Yeah, definitely)

Sorry for the typo, it is fixed now.

Is it amazing???

It means if and only if.

I'm quite certain that it can be solved by algorithm Boruvki

I used tries, but it passed with just 150 ms left, will probably give TLE on hacks. Also, weirdly, none of test cases have n = 1.

Divide and conquer, or meet in the middle. Whatever you wanna call it

Meet in the middle technique:

Divide the array into two smaller ones of size n/2. Now you can brute-force all the possible combinations of values you can get by choosing subsequences of each array. Store them in two sets. This step takes O(2^(n/2))

Now iterate through the first set. Our goal is to get the closest number to m-1 (which is the best answer possible). So, for each number x in the first set, we can binary search for the best value y of the second set that gets us closer from m-1. Keep track of the maximum x+y gotten, that is the answer

I solved it using Meet in the middle + Binary search

Split the n numbers to 2 sets, find all the possible subset sums for every set, that can be done in O(2^(N/2)) for each set

sort one of them and iterate over the other set, for every element, binary search for the value that gives the max value % m

Binary Search

I used simple greedy. Mark positions of occurrence of each letter in vector v[ch]. Additionally, add -1 and n to it (indexing begins from 0). Now calculate the max distance between 2 elements in each v[ch]. Take min of all these maximums.

For every letter from a-z that exists in the string find the maximum k such that any substring of length at least k contains that letter. Or in other words find the maximum difference between two contiguous occurrences(or boundary) of that letter. Report the minimum k as the answer.

I did without B.S.

what I did is:

For any String minimum k value goes to l/2+1 where l is length of string...

For ex:abcde k=3 and abcd k=2

I checked for all character,found maximum of distance b/w them..And took minimum

That will be final answer...32173323

When it's your turn for preparing editorial, you will know why.

For this contest open hacking phase is still going on. They can not post editorial before it finishes.

Yeah you should be good

Left at derangement

The thing is that Edvard doesn't take any part in preparing Educational Rounds now. He doesn't send me any problem proposals or anything. So the fact that we have prepared a problem that you sent him is just a coincidence.

Derangement

1) We always can get sequence 1, 2 ... n

2) Then calculate answer for each k from 1 to k (from input). We can swap C(n, k) sets of positions. Numbers of swaps for each set is !k (subfactorial).

ans = 1 + ∑ C(n, i) * !i (i from 1 to k)

Да, но позже Да, можно, заходишь во вкладку статус, дальше внизу страницы видишь Фильтр статуса, выбираешь нужный язык, вердикт и статус и нажимаешь применить

Let's say that you have a lot of groups and want to merge them together with the minimum answer. You know the very base case that every node in a separate group, but how we will continue a lot of xor problems solve by trie store binary bits instead of strings.

Trie or prefix tree store one copy of every same prefix of the number, here we will put every number from the most significant bit to least significant bit. start traverse through trie all the numbers will have in common the very first node in the trie every node we will have 2 choices in this node (1 or 0) we need to merge the 2 groups by each other, every group of them will connect internally by each other later by the same way and its very obvious because every time we move in the trie the value that we add to the answer will decrease. We will connect the 2 groups by a search for every number in the first group and try to find the minimum number to it in the second group so now we connect the first group to the second group, but internally nothing happened we will move to the first and second group and do the same algorithm.

Algorithm: when traversing the trie if there is one edge only(1 or 0) you will go through it without adding anything to the answer if there are two edges(1 and 0) you will go through the two edges with adding value to the answer.

Value: as we said before we need to connect the two groups together with the minimum value, for every element in group 1 we will find the minimum element for him in the second group and take the minimum one.

Simplest code: here

Sorry for poor English and poor explanation :'D .

Your function fun() has two loops which will take max k*n steps and hence the TLE. You can make it run in linear time by maintaining the frequencies of letters in each subarray of size exactly k. My submission based on binary search: http://mirror.codeforces.com/contest/888/submission/32166677.

here