Okay, welcome back. In this video (It's too long and I didn't publish it, but some later questions are done in live streams), I'm going to go through the Bronze group contest of the U-Cycle, the first contest in 2026. I didn't perform well during the actual contest, so I'm going to review the problems and see if I can solve them all now.

For this video, I’m trying out speech-to-text and might use some AI to summarize it for my blog later. I want to keep the recording short, just to remember the experience. If a video is two hours long, who would watch it, right? It’s more convenient for me to just record my thoughts and let the software handle the transcription. I don't care too much about the post-processing.

I may post separate videos for the three contest questions. This is the first one: "Chip Exchange."

I glanced at the solution earlier and it seemed like there were no loops—just pure math. I thought it wouldn't be too difficult. During the actual contest, I didn't perform well. It was late, I had just finished school, and my brain wasn't working properly. Also, using Java was a problem; it takes too long to write compared to C++. There's a lot of overhead and cost of abstraction in Java.

Anyway, let's begin properly. I’ll create a new file. The problem is "Chip Exchange." I'll write the main function first and set up fast I/O in C++.

I need to include the necessary headers and synchronize with stdio for speed. I'll define a solve function to handle each test case. The input format: T (number of test cases), then for each test case, five numbers: A, B, CA, CB, and FA.

The numbers can be up to 10^9, so I need to use long long. I'll read T and then loop through each case, calling solve(A, B, CA, CB, FA).

Now, the logic. The problem: We start with A chips of type A and B chips of type B. We can receive X additional random chips (each could be A or B). After receiving them, we can repeatedly exchange CB chips of type B for CA chips of type A. We need to guarantee that we end up with at least FA chips of type A in the worst case (i.e., for the worst possible distribution of the X random chips). We need to find the minimum such X.

Initially, I thought about iterating over possible values of X, but that’s not feasible since X can be huge. There must be a formula.

Let me reason. After receiving X chips, let b be the total number of B chips we have (initial B plus some from the X random chips). The number of A chips we can get via exchange is floor(b / CB) * CA. The total A chips we end with is: initial A, plus some A from the X random chips, plus the exchanged amount.

The worst case for a given total chips total = A + B + X is when the number of B chips is such that we get the minimum final A count. We need to ensure that even this minimum is at least FA.

I realized the function for final A count in terms of b (the number of B chips after receiving X) is a step function. It decreases when CB > CA and increases when CB < CA. The worst case will be at one of the "steps" in this function.

After a lot of scribbling and confusion (and taking a break for dinner), I was still stuck. My code was getting messy with special cases for CB < CA, CB == CA, and CB > CA. I tried to compute the worst-case point by finding the intersection with a line, but it wasn't working. My submissions were getting Wrong Answer or Time Limit Exceeded.

I spent over two hours on this. It felt too long for a Bronze problem. I decided to look at the official solution.

The solution idea is clever: Instead of thinking about the minimum X to guarantee success, think about the maximum total chips (A + B + X) for which it's still possible to fail (i.e., we cannot reach FA A chips). Then the answer is that maximum failing total plus one, minus the initial total (A + B).

So, find the maximum NA (number of A chips) and NB (number of B chips) such that NA + NB is maximized but we still cannot achieve FA A chips. Then X = (NA + NB + 1) - (A + B).

When can we not achieve FA? Let NA and NB be the counts after receiving chips. The final A count is NA + floor(NB / CB) * CA. For this to be less than FA, we need NA + floor(NB / CB) * CA <= FA - 1.

To maximize NA + NB under this constraint, we should:

1. Make NB as large as possible but still have floor(NB / CB) be minimal relative to NA. The worst is when NB ≡ (CB - 1) (mod CB), i.e., NB = k * CB + (CB - 1). This maximizes NB without increasing the exchange gain.
2. Then, set NA as large as possible given the constraint: NA = FA - 1 - floor(NB / CB) * CA.

But we also have the initial A and B. We need to express NA and NB in terms of initial values plus possible additions. Actually, we are searching over all possible NA >= A and NB >= B that are reachable by adding some chips (i.e., NA - A and NB - B are non-negative integers).

However, the solution simplifies further. It finds the maximum I (a non-negative integer) such that: NA = FA - 1 - I * CA NB = (CB - 1) + I * CB and NA >= A, NB >= B. Then the maximum failing total is NA + NB. If no such I exists with both NA >= A and NB >= B, then we can fail with just the initial chips? Actually, we need to check the initial condition separately.

I coded this logic: For each test case, compute the answer as follows:

1. If with initial chips we already have A + floor(B / CB) * CA >= FA, then answer is 0 (we already guarantee FA).
2. Otherwise, find maximum I such that: NA = FA - 1 - I * CA >= A NB = (CB - 1) + I * CB >= B This I is bounded by (FA - 1 - A) / CA and also by non-negativity. We compute the minimum I that satisfies both lower bounds? Wait, we want to maximize NA + NB, which is (FA - 1 - I*CA) + ((CB - 1) + I*CB) = FA + CB - 2 + I*(CB - CA). So if CB > CA, we want the largest possible I. If CB < CA, we want the smallest possible I. If CB == CA, then I doesn't affect the sum. So we need to consider the range of valid I (where NA >= A and NB >= B) and pick the I that maximizes the sum given the sign of (CB - CA).

But the official solution was very short. Let me implement it step by step.

We compute I as follows:

• Let minI be the smallest integer such that NB >= B, i.e., I >= ceil((B - (CB - 1)) / CB). But since NB = (CB - 1) + I*CB, we have I >= ceil((B - (CB - 1)) / CB).
• Also, NA >= A implies I <= floor((FA - 1 - A) / CA). So valid I exists if minI <= maxI. If no valid I, then the worst case is just using initial chips? Actually, we need to check the case I = 0 separately? Let's think.

The maximum failing total is NA + NB for the I that maximizes the sum. If CB > CA, we take largest I (i.e., maxI). If CB < CA, we take smallest I (i.e., minI). If CB == CA, any I gives same sum, so we can take minI (or maxI). Then answer = max(0, (NA + NB + 1) - (A + B)).

But we must also consider the possibility that we cannot fail at all even with no additional chips? That is covered by the initial check.

I implemented this and tested with the samples. It passed the given examples. Finally, after all that struggle, the solution was much simpler mathematically. I spent way too long overcomplicating it.

Here is the extension based on the provided transcript and your initial summary:

After finishing “Chip Exchange,” I moved on to the second problem: “Cow Splits.” This one tripped me up badly during the contest. The problem gives a string of even length made only of the letters C, O, and W. You can remove any subsequence in one operation, and you want to split the string into two identical halves using as few operations as possible. If k = 0, you must use the minimum number of operations; if k = 1, you’re allowed at most one extra operation.

At first, I thought it was simple:

• If the string is already two identical halves, answer is 1.
• If not, maybe try removing all of one character (e.g., all Cs) and see if the rest forms a valid double string. That would give 2 operations.
• Otherwise, just do 3: remove all Cs, then all Os, then all Ws—each half will be empty, which trivially matches.

But this greedy “remove one character type” approach failed on test case 3. I spent nearly an hour trying to debug it, even attempted a brute-force DFS for small inputs, but realized that wouldn’t scale to 10⁵ characters.

Eventually, I gave in and checked the official insight. The key observation is: if the answer isn’t 1 or 3, it’s always 2—and there’s a constructive way to prove it.

Split the string into two halves: left and right. Now, slide a window of size 3 across both halves simultaneously. For each triplet (L[i], L[i+1], L[i+2]) and (R[i], R[i+1], R[i+2]), there must be at least two matching characters in the same positions (by pigeonhole principle, since only 3 letters exist). Use those matching positions in operation 1, and the rest in operation 2. This guarantees a valid 2-operation split.

I implemented this by comparing substrings of length 3 from both halves and building two output strings accordingly. It passed all tests in under 15ms—far simpler than my earlier over-engineering.

Finally, I tackled “Photo Shoot.” Given an N×N grid and a K×K kernel, you can add any integer value to every cell in a K×K subgrid. The goal is to find the maximum possible value of the smallest K×K subgrid sum after optimal updates.

I initially confused coordinate mapping between the original grid and the convolution-like sum grid. After several indexing errors and a spilled drink, I realized:

• Each update affects a contiguous K×K block.
• The final value of any K×K subgrid sum is its initial sum plus the total of all updates that cover it.
• To maximize the minimum subgrid sum, we should apply updates uniformly—but the problem allows arbitrary updates, so the optimal strategy is to compute how much each potential update location contributes to each subgrid, then greedily raise the lowest ones.

However, the actual solution is simpler: simulate the effect of updates in reverse. Maintain a 2D difference array to apply updates efficiently, then compute the final sums and take the max of the mins. With fast I/O and careful bounds (N−K+1 for the sum grid), it ran in 75ms.