We can solve problem E in using two pointers: find the smallest range where product of numbers inside this range is divisible by k, then res = res + (n - end) * (start - previousStart) 28156236

for problem G we can build a flow network with only O(n) edges, so normal minimum cost max flow would work:
So we'll have 4 vertices for each array position, 2 of them are connected by an edge with cost -1, if this edge is part of the answer, then we would use this array position for one of the sequences, but we'll need to have a path from ai to every aj such that i < j and abs(ai — aj) = 1 or abs(ai — aj) % 7 == 0, for every array position i have an additional vertex which has a path to all those j > i such that ai = aj, we only need to connect this vertex to the first j which has the attributes mentioned, so we would only need to connect ai to the first j's additional vertex that is ai — 1, and the first j's additional vertex that is ai + 1 to have path to all such j's that aj = ai — 1 or aj = ai + 1, a similiar thing can be done for the modulo 7 part, with another additional vertex for every vertex (it can also be reduced to 7 overall, because there are only 7 modulos), you can see my code for better understanding

In problem G, am I mistaken by saying that almost everybody who solved it used a compressed graph with O(N) edges? That's what I understood from most of the sources: that for each i they add just about 3 edges backwards to the last j with +-1 different value or the same residuum modulo 7. Is that a correct approach? I haven't thought of some counter example but I cannot prove the correctness of this approach either...So could someone shade some light on this issue?

PS: Ok, it seems Reyna said about the same thing while I was typing the comment, so nevermind

For E, I built a sparse table of products of segments modulo k. It seemed simpler than fiddling with vectors of prime powers.

In E i used two pointers and queue, which can return (product of numbers inside) % k, similar to queue for finding min/max. Solution O(n) 28158111

I solved E with factorisation and then binary search (fix right border, binary search for left border). I am getting Runtime error on case 16 when n = 100000 and k = 1000000000. I tried some smaller n and k = 1000000000 and it works oK. Any ideas ?

http://mirror.codeforces.com/contest/818/submission/28159818

If I'm not mistaken, my solution to F is O(1) : Solution I guess it's technically O(Q).

For Problem C,I have something puzzled.There is one sentence in problem,"Sofa A is standing to the top of sofa B if there exist two such cells a and b that ya < yb".Why ya < yb ?Should it be ya > yb?

In Problem C Editorial , shouldn't we decrement the result by 1 when x1i!=x2i instead of when x1i=x2i

For problem D, if cntA(n) = cntB(n) = 0 then which case should it be classified as? From the problem statement it appears that both Alice and Bob would win, but that sounds like a draw to me.

In Problem E I built a segment tree for modulo K so I can check if a segment [L..R] is divisble by K

It looked much easier that factorizing numbers and such

My Code

Could somebody take a look into my solution for E? I can't understand why I got WA :( (Counting prime divisors in k first, then moving R until we have not enough divisors from k in current segment). Thanks in advance.

http://mirror.codeforces.com/contest/818/submission/28168045

EDIT: Oops, my fault.. I took sqrt(10^9) = 10^3 somehow..

Why do we have edges in connected component? Isn't it just ?

So,can I ask for the hack to problem A?

and fill the unvisited children with remaining values to form the permutation.

It doesn't make a difference in which order we place them?

For problem F why is building the graph as suggested optimal? How to prove it?

"Firstly you need to learn to factorize any number in no more than O(logn)." How can this be done ?

I have a question for Problem F: since k * (k-1) / 2 is increasing and n-k is decreasing, wouldn't the maximal value of the function be at the index where k * (k-1) / 2 becomes greater than n-k? If that is true, could we just binary search to find the point where k * (k-1) / 2 > n-k, and get our answer from there? Submission: 28155258 (Sorry not so familiar with ternary search so just asking)

There is another (shorter and simpler than other solutions I saw) solution for E (it looks like O(Nlog(MAXN)), but without binary search or something like this):

We can start iterating through the array and stop when the product of prefix mod k is 0. Fix last index as curEnd.
If we have curEnd == n-1 and mod is not 0, stop iterating and print answer, because there are no more subarrays that are divisible by k.
Then go left from curEnd until the product of elements mod k is 0.
Now we have len of subarray that is divisible by k.
We can add value (curEnd - len + 2 - start) * (n - curEnd) to the answer.
In other words, we add all possible subarrays that contain our subarray (that is, array [curEnd-len+1, curEnd]).
Then update start (it's 0 at the start of algorithm) to curEnd - len + 2 (start is the first index of subarray that we're gonna check on the next iteration, next value of start will be the second element of just found 'good' array, so we won't count it twice or more).
Then repeat the same algorithm, but starting with the new start.

So, why this solution is O(Nlog(MAXN))?
The worst case is when k = 1<<29 and the array consists only of 2's (i.e. [2,2,2,2,2...,2]).
In this case start will be 0 at the first iteration, 1 at the second iteration and so on, but on every iteration there will be len=29 (this is log(MAXN=1e9)).

Link: 28180910

Does anyone know the proof for why in problem F the optimal solution will be achieved by having one large component and simply connecting the remaining vertices with bridges to it? If it's rudimentary, then let me know please so that I can think about it for some more.

Problem D statement is so difficult to understand

For 818F - Level Generation, there can be an O(1) solution.

Here is a mathematical solution that does not require the ternary search. It only uses two values of k.

Let, x be the variable k from the tutorial.

x = sqrt(2*n-1);

n_minus_x = n - x;

ans = min(x*(x-1)/2, n_minus_x) + n_minus_x;

Another try

x = sqrt(2*n-1) + 1;

n_minus_x = n - x;

ans2 = min(x*(x-1)/2, n_minus_x) + n_minus_x;

ans = max(ans, ans2);

Note:

• The min() is used so that the non-bridge edge count does not exceed bridge count.
• sqrt(2n-1) is derived from a formula similar to the one mentioned in the tutorial

Code: 89187360

Hope it helps someone.

Edit:

Derivation of the formula:

x*(x-1)/2 <= (n-x)

-> x <= sqrt(2n-1)