Another good thing about atcoder is short and to the point problems.

It's easier than Div3 on problems A and B, but sometimes as difficult as Div3 on problems E and F, the overall difficulty is slightly lower than Div3, personally, I believe it's going to be a interesting contest for people with rating below $$$1450$$$ on CF.

UPD：I was wrong.

The early problems on Atcoder are much easier than any problems on CF.

However, in my opinion, the last problem (F) on ABCs is typically substantially harder than the last problem on CF Div3. I think it's harder for me to solve all problems on an ABC (I frequently don't) than all problems on a CF Div3 (I typically do, small sample size though), and since the earlier problems are harder on CF, that implies the later problems are harder on Atcoder.

In summary, ABCs are definitely easier than CF Div2, but overall harder than a CF Div3 (if you aim to solve all problems). If you're only aiming to do the first few problems, ABCs are easier than CF Div2 and Div3 both (the first problem on last week's ABC was "input X and output 3X^2").

on the standing page, the last row called Accepted / Tried
total # Ac over the overall submissions

Concatenate $$$S$$$ with itself until it has a greater length than $$$S+T$$$. Then, using a string matching algorithm, check which positions in the first $$$S$$$ could be the start of an instance of $$$T$$$. For each of these positions, use the length of $$$T$$$ to figure out which position, modulo the size of $$$S$$$, you'd reach in the concatenated $$$S$$$ after placing a copy of $$$T$$$ starting here.

Afterwards, the problem reduces to longest paths in a directed graph. Our nodes are the positions in $$$S$$$ and our edges connect each valid position at which we can place a copy of $$$T$$$ to the position at which the next copy of $$$T$$$ would start. If this graph contains a cycle, the answer is $$$-1$$$. Otherwise, we have longest paths in a DAG, which is a relatively standard problem.

We have to use the divisibility rule of 13 i.e the difference of alternate chunks of 3-digit numbers from the unit place if divisible by 13 than the number is divisible by 13. so we have the form of (A*100 + B*10 + C)%13 = 5. Here A,B,C consist of sum and difference of ? which can take values from 0-9. Sorry this is what I was thinking during the contest but the idea was too tough to implement.

Submission

According to the above explanation.

Let $$$dp[i][j]$$$ be the number of length-$$$i$$$ numbers that match the pattern's first $$$i$$$ digits and are $$$j$$$ mod $$$13$$$. Then, we can transition by iterating over every possible digit in the $$$i+1$$$'st position, noting that appending a digit $$$k$$$ to state $$$(i, j)$$$ gives a remainder of $$$10j+k$$$ mod $$$13$$$.

Actually this was a simple digit dp problem. You should explore some problems like that.

If you think this problem was difficult, then look into this codechef problem. During contest time I implemented it and got it accepted, I had to use 6-7 parameters to define a dp state. And surprisingly the editorial also used same no. of parameters.

Another example of digit dp problem: Atcoder Educational dp contest, problem S

I didn't even solve D

No

No partial scoring

You will not get partial score

How did you solve F?

DP. States are position and remainder.

Go index wise in the string starting from the left end. Think what will be the contribution of the current number formed till now in the modulo 13. DP[i][j] — stores the number of numbers that can be formed using length up to i and have modulo j with 13.

Obviously, i -> (0..n-1) and j (0..12) Since mod 13 only 12 values are possible.

Now If at S[i] there is no question mark then the transition is simple. the new number formed is just the previous * 10 + (s[i]), And we have 12 possible previous so check for all of that. Now do % 13 of this and increase the count of the new mod in i. Otherwise, if there is a question mark just do this for all digits from 0 to 9. And update.

Finally, the answer is dp[n-1][5]

I won't be posting an editorial today--I was working on one after I finished the contest, but I accidentally hit my backspace key just before I finished, causing me to leave the page and lose all of my work. Unfortunately, I don't have time to write it up again.

I didn't participate in the contest this morning, but I'm working on the problems for practice now and might write an editorial (if nobody else has and I actually solve all the problems, haha, but I heard today was a hard round!).

Just recursive dp(dp[i][j])

I think so.

Just recursive dp(dp[i][j])

"abb" "bab"

Dont know about disqualification, but they will tell to improve your english for sure..

I solved this with recursive dp: dp[i][j] is if it is impossible to reach i,j(point with coordinates (i;j) then dp[i][j] will be someting like -1. if not, it will be the minimal number of score needed to reach i,j and the point from witch we went to i,j. this may fail in time, so I used rands and probalities and my solution worked with a big chance correct and with a big chance to be fast at test.

I store the total number of monsters in a variable say res.Then I check for each i if B[i]>A[i] then we update A[i+1] with A[i+1]-min(A[i+1],(B[i]-A[i])) because ith hero can defeat some of the (i+1)th position monsters. and for each i we have to update final A[i] value.that is max(0,A[i]-B[i]). Finally we calculate the sum of total number of monsters left say res1. So our answer is (res-res1).

I also thought of the same idea but did not implement it as I thought it will TLE :|

What is the reason for concatenating s with itself until it's length is greater than |S|+|T|?

Thanks a lot

you can use Atcoder's old server https://abc135.contest.atcoder.jp/ it's kinda fast

I also want the data of problem F. I got WA on seven tests and I don't know what my mistake is.

Btw your regex is backwards