yes, just go to the problem section i.e.here and click on any tag written along with the name of the problem.

simpler solution: — just google it... like "codeforces problems tag"

Go to http://codeforces.me/problemset/tags/tag1,tag2,..,tagn. So to see all problems about binary searching and graphs, go to http://codeforces.me/problemset/tags/binary%20search,graphs

or use this link: LINK ( I don't know does it have all tags or no )

why some people downvoted him? he just asked a question :(

// now he is getting upvotes :))

Because this contest is unrated :V , I think.

You can find all unlucky years in log(r)^2, then just sort them and find maximum length between all pairs of neighbours in sorted array. Also check for segments starting in l, and ending in r.

for each edge find when its in the graph, after that something like this

Which problem did you plan to use Persistent IT on ?

You can use SQRT decomposition to answer the queries rather than using complicated Persistent IT

What is IT ? Interval Tree ?

Calculate all the distances from Alice and Bob, and find the node where the distance from Alice is maximum and greater than the one from bob, multiply that distance by 2 to get the number of moves.

Notice that Alice will walk down the tree to wherever Bob is, and that the number of steps it will take is just the depth of that path in the tree. That means that Bob should try to get to a node that has the greatest depth.

If the root node is at level 0, and Bob's node is a level x, the Bob can walk up at most nodes up before choosing to go down the path that has the greatest depth. We can find the depths of the paths using a dfs.

My Submission: 27589077

Solve SPOJ problem DQUERY first, online using persistent segment trees.

For each i, compute b_i the position of k - th number that equal to a_i to the right (if there's no such position, b_i = n + 1), then for each query, count from l to r how many number greater than r.

PS: My solution

Find out all the possible value of x^a + y^b and then fond the max range between l and r.

You can copy the code of a couple of high-rating coders, run this case and, if their answer is the same, it is very likely that they are correct. My program outputs 16, but it might have a bug.

Answer is 16. Bob will walk down to node 16 and wait there. It will take Alice 8 moves to get there, so the total number of moves is 2 * 8 = 16.

The answer is 16 according to my code.

Взломай точно так же еще раз, мне помогло

Something like:

5 3
1 2
2 3
2 4
4 5

Answer should be 4.

I also got WA at test 8. I found one possible test case below.

10 2 7 1 8 2 1 3 10 4 7 5 3 6 5 8 6 9 7 10

FeelsBadMan

(Got E accepted 10 minutes after the contest).

PS: Btw, is there a fast way to insert image from computer?

I think it was a simple greedy type solution,using DP probably wont help at all,also the constraint were not DP type,I may be wrong though..

yes same problem.

Sorry, this test is removed now. It was added after contest ended but it was wrong.

It was possible if u resubmit until education round 19/20. Now not possible

I submitted your code using GNU C++11, and it was accepted. I don't use C++ normally so I'm not sure why. I hope this helps you discover the problem.

Segment Tree.

(AFAIK) You can get log(n) time per query with offline processing and segment tree. See SPOJ KQUERY problem for reference. (not giving details as you asked only the name of the data structure :P)

You can do it easily with merge sort tree, a variation of segment tree, but time per query there will be (log(n))^2

Start with the easier ones first. You anyway have to do them at some point in the contest. So why not to finish them up asap and then devote your time in the tougher questions.

The rest will be {9, 8, 2, ...} and you can't take 2 after 8.

Happens to me a lot too, I just refresh the page over and over until it works lol