Can we do div2B in O(n)?

Написал решение как у автора по задаче С тл 18

Div2B is a beautiful problem!

Typo in the tutorial of 1137D — Cooperative Game.

"Let r denote the distance from finish vertex to the one fast and slow met."

The r here should be x.

It seems too many points for Cooperative Game problem. It is just the search a cycle in a list. Very known problem!!

I have submited a solution of Museums Tour problem. It has Run Time Error on test 16 with try http://codeforces.net/contest/1138/submission/51133791. I very carefully look at my code and don't understand what problem is. Can tell me what to do to find error.

Another solution for F with treaps:

Lets have an implicit treap of the order of the vertices. Initially we build it in by simulating the burning process with heap or set.

"When" and "Compare" queries are now easily done in by simply finding the position of a given node (we keep pointers for each vertex id to its corresponding node in the treap).

For "Up" query we notice that we need to extract nodes on the path from the last updated vertex (initially assume this is n) to vertex u, then reverse their order and finally insert this new reversed subsequence to the end of the treap.

The problem is that we are not sure that these vertices are consecutive in the treap. But we can prove that the number of consecutive segments is amortized . The proof is similar to the one for link-cut tree's complexity. So now we have a straight forward by doing binary lifting and "when" query for finding the consecutive segments in the path. I had this solution during the onsite version, but unfortunately it's too slow. Fortunately we can do this finding in by going through the treap and keeping "next" and "prev" pointers in each treap node. This way the complexity is and without optimizations passes in ~3 sec.

Here is the code if someone is interested: 51142776

Also if I'm not mistaken by replacing the treap with splay tree, the complexity will become just .

If you use Google properly, you may solve div1D in one minute.

I don't think problems like this should appear in Codeforces.

I spent my time and effort just to find that my friends all used Google to solve it(they told me after the contest).

It's not fair.

Can someone describe solution for Div1C. "Museums Tour" in more detail? I can't understand why my submission got WA 79 on this problem.

HELP! Why I got RE on test 73? Submission: 51210282

I wonder why my div1C code gets either TLE18 using array analog link list to storage edges, or MLE16 using vectors :( TLE18: https://codeforces.net/contest/1137/submission/51214068 MLE16: https://codeforces.net/contest/1137/submission/51253929

simple o(n) solution for the B problem we need to check for 3 formulas

c= clown(10) b=both(11) no=none(00) a=acrobates(01) Aa=allacrobates Ab=allboth Ac=allclowns Aa=allacrobates
(1) c+a+b+no=n/2; // first condition all the numbers in the first half should be equal to n/2
(2) c+b=(Aa-a)+(Ab-b); c+2b+a=Aa+Ab; // second condition all the clowns should be equal to the acrobates
(3) no-b=n/2-Aa-Ab; //subtract the first equation from the second one to get the third one which is the one we will solve with 

step1 :
first of all use a loop to fix (no variable) in the third equation and find b as the right hand side is given in the third equation now we have no-b=(n/2-Aa-Ab) (no is now known as we used loop for it now find b and check for validation through the 3 equations and b>=0  and save no and b in temp variables then break the loop see the code for more details )  

step2: 
so after finding (b) variable in the same loop and checking if it's valid now use the second or the first formula (number 1 or 2 ) to find c+a as you now found (no and b variables ) put them in this formula  (c+2b+a=Aa+Ab;) for example to find (c+a) you can also use the first formula 

step3: 
after this use another loop (not nested)  (to fix a or to fix c loop in it as no variable in step 1) and to find the other variable a or c now you have (a and c and b and no variables) you have the four variables just print the answer check the link below for the solution link 

https://codeforces.net/contest/1138/submission/51265926 link to submission

Can anyone tell me what's going wrong here?? Problem C Code

How to solve 1137B — Camp Schedule by hashes?

Yes!!! Check below solution. http://codeforces.net/contest/1138/submission/51394343

For DIV1 problem B,Why I WA on Test 71,using the fail array(may like Aho-Corasick automation)to calculate the max same prefix and suffix. What is the hacks in Test71. my code is as following https://codeforces.net/contest/1137/submission/51554217

For Div1.C,why do we have to compute in SCC? We can stop our tour in an arbitary city besides 1.

F :

At the beginning, burns out the leaf

this sentence doesn't really make sense, can you please fix it

Alternate simple $$$O(q\log{q})$$$ solution for div-1 E:

It's easy to see that only the leftmost car added in any add operation matters. Also, whenever type-1 operations are performed, the optimal car always becomes the leftmost (first one).

So now we can basically solve for all segments of queries between type-1 queries independently. Let's look at any such segment. It contains only type-2 and type-3 queries.

Let's maintain a set $$$S$$$ of cars which have the potential to be the optimal cars (min convienence value).

What will this set $$$S$$$ be like? It will contain the first car and leftmost cars added in some type 2 queries. Also, for any two adjacent elements $$$S_i$$$, $$$S_{i + 1}$$$ in this set (when sorted in ascending order of index), we will always have $$$convenience_i > convenience_{i + 1}$$$. (easy to see why this is true). The optimal car will always be the car with the greatest index in $$$S$$$.

How do we maintain this set?

• Type-2 operations: simply add the leftmost added car into this set.
• Type-3 operations: The main issue is that when we perform type-3 operations, the condition $$$convenience_i > convenience_{i + 1}$$$ might be violated for some $$$i$$$, as elements with greater index recieve greater increments.
  
  Observe that, for any $$$i$$$, whenever $$$i + 1$$$'th element is inserted, from that moment onwards $$$convenience_i > convenience_{i + 1}$$$ gets violated only when sum of $$$s$$$ across all type-3 queries from that moment becomes $$$\geq \left\lceil \frac{value_{S_i} - value_{S_{i + 1}}}{S_{i + 1} - S_i} \right\rceil$$$. It's easy to simplify this further and say that the $$${i, i + 1}$$$ pair will become invalid at time $$$T$$$ when $$$\sum_{t = 0}^{T} {s} \geq \sum_{t = 0}^{P}{s} + \left\lceil \frac{value_{S_i} - value_{S_{i + 1}}}{S_{i + 1} - S_i} \right\rceil$$$. With this simplification we can solve this task with a simple priority queue (store this threshold value for all $$$(i, i + 1)$$$ and greedily keep removing stuff during type-3 queries). Note that the calculation of value for any point can be done in $$$O(1)$$$ just like the way in the editorial.

Implementation: link