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You know the author does not bother about contributions when he/she does not attach nice solution codes with the editorial. (Although the explanations are quite good).

UPDATE: Now they are added

Sorry for that :(

The solution codes will be added shortly.

They were added.

Thank you for the timely editorials stanislav.bezkorovainyi.

too stupid to remember the meaning of problem D ....

Did anyone solve E using binary search? I tried and get TLE, reason was modulo operator of get hash function.

I got AC with O((nm)^2) 46184585.

46184740 after a little change, AC 420ms !!!

maybe the data is random,so you can AC it

I solved E with decimal search.

Can anyone help me with a WA on problem E? 46186783

I am doing the anagram hashing by assigning a prime to each letter of the alphabet and multiplying them modulo a big prime. I tried different primes, so it's probably not a collision.

OK, I found it. Previously I thought when I "disable" a row, I can just subtract one. But the palindrome centered on the disabled row can be of course bigger.

But I still had WA on later tests. Turns out scientific notation (like

`1e17`

)is for floating point constants. So my const prime wasn't prime, because`double`

can store integers only up to 1e15.Interesting note: on problem D, for n larger than 35 you can just output "YES (n — 1)" in O(1).

Note:for n larger than 31 ,you can just output "YES (n-1)",,,-_-

Wow! this Editorial is so detailed. Thank you.

nice solutions and editorials for me! Thank you!

So overall, the complexity for E will be 26×n^3 or n^3 only ?

26*n^3 passes, but you can get rid of this factor. When you increment a letter's count for a row, you can check if you increment the count from odd to even or from even to odd. So alongside eg.

`letter_count[n][26]`

you can keep`odd_occurences[n]`

Alright, thank you !!!

I solved F with an overly complicated sqrt decomposition in 46236775. I feel so silly.

I solved problem F using a normal segment tree in O(q*logn*logk). 46262923

stanislav.bezkorovainyi your editorials are very detailed and well explained, for a newbie like me detailed editorials matter a lot. Thanks!

Thank you!

It feels so good when community appreciates your job (^-^)

Can you explain/prove this equation for me?

W(a, b, c, d) = w(c, d) − w(a − 1, d) − w(c, b − 1) + w(a − 1,b − 1)

Maybe one day I'll learn how to manage pictures in codeforces' comments, but for now here is the link:

https://drive.google.com/open?id=1NhVlMIdhtSjGo4Djhygypd_8U4-3U0CC

Hope the proof is good and my English is not that bad :)

Thanks, so straightforward

You are welcome)

Can anyone help me with my solution for D. I am not able to figure out what my mistake is. Link to My code . TIA

best editorial indeed

Seriously, what the f###? How can a div2 round consist of Manacher algorithm as E and then persistent segtree as F??? In div2 rounds, focus should be on thinking problems and atmost lazy-segtree is allowed. A very bad round.