I solved 4 problems for the first time in Educational Round. I'm very happy!

but newbies can become Grandmaster

Just keep solving the prolems in 800-1200 range atleast for one more month to build a solid foundation. Then move ahead. Your practice regime looks good. Just be at it..

It's a good way to begin with basic algorithms. When my rating was below 1400, I started with partial sums and differences, greedy, binary search and some constructive problems and math problems. The algorithm which left me the greatest impression was double pointers because that was the very beginning of my experiences in Algorithm Contests. It shocked me a lot and made me feel the beauty of algorithms at that time. For me, I have better skill in mathematics than coding, so I always practice harder problems in stuffs like permutations, number theory, etc. Algorithms like DP, searching, data structures (For example: Binary Indexed Trees(BIT), Segment Trees, Disjoint Set Union(DSU)) are also fundamental. I learned them after I became Specialist, but I'm still not really skilled at them yet:C
Also it's recommended to practice in contests. You can participate in every div4, div3, div2, div1+2 and edu if possible. Atcoder has a type of contests named ABC and begin at 20:00(UTC+8) every Saturday, which are very suitable for beginners. Sometimes you can do Virtual Participation in div3 and div4 rounds before. When your rating gets higher, it could be div2. Sometimes you might encounter a situation that you have the ability to solve some problems, but don't have enough time to pass in the contest. This time don't forget to make more trials after the contest, it will help a lot if you get Accepted.
Different people might have various learning process from each other. Hope you will enjoy your time in Codeforces and become more skillful in the future!

always like that in edu rounds

[deleted]

It's roughly the same thing as testers not participating in a contest that they tested from their main account, through their alt account: mainly trust-based.

don't worry,you'll soon find out,no matter how hard is it,just to remember keep calm and carry on!

div.2 is hard for beginners, try div.3 or div.4 contests. by the way i believe you can solve at least 1 question. but don't push yourself to solve all questions. keep calm and concentrate on first two problems. if you could solve them that is really great and would give you a nice start rating.

Oof, you wrote 5+ educational rounds without knowing the difference?

currently its available for div3/4 and edu rounds only.

Balanced?

E<<<<<<F OK?

All cheaters should be banned after cheating for once.

And we need roll back.

4*

my solution for each t is: find the optimal subsequence within s(optimal meaning that the last letter used is at the lowest index possible), then for the remainder of the string s, just figure out the smallest string which isnt a subsequence of that remainder of the string. i think the idea is correct, and that i messed up in implementation, but its weird that so many people also had wa on test 6, so i dont know

For me, it was related to the transition between DP states. You can compare my last 2 submissions.

You can use a array of set to store the position of every char to check if a string lie as a subsequence or not in NlogN then two cases are possible:

• the string t is not found as a subsequence in s straight away the answer is 0
• if the string t is exhausted you will have the last position in s upto which the prefix substring is capable of giving t now we can try out all possible k characters one by and the raising the required pointer to that index, basically trying out all possible cases , we can optimise this with dp

My submission link : submission link

although I am not sure it will pass hacks or not cause the time consumption is really on edges

I think this guy is using AI to write his code too 317605605.

I think I found him on Linkedin, same name and same Technocrats Institute of Technology: link

report this yt channel

Getting to Master is too hard now as I need to have a REALLY good performance LOL.

you should binary search over amount of elements being deleted ( say k ) at each step of binary search, you check if deleting k smallest elements helps.

to check, you should use the fact that any array with sum S and N elements. can converted to N first primes iff S >= sum of N first primes

Try to convert the largest element of array into smallest prime number, it will give you additional coins for future.

so make an increasingly sorted array of the first n primes,and sort a decreasingly, then just find the first element where the prefix sum of the primes is larger than the prefix sum of a, those are all the elements that can stay, so the answer is n-(index of that element). no need for binary search

It made me remember the exact same problem :)

I find it very interesting how the brain works for people to remember a problem from over a year ago.

you can optimize your approach by precomputing the result using DP

for each index $$$i$$$, what is the minimum number of allowed letters that I can add such that it won't be a beautiful string?

precompute this using DP and your answer should work well

Greedy works. you used binary search to find the string 't' last index right? now you can again use binary search to calculate number of operation required.

come from last on string s, put all index to some list when all k character are visited. now binary search on that list.

Same thing happened with me . I was able to figure out my error just after the contest :(

You checked the first 1e7 numbers so that doesnt contain 4e5 prime numbers needed to calculate the prefix sum

because you need about 6e6 numbers to get 4e5 primes

Your code doesn't generate 1e6 Primes, it only generates all primes less than 1e6. Your code roughly generates ~78000 primes.

okay the i went like this,as the size of array is 4*(10^5) that means we need atmost this number of primes and i used cpp to find that number in vs code.

for (int j = i*i; j <= N; j += i)
    is_prime[j] = false;
//================

int N=1000000; //1e6

Doesn't that mean that in some cases you do not mark enough values as being composite?

2.5 maybe

When s[0]=='B' and s[n-1]=='A',that is #1 in Bob，#n in Alice.

As long as #n-1 in Alice, Alice will win since only #n can beat #n-1 but Alice owns it.

800-1000