Sumitomo Mitsui Trust Bank Programming Contest 2019 has just finished, and this is an unofficial editorial.

Thanks to my friends oolimry and shenxy13 for helping write some of the editorial.

A — November 30

You can solve it simply by checking for each end date of the Gregorian calendar. However, note that as the second date directly follows the first date (a fact which I think is not clear in the English translation), we can also check whether they're in different months, or whether the second date is the first day of a month. This can be done in constant time.

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B — Tax Rate

We note that there is monotonicty in this question. $$$\lfloor 1.08(X+1) \rfloor \geqslant \lfloor 1.08X \rfloor$$$. Hence, we can binary search the answer. When we binary search the value of X(the answer), if $$$\lfloor 1.08X \rfloor = N$$$, then we have our answer. Otherwise, we can search higher if $$$\lfloor 1.08X \rfloor > N$$$ and search lower otherwise. If we find that no number gives us desired N, then it is impossible.

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C — 100 to 105

We can simply do a 0-infinity knapsack with weights 100,101,...,105 and check if some value is reachable. We get a time complexity of $$$O(N)$$$.

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D — Lucky PIN

First, we note that there are $$$O(N^{3})$$$ subsequences of the string, so generating all of them and using a set to check for number of distinct subsequences is TLE. However, there are only at most 1000 distinct such subsequences, from 000 to 999. We can linearly scan through the string for each of these possible subsequences to check if it is actually a subsequence of the string in $$$O(N)$$$. Thus, this can be solved in $$$O(1000N)$$$, which is AC.

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E — Colorful Hats 2

Firstly, we can imagine there are 3 imaginary people standing at the very front, each with a different colour hat. For each person, we consider how many possible people could be the first person in front of him with the same hat colour. If the current person has number X, then the number of ways is:

(no. of ppl with X-1 in front) — (no. of ppl with X in front)

This is because those with X in front of him must be paired with one of the X-1 already, so this reduces our options.

Implementation wise, we can store an array which keeps track of the number of people with no. X who are not paired yet. Initially, all values are 0, except at index -1 with the value of 3. Then when processing the current p[user:AtillaAk]erson X, we multiply the answer by arr[X-1], decrease arr[X-1] by 1, and increase arr[X] by 1.

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F — Interval Running

Firstly, if $$$T_{1}A_{1}+T_{2}A_{2}=T_{1}B_{1}+T_{2}B_{2}$$$, the answer is infinity.

Else, WLOG, we let $$$T_{1}A_{1}+T_{2}A_{2} > T_{1}B_{1}+T_{2}B_{2}$$$. If $$$A_{1} > B_{1}$$$, then Takahashi and Aoki will never meet each other. The answer is 0. Now, we have solved all the corner cases. We shall move on to the general case. We shall call the period $$$T_{1}+T_{2}$$$. Now, we shall find the number of periods where Takahashi and Aoki meet each other. If we do some math, we get the number of periods to be $$$\lceil \frac{T_{1}(B_{1}-A_{1})}{(T_{1}A_{1}+T_{2}A_{2})-(T_{1}B_{1}+T_{2}B_{2})} \rceil$$$.

The number of times that Takahashi and Aoki meet each other is $$$2periods-1$$$ since every period they meet each other twice when Aoki overtakes Takahashi and Takahashi overtakes Aoki. We need to minus 1 since we do not count the first time they meet each other at the very start. We submit this and... WA.

Yes, we need to think about the case where $$$\frac{T_{1}(B_{1}-A_{1})}{(T_{1}A_{1}+T_{2}A_{2})-(T_{1}B_{1}+T_{2}B_{2})}$$$ is a whole number. In this case, we need to add one, as there will be one more time where Aoki will meet up with Takahashi but never overtake him. And now we get AC. (Thanks to Atill83 for pointing out the mistake)

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