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my solution for div1F

1) first let's sort princesses as in editorial

2) then if we take princess i, we make edge from ai to bi

3) then we can add new princess, if and only if graph, which we got, has no component, that has n vericles and > n edges, so you need only check it with dsu

As example of this solution you can see this submit(http://mirror.codeforces.com/contest/875/submission/31437228)

My solution for DIV 1 D:

Fixing the left end of subarray, we count the required number of right ends. Now, if we fix one index and start taking prefix or from that index, we see its changed at most O(log(10⁹)) times.

So, we consider each range of interval, where prefix or is constant. In each of these intervals, we can calculate the first index where prefix max is not less than prefix or ( prefix or is constant) by RMQ and binary search, we can add all the subarrays in this interval before this index.

O(nlogn(n)log(10⁹))

A little bit optimisation and I got AC with 982 ms. 31408332

In problem Div2E we also should take in mind that if s[i] is a prefix of s[i + 1] and len(s[i]) != len(s[i + 1]) then it is impossible to get the correct capitalization of letters.

Yet another approach for Div1D

Create sparse tables for operations or and max. Then calculating the largest segment where i th element is maximum and the number of segments that satisfy could be done with binsearch.

That's

http://mirror.codeforces.com/contest/875/submission/31414300

"Если x - y делится на k, то x и y дают одинаковый остаток при делении на m."

Сомнительно.

DIV 2 D, TLE with cin/cout 31455585 but AC with printf/scanf 31455839 with execution time reduced from 1000ms to 124ms !

can anyone please explain me the 876B — Divisiblity of Differences editorial

I don't get the editorial explanation of 875D — High Cry , can someone clarify that approach...what is "go" and what is it's purspose, Also if someone submitted a solution using the editorial approach, please post your submission.. thanks :)

I think the problem High Cry can O(n) and get AC . Here is my solution http://mirror.codeforces.com/contest/876/submission/31447193

In DIV2-E can anyone please help me to understand that if we add directed edges than how can we update all the nodes connected with curr node(if it is somewhere inside the tree)! and also if we do dfs each time then how it fits the time limit !

31392967

Div 1E: Could someone please provide the proof for this implementation?

I've got an alternative solution for div1A. Let's represent n like sum (1+1)d[0] + (10+1)d[1] + ... +(10^8+1)d[8].

Now we can see that we can greedily get all d[i] starting from 8. d[i] = n mod (10^i+1) or d[i] = n mod (10^i+1) — 1.

A recursive solution works in O(2^log10(n)).

31428143

my solution to DIV2-F : I first created next and prevs arrays (the next greater element and the prev greater element) as mentioned in editorial. After that i store all the indices which have that particular bit set, then for each index i just iterate over all 29 bits(except those which are set in current number) and do a binary search on each of them to find if there exists a position which has the required bit set and then just taking minimum among all those and after getting the first valid position i just calculate the answer using basic combinatorics!

my implementation : http://mirror.codeforces.com/contest/876/submission/31474937

I've got a tiny brute force solution for Div1D High Cry here.(AC with 62ms) http://mirror.codeforces.com/contest/876/submission/31465112

Can someone explain their approach (in detail) for Div1 D 875D - Ор выше гор ? I am not able to understand the editorial.

Well, E can also be solved without using graph SOLUTION LINK: http://mirror.codeforces.com/contest/876/submission/31502316

I have O(n.log(1e9)) solution for 875E - Курьерский клуб

check my solution here : 31480382

.

875C When we are comparing adjacent strings.

1.If s i, k > s i + 1, k, you capitalize s i, k and not capitalize s i, k + 1.
2.If s i, k < s i + 1, k, both these letters should be capitalized or not capitalizes simultaneously.

I have doubt for 2nd case : there may be solution for s[i][k] is capitalized and s[i+1][k] is not capitalized.

Another way to solve DIV1 C. First for each consecutive pair of strings, If first string is prefix of first then it's already sorted, else if second is prefix of first then answer does not exists. Now in last case, first index where first and second string differ is the only pair of character on which order of these two strings depend. For string number i-1 and i, character on i-1 should be less than ith one. we can build a directed graph, denoting these pairs as edges, where each edge goes from character which should be smaller than where it goes to. Finally perform topological sorting on the graph, travel in reverse order in topo sort, for each node check each of the outgoing edge should go to the character smaller than current one, if it doesn't then capitalize the current character and check again. if it doesn't satisfy after capitalization then there is no answer. Once you have checked for each node in the sort, perform the changes in the original string you did while iterating and check whether all of the string are sorted or not. If any pair fails print "No" else print the answer. For Capitalization sign can be reversed, if sign is different then character which should come first should be -ve else use absolute. 112439293