was quite clever.

Problem B: “Vending Machine”

By placing a vending machine, you can cover doors within distance $d$ from the installation point. The goal is to find the minimum number of vending machines required to cover all doors.

Initially, I misread the problem, thinking that “a vending machine must be placed between every single pair of doors,” which caused a bit of a time loss.

Let the positions of the doors in ascending order be $x_0, x_1, \ldots, x_{n-1}$ . If the leftmost door not yet covered is $x_i$ , placing a vending machine at $x_i+d$ allows us to cover up to the right edge at $x_i+2d$ .

Therefore, it can be solved using the following greedy approach:

Set $i=0$ and the answer to $0$ .
Using $x_i$ as the left edge, place a vending machine at $x_i+d$ .
Skip all doors $x_j$ that satisfy $x_j \le x_i+2d$ .
Increment the answer by $1$ , and if there are unprocessed doors, return to step 2.

I’m glad I was able to move to the correct strategy quickly once I realized my misreading.

Problem C: “Water Remaining”

My poor reading comprehension skills were on full display here, and I couldn’t understand the problem statement at all. After racking my brain, my thoughts got locked into the idea that “some kind of convexity must be involved.”

After the contest, I talked to members of other teams and found out that it could be solved simply by performing a process once from the left and once from the right: “looking from above, remove water from each cell until it hits a wall.” I felt truly miserable.

This was entirely a failure of analysis on my part. I feel very sorry for my teammates.

It felt so difficult that I thought this year’s “heavy implementation” slot had moved to Problem C. In reality, it was just my analytical error.

Problem D: “Frequency Sequence”

I was hardly involved in this problem, but it seems my analysis was mostly correct. If I had finished analyzing Problem C smoothly, I think we might have been able to solve this too…

Regarding the content, it has a structure like a group sequence, and it seems case handling is required based on the relationship between $s^2$ and $k$ .

Problem E: “Shopping Master”

The more I thought about it, the more it seemed like a greedy approach would work.

Let $U$ be the set of all bottles and $G$ be the set of bottles with gems. The set of bottles without gems is the complement $U \setminus G$ . Also, let $c_{\min}$ be the minimum cost of a bottle with a gem.

During the contest, I considered the following cases:

If $|U| \le |G|$ : By buying one bottle with a gem, you can get all bottles, so the cost is $c_{\min}$ .
If $|G| = 0$ : Since gems cannot be used, the cost is the sum of all bottle prices.
Otherwise: The sum of $c_{\min}$ (to secure the number of bottles that can be bought with gems) and the costs of the bottles bought with coins.

It turns out that if I had refined the “otherwise” case a bit more, I could have reached the correct answer. It is a deep regret of mine that I couldn’t allocate enough time to this.

Problem F: “Optimizing a Map Application”

I thought about using BFS, but since it’s an $n \le 10^9$ square grid, I withdrew.

I’m sure there’s some absolutely clever method out there…

Closing Thoughts

I didn’t even get a chance to look at Problem G or beyond.

Before I knew it, next year will be my last year to participate in ICPC.

I’ll be a first-year Master’s student (M1) next year. I want to go to Yokohama next year. I intend to work seriously hard this coming year.

Including proper team practice and all that…