The the "educational" name may change!

Let's make educational rounds rated like atcoder. In atcoder beginner contests are rated for beginners. More beginners will participate to rounds.

I think 6 or more problems will be , (He said :: Task F may end up beneficial for many reds )

Like they were usually held, there will be 6 tasks for 2 hours. I'll update the blog with this :)

maybe this Propose a Problem tab

NOO

"You will be given 6 tasks and 2 hours to solve them." Which part did you not understand?

It is unrated contest. You can improve your skills and become beneficial with some good problems.

You are expert. And you don't know that educational rounds are not rated.

I think that rating Educational Rounds would be like giving points to soccer teams for what they do during practice sessions.

I believe the main purpose of Educational Rounds is to learn important techniques and algorithms and the people that take part in them are either like "I'll participate and see if I learn anything new" or like "I'll take part until I get bored". Rating such an irregular contest would be a mess.

educational means unrated. you wont lose or gain rating.

Don't ask such questions during the contest!

I didn't need to generate prime numbers for A. I just generated the numbers that divides N that are less than or equal to sqrt(N). Every other divisor can be generated from these numbers.

If query K is less than the size of the divisor array just print the divisor. If K>divisors.size() and K<=2*divisors.size() than you can generate the divisor by dividing N to K-divisors.size()th divisor from the end in the array. If K>2*divisors.size() print -1.

You should have implemented it in and it would have been much faster.

My N*log(N)*log(N) code passed :)
Here is link.

I am refreshing every 5 seconds.I reallllly want to know the solutions of D and E

UPD: problems already in practise.

The answer is out of the range of int in C++

No. LCS will be N^2 which is too much for the given time limit

actually it doesnt have anything in common.

My solution is O(n). Let dp[0][i] be the length of the longest prefix of b that is a subsequence of a[1: i], and dp[1][i] be the length of the longest suffix of b that is a subsequence of a[i: len_a].

We can easily see that if a[i] equals to b[dp[0][i - 1] + 1] then dp[0][i] = dp[0][i - 1] + 1, otherwise dp[0][i] = dp[0][i - 1]. Do the similar thing for dp[1][i].

We will find the position that dp[0][best] + dp[1][best + 1] is maximum, then print the first dp[0][best] characters and last dp[1][best + 1] characters of b.

Source code: 24121085

Let P_i be the set of indices of A which form the subsequence for B[1..i]. Similarly, let S_j be the set of indices of A which form the subsequence for B[j..m], where m is the length of B.

That is, store the indices of A for the longest prefix and suffix of B that is a subsequence in A.

Iterate over indices of A and try to maximize the length of P_i + S_j, which is equivalent to minimizing the length of the segment erased from B which is [(i + 1)..(j - 1)].

Submitted Code

on codeforces educational doesn't mean simple.and don't worry,after just 1 year i didn't know much.You have to patient and you will get better

These problems aren't simple,just the ideas maybe are well-known and i think this helps a lot to build your set of ideas when you solve a problem.

I got hacked on a test like this:

aa
a

a aba

Sometimes works I guess

Good day to you,

I got many of hacks with A=10^5*'c' B=9*10^4*'c'

I.e., string full of "same" characters and one shorter string {two short strings might work too}

For problem C . I have a input

axbxcxdxexfxg

abcwwdewwfg

why ans is abcfg ? shouldn't it be abcdefg ?

because :

abcdefg we have to remove 4 characters

abcfg we have to remove 6 characters .

In this problem they said removing minimum characters . So It should be abcdefg . why abcfg ?

Can anyone explain ?

Good day to you,

well you have two strings, and you have to remove EXACTLY ONE consecutive passage [i.e. "K" consecutive characters] form second one, so it becomes sub-sequence of first string {i.e., all characters from rest of second string appear in first string [not necessarily consecutive] in same order}

HINT: As you can see [from statement], you remove the part from middle, suffix or prefix so (if anything) only what can remain is: "suffix", "prefix", "suffix+prefix" or "whole string"

Good Luck & Have nice day!

Only 3 accepted :/

Yeah, don't worry, we will rejudge it soon!

Compare function must return false on equal elements. I challenged so many solutions with this error...