Hello Codeforces Community!

We would like to invite you to our next monthly contest and we’re sure you’ll enjoy — the October Lunchtime 2018 sponsored by Sharechat! Along with interesting problem sets, we have exciting full-time job opportunities by ShareChat for professionals across the globe. More details can be found on the October Lunchtime contest page.

Joining me this time on the problem setting panel are:

Problem Setter: kingofnumbers (Hasan Jaddouh), PraveenDhinwa (Praveen Dhinwa)

Problem Tester: teja349 (Teja Vardhan Reddy)

Editorialist: taran_1407 (Taranpreet Singh)

Statement Verifier: Xellos (Jakub Safin)

Russian Translator: Fedosik (Fedor Korobeinikov)

Mandarin Translator: huzecong (Hu Zecong)

Hindi Translator: Akash Shrivastav

Bengali Translator: Imran Hasan

Vietnamese Translator: VNOI Team

### Contest Details:

**Time:** 27th October 2018 (1930 hrs — 2230 hrs). (Indian Standard Time — +5:30 GMT) — Check your timezone.

**Contest link:** https://www.codechef.com/LTIME65

**Registration:** You just need to have a CodeChef handle to participate. For all those, who are interested and do not have a CodeChef handle, are requested to register in order to participate.

**Prizes:** * Top 10 performers in Global and Indian category will get CodeChef laddus, with which the winners can claim cool CodeChef goodies. Know more here: https://discuss.codechef.com/questions/51999/how-do-i-win-a-codechef-goodie. (For those who have not yet received their previous winning, please send an email to winners@codechef.com)

Good Luck! Hope to see you at the contest!

ACMIND18 part 2?

We'll see

I want to be a Um_nik but I stop myself here. Hope one day I will give my opinions like him.

The real question though is that if you have never solved the third problem in any contest ever, does there exist a contest not like ACMIND18 for you?

There is no need to be Um_nik teja349. The comment by jtnydv25 hits it home.

I have seen many times that Jatin goes after the toughest question in every Div1 contest. No matter how tough the question is. This really shows the mentality these red coders have. They seek challenges and solve them. Normal people think of a problem as some kind of a curse. And they just want to get over with it.

Ratings for Encoder are not updated.Will it happen before the contest?

How to solve third problem(div. 1)?

A bit hard to explain!! Here's what I did

Count all -1's (let it be CT) and subtract the non -1 elements from S.

Then generate

all distinct sets (multisets) of size CT having sum S. The number of such sets is small under the given constraints (order 10^3) and all such sets can be generated using simple backtracking.For each such set count the number of ways you can rearrange the elements... Simple combinatorics!! (Maintain count of each element in the set)

To each set append the non -1 values and find the ans (lets call it Temp_ans) for each set(brute force).

Final ans=Summation( (ways*Temp_ans)%MOD).

Link to my solution: https://www.codechef.com/viewsolution/21244428

is this the expected solution ??

yes

A simple doubt: how you estimated no. of distinct sets will be order of 10^3 ? or Is there any formula to calculate no. of distinct solutions ( By distinct it means if eq'n- x+y+z = 8 then 1,2,5 and 2,1,5 and 5,1,2 should be treated same and counted as one)

I don't think there exists an explicit formula, but wolfram alpha can calculate it. The order may be higher than 10^3, but then

Nwill be small. For example forN= 10,S= 50 andA_{i}= - 1 there are 17000 partitions, but the complexity of the solution isO(XN^{2}) whereXis the number of partitions. So basically ifXis big, thenNmust be relatively small, and vica versa, so all in all it will pass.Thanks

"So basically if X is big, then N must be relatively small, and vica versa"Can you tell me why? How are number of partitions and N related?

The number of partitions will be big if

Sis relatively big and the number of parts is relatively small. The second means that there are few -1 s, and the first means that there must be few non -1s because every number is positive.I have a polynomial solution.

Let's look at the following solution to find the niceness of a given array (no - 1

s) .. loop for the gcdgand then count the number of pairs of elements withgcd=g(let it beans[g]) .. addg*ans[g] to the answer. However, finding the number of pairs of elements withgcd=gsounds tough .. the idea is to find the number of pairs of elements such thatgdivides both (much easier problem) and then subtractans[2 *g],ans[3 *g], ... from it. Which means that we find all pairs with gcd multiple ofgand then subtract those withgcd! =gto end up with those withgcd=g. How to adjust this to our problem ? Instead of counting the number of pairs with gcd multiple ofgin the given array, we want to find the number of pairs with gcd multiple ofgin all possible arrays .. the rest of the solution is the same. Notice that it's sufficient to know the count of multiples ofgfor this. If all the elements are - 1, we can calculatedp[g][n][c][s], the number of arrays withnelements,cof which are multiples ofg, and their sum iss. The transitions are simple: just bruteforce every next element and change the state based on it. We'll add to our count for allc≤n, but what about the elements that aren't - 1? well, they only change the startingn,s, andc(we'll reduce their count fromn, their sum froms, and the count of multiples ofgin them fromc). The complexity is upper-bounded toO(n^{2}s^{3}+t(sn+slog(s))) but, of course, it's faster. codeBonus for third problem:

Instead of given formula try to solve problem for range formula(instead of pairs), like this:

During the contest I spent my lot of time to this problem. I debugged my code for 1 hour and I could not found any mistakes. My code giving 21 for second test. After 1 hour I realized that problem asking for pairs not ranges :)

I didn't read tasks for Lunchtime, but this task can be solved with binary search + segment tree in

O(nlog^{2}(n)·logX_{i}):For each position

i, it will be at mostlogX_{i}postionsjwith different valuegcd(X_{i}, ...,X_{j}). and values will go decreasing. After it, just do binary search + segment tree to find all positionsj. Probably it can be done without binary search without onelogn.Do you have something better?

Try to read problem, because

N≤ 50 andA[i] ≤ 50 :)