As an experiment the Educational Codeforces Round 33 will be rated for Div. 2. ×

fcspartakm's blog

By fcspartakm, history, 16 months ago, translation, In English,
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16 months ago, # |
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Has anyone tried solving D using a binary search?

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16 months ago, # |
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Is it possible to solve D using ternary search on the number of times we repair the car?

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    16 months ago, # ^ |
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      16 months ago, # ^ |
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      can you please explain your solution ? what is cnt ? and what exactly is the function you are applying the search on ?

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    16 months ago, # ^ |
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    I don't think so...At first I tried to use ternary search,but for the two chosen points the answer may be the same,causing it impossible to calculate the most optimized result...

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      16 months ago, # ^ |
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      Such cases happen if and only if t+a*b==k*b, where the graph of the function is horizontal line. So it works in this problem.

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        16 months ago, # ^ |
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        You are right%%%

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        16 months ago, # ^ |
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        But then what tests cause ternary search to fail? Or is it just other bugs in their code?

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          16 months ago, # ^ |
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          Don't forget to check whether it is better to ride car through all road, because such cases cannot be included in ternary search. Without this checking I got WA on test 15.

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            16 months ago, # ^ |
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            Shouldn't the check for that be

            res=min(res,a*d+d/k*t);
            

            Since we require d/k repairs? I see you have

            res=min(res,a*d+(d-1)/k*t);
            
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              16 months ago, # ^ |
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              There is no need to repair the car when it breaks down at the post office.

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                16 months ago, # ^ |
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                quality: Yes, but i think d%k==0 case will be covered in ternary search (interval; 1,d/k). It is when d%k!=0 that if we cover entirely by car that we will need d/k repairs. Isn't it?

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                  15 months ago, # ^ |
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                  So there is no difference between the two kinds of checking above :)

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    16 months ago, # ^ |
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    Still not hacked: 19492314

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      16 months ago, # ^ |
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      would you mind explaining your solution ? I just don't seem to get the function u r applying the ternary search on. Thanks in advance :D

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        16 months ago, # ^ |
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        The function I'm applying ternary search on is a function that takes and returns the required time if the rest of the distance was walked. Note that when I leave the car I don't need to repair it, that's why I have ret +  = (x - 1) * t, not ret +  = x * t.

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      16 months ago, # ^ |
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      I think your solution may be correct.As quailty says,the results of tho adjacent points may be the same only if t + a * k = b * k.In such conditions your solution could still work correctly. P.S I tried to use random testcases to hack this solution,but in 4e8 tests your solution remains correct...

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    16 months ago, # ^ |
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    16 months ago, # ^ |
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    I think the author's solution is much more simple than applying ternary search. Remember, a is always less than b, so that cuts back a lot of the casework you have to do.

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      16 months ago, # ^ |
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      The only thing that it "cuts back" is one simple if statement: if(b <= a) cout << b * d;

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        16 months ago, # ^ |
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        The thing is you only have to consider driving the car once or drive it almost the whole way, which greatly simplifies the problem. So, you don't have to consider the case where you drive the car between 1 and d/k times :)

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        16 months ago, # ^ |
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        a < b by task statement.

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16 months ago, # |
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Please help... Why the solution of Problem B is working? what is the theory behind it?

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    16 months ago, # ^ |
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    The tutorial is confusing or wrong. You should add cnt[curr]*cnt[a-curr] or curr[a]*(curr[a]-1) if a=b.

    How I would word it:

    You are counting the number of times each number appears in the sequence. Then for each power of 2, p, you look for each number n that appears in the sequence to see if p-n is also there. If it is there, then there are (p-n)*n pairs you can make from those two numbers, except if p-n=n, in which case there are only (n)*(n-1) pairs you can add.

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    16 months ago, # ^ |
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    errr... I think it is pure math... you can see my solution below 19500454

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    thanks a lot.

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16 months ago, # |
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In problem B, we must remember to avoid counting pairs of the same element. If the current value ai, which we process, is the power of 2, we need to add cnt[2j - ai] - 1 to an answer (look at the second test). At the end, we also have to divide our reulst by 2 because we counted all the pairs twice.

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16 months ago, # |
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For D consider the function modulo .Now we want to find .If we fix x modulo k then the function is increasing or decresing,depends on so we can take the min x and the max x and it remains to find value of function at only 2 × k points,the first k points and the last k points of the interval 1..d.

Complexity O(k).

19488570

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    16 months ago, # ^ |
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    stupid:D

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    16 months ago, # ^ |
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    You're totally over complicating the solution. There is a simple O(1) solution.

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      16 months ago, # ^ |
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      It is not important how complicated is solution while i get AC with small number of tries.

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    16 months ago, # ^ |
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    The in your function is it double or integer division please ?

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      16 months ago, # ^ |
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      double,i transformed to modulo ,you need the second form only to observe what points you need to check,you can see that i used the first form in my source.

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16 months ago, # |
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I'd like to explain my solution of F.

  1. Sort all t-shirts by decreasing quality then increasing price.
  2. Sort all customers by their budgets.
  3. For each t-shirts T, find the next t-shirt one may afford if they cannot afford T, call it fail[T].
  4. Call go(all customers, all t-shirts, 0, 0).
  5. Print answer.

And go(customers, tShirts, ans, spent) does the following things:

  1. If customers is empty, return now.
  2. If tShirts is empty, set the answer of all customers to ans and return.
  3. Let C1 be the customers that can afford spent + cost of tShirts[0] and C2 be the remainding ones.
  4. Call go(C1, tShirts[1]..., ans + 1, spent + cost of tShirts[0]).
  5. Call go(C2, fail[tShirts[0]]..., ans, spent).

EDIT: It's hacked.

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    16 months ago, # ^ |
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    it's hacked:D

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    16 months ago, # ^ |
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    Doesn't seem true to me.

    There should be a way where almost all M of customers end up in different recursion calls (we need only t-shirts to do that I guess). And every single one of them can end up taking all of t-shirts except of first ones. So I think there is a test (which shouldn't be hard to create) where your solution is O(MN)

    UPD:

    Not completely correct, I'll try to hack you.

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16 months ago, # |
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For question 3 I used the following n*m approach which timed out at test case 13.Can I improve it? http://codeforces.com/contest/702/submission/19495043 here's the submission code

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    16 months ago, # ^ |
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    Yes, surely you can. You are currently using two for loops that give the complexity O(n * m). You can use binary search to find the closest tower for every city. This reduced the complexity to O(n * log(m)).

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16 months ago, # |
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thank you for problem E , it's so clever

first time to solve something like this

any similar problems ??

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16 months ago, # |
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Problem C using the Two-pointers approach: 19503041

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    16 months ago, # ^ |
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    can you please explain your algorithm.

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      16 months ago, # ^ |
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      it's so simple I guess; all you have to do is find the Maximum shortest distance between two towers and a city in between, so as typical Two-pointers algorithm suggest you have to indices l and r :: l for the cities array and r for the towers array, then you move the r index while city[l] > tower[r] by then we have tower[r] >= city[l] so we check the right distance between them which is tower[r] - city[l] under the condition that r < m and the left distance i.e. city[l] - tower[r - 1] also under condition that r > 0 and get the minimum out of these two i.e The shortest range that the city[l] can be covered with; achieved by the comparison of the distance of those two adjacent towers and then you have to maximize this property as much as possible!

      The technique is simple and very intuitive, I recommend this one too, to sharpen your skill 51C - Three Base Stations

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16 months ago, # |
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Why this has been speed up ( almost half of time ) just by sorting the array ?
19503212 19503179

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16 months ago, # |
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I'm very curious to hear about problem F... :)

It seems that some solved it using treap, right ? It's not something you see everyday ! Can anyone give us some clues on how to solve it ?

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    16 months ago, # ^ |
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    I came up with following method during contest:

    1. First, we can sort all t-shirts and get the order of them. Then, quality isn't important anymore.
    2. Let dp[i][j] denote the number of t-shirt one will buy if he has j dollars and is going to consider i-th t-shirt(in the sorted order).
    3. Then, dp[i][j] = dp[i + 1][j] if j < ci and dp[i][j] = dp[i + 1][j - ci] + 1 if j ≥ ci. (Now, ci is the cost of i-th t-shirt after sorting)

    Then, we can maintain the dp value backward. But, maximum j can be 109. Thus, we need some copy-on-write structure.

    Considering the transition of dp, for i-th dp value. It's actually the concatenation of [0, ci) and [0, 109 - ci] of i + 1-th dp value. Then, we need add more 1 in the range of [ci, 109](lazy-tag can work).

    Thus, it requires a copy-on-write structure which can maintain duplication, move, and range addition. Finally, we can find the answer in dp[1][bi] by querying the 1st dp value in that structure.

    Treap seems to be that required structure. However, I never implemented a copy-on-write treap so far, not to mention a duplication.

    Look forward to easier solution!

    UPD: It seems that LHiC solved it with this method.

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      16 months ago, # ^ |
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      Are you sure that the data structure you described even exists? A treap which has split, merge and duplicate operations is questionable because if you merge a tree and its copy, the priorities are not random anymore.

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        16 months ago, # ^ |
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        Yes, it does exist. A treap with duplication can actually be maintained, but, in a different way.

        When merging two normal treap, one will decide which root of sub-treap will be root by their pre-given random value. However, when duplication, the pre-given random value won't be random anymore. Thus, instead of giving a random value at the very beginning, now, we decide which one will be root by their size in this way:

        merge(treap a, treap b){
          # if a or b is null, do something first
          ret = random() mod (size(a) + size(b))
          if ret < size(a): # root of a is new root
            a.right_child = merge(a.right_child, b)
            return a
          else: # root of b is new root
            b.left_child = merge(a, b.left_child)
            return b
        }
        

        When merging, the probability of choosing the root of each treap is proportional to its whole size. It can be proved that time complexity will also be .

        Also, as I mentioned above, LHiC got accepted with this solution at 19527903.(Although, it's just in time.)

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          16 months ago, # ^ |
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          Thank you

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          16 months ago, # ^ |
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          I can't understand from the code, how do you perfom duplication.

          Could you describe it in more details?

          Thanks in advance.

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    16 months ago, # ^ |
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    I have another solution that also uses a split-merge treap (different than eddy1021).

    The core of this problem is to efficiently simulate the greedy buying process. Let's store all the query values in a treap. Each treap node will store information about the current value of the node, and the number of t-shirts bought by it. When we meet a t-shirt that costs c, everything in the treap with value  ≥ c should be decremented by c and then their number of t-shirts bought should be incremented by 1. Then we should somehow fix the treap so that the keys are in proper order again.

    First, split the treap by keys  < c (treap L) and  ≥ c (treap R). Apply a lazy key decrement and then a lazy t-shirt increment to treap R. Now, we can do something like mergesort. While the smallest value in treap R is smaller than the largest value in treap L, we extract it from R and insert it into L. When no such elements exist, we merge L and R and move on to the next element. At the end, the answers are in the treap nodes.

    The total time complexity is , because we only need to move the smallest element in R into L at most times. To show this, let's consider when an element from R would be moved into L. We know that in the beginning, before the lazy updates, all nodes in R are greater than all nodes in L. After we updated, the smallest nodes b in R is less than the greatest nodes in L, a. Recall that the price of the current t-shirt is c. Before the update, we had a < c ≤ b and after the update we have b - c < a. Combining these, we have b - c < a < c ≤ b, adding c to all terms, b < a + c < 2c ≤ b + c. The important part is b < 2c, since that implies and so the value of b after the update will decrease by a factor of at least half every time we move, guaranteeing us that a node can only move times.

    If you think I made a mistake somewhere, feel free to hack me at 19506886.

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      16 months ago, # ^ |
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      So, you don't actually need to make the comparison with the greatest element in L, just splitting R in  ≥ 2c and  < 2c, and then inserting all the elements in the second treap one by one into L would also give you the same complexity.

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        16 months ago, # ^ |
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        it seems completely strange, since I agree with proof, but I get TL... Someone can tell me where is my mistake? Thnx

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          15 months ago, # ^ |
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          To generate random priorities, you are putting values from 1 to K in a vector and then shuffling it. I did the same and got TLE on test 62. I removed it and assigned priorities randomly, it gets AC. Don't know the reason why putting priorities in a vector gives TLE.

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            5 weeks ago, # ^ |
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            Because random_shuffle() uses rand().

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16 months ago, # |
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Hello ,for D I used binary search. I found how many stops there are at most, and I set two variables l, r as l = 0 and r = stop then I used a boolean toLeft(to hold what was the last direction I headed towards) and then I calculated that time for 'mid' . But my solution got hacked. my submission is here : Can you guys tell me what I am missing ? http://www.codeforces.com/contest/702/submission/19497043

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16 months ago, # |
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What test hacks a ternary search in problem D?

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16 months ago, # |
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I have a different idea to solve E which complexity may be O(n).

Each point has only one edge,we can shrink point. This can be done in O(n).

After this, we get a new graph. The new node is a node from original graph or a circle. And we find that each node in this new graph will enter a circle finally.

Suppose u->v1->v2->(circle), k arcs in the circle can calculate in O(1): convert the n node circle to 2*n,cal the prefix sum.

Now we just need to cal the node not in circle dfs v2 then v1 then u, also maintain the prefix sum. So each node res can be done in O(1)

I have not code it, Is that right?

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    16 months ago, # ^ |
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    what about min ?

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      16 months ago, # ^ |
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      what did he meant by prefix sum?

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      16 months ago, # ^ |
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      You can use sparse table for it I guess.

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        16 months ago, # ^ |
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        I don't think it helps here , even you put nodes in an array in some order , path of size k for one node will not contain a contiguous subarray on it always.

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16 months ago, # |
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On problem B editorial, I think the part "Then we need to make cnt[ai] –  and add the value cnt[ai  -  cur] to the asnwer." would be better as "Then we need to add the value cnt[cur - ai] to the answer and increase cnt[ai]". I don't know if I'm correct, but it was the way I understood the solution. I'm considering cnt[x] starts empty before iterating. Maybe you meant the oposite, and cnt[ai] should be decreased before adding cnt[cur - ai] to the answer, but it was not clear from the editorial statement.

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    16 months ago, # ^ |
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    I believe you're right. Because i < j, so when we're working with a[j], we'd better increase it last

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    16 months ago, # ^ |
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    Yes, it is not clear from the explanation, but the phrase "Then we need to ..." implies to some extent that cnt is filled in beforehand. Nonetheless, there is a typo in the explanation: cnt[ai-cur] should be cnt[cur-ai].

    PS: I couldn't come up with a way to make bold, italic and mathematical expression at the same time. Can you help me with that?

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16 months ago, # |
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How to solve 702C — Cellular Network use two pointers in linear time?

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16 months ago, # |
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I wrote solution to problem B using two pointer technique(C lang),if anyone is interested, check it out. http://codeforces.com/contest/702/submission/19515728

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    16 months ago, # ^ |
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    can you please explain your algorithm.

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      16 months ago, # ^ |
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      Hello, I first store all powers of two in an array and then for every of its members i check how many different combinations add up to that number.If for example i have sorted array that contains these elements " 1 2 3 4 5" and i want to find two numbers that give number 5(3+4) i would do it like this : First i put "pointer" on last element and on first element and then you ask if their addition is bigger then the value you are looking for,in this case 5+1>2+3 so we decrease our right pointer(meaning that new value has to be lower then the previous one), in case the value is smaller we shift our left pointer to the right untill we get to 2+3.Now the way i solved it is that i stored all numbers in a set and when i get for example to some result i just multiply the amount of numbers there are from left side and right side, for example 1,2,2,3,3,4,4,4 (4*2 = p.o2) i just do 2*3 = 6 because there are 2 (twos)and 3(fours).Now the problem comes if both are power of two for example i look for number 8 and i have this 4 4 4 4 that means that the result of this should be (n-1*(n))/2 which explains my "calc" function...

      I hope i explained enough =)

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16 months ago, # |
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how to slove E?

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    16 months ago, # ^ |
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    The idea is parent jumping : You calculate the answer for the 1st parent, then the 2nd parent using the answer for 1st parent, then the 4th parent answer using the 2nd parent answers, 8th parent using the 4th parent answers etc... You can see my code for more details (it's very short and simple)

    Similar applications of this idea include finding the LCA and IOI 2012 — Scrivener.

    My code : 19509132

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    16 months ago, # ^ |
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    19516878

    implemented editorial!

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(E):amazing question and tutorial!

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How to solve problem D using ternary search?

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    16 months ago, # ^ |
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    You can do ternary search over the number of times the car will be repaired.

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I wonder if we could solve F by taking advantage of the following fact:

Consider that the shirts are sorted by decreasing quality and increasing cost, a shirt must be picked after a shirt that has the LEAST quality among the shirts which has a higher quality and lower cost.

By this, we can build a prefix sum array and a sparse table of minimum value (or BIT, segment tree) of the sorted array to solve the problem. We use binary search on the sparse table to find the first t-shirt that the customer could buy, then use binary search on the prefix sum array to find the last t-shirt the customer could buy in a row until he doesn't has enough money, then use binary search on the sparse table to find the first t-shirt that he could buy with the remaining gold, until there no longer exists a viable shirt for the customer.

However this approach seems to be extremely weak against some testing case like this (sorted in quality): $1,$100,$2,$101,$3,$104 (All customer carries $99) .... I wonder if there exists a way to optimize this approach to take care of such case?

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16 months ago, # |
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Anyone solved 702B(B) with BinarySearch?

1.We have to count 2^x (1<=x<=30) 2.We have to access from idx 1 to N-1 3.Sort ( NlogN , quick sort ) 3.if we choose arr[0](let arr[0] = 30) , we have to search number (number + arr[0] = 2^x in the arr from arr.begin()+idx+1 ~ arr.end()) if we find the number, change the number -1e10. and Search again till the 'number' doesn't appear. if we can't find number, let's go to the next index(idx) 4. Repeat

NlogN (sort) + 30*logN(binarySearch)*N(from idx 0~N-1) = O(NlogN)

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16 months ago, # |
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What does mean the expression " residuary t-shirts " from problem F ?

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16 months ago, # |
Rev. 2   Vote: I like it 0 Vote: I do not like it

SOlved D with smallest solution ( 3 days ago) : solution

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16 months ago, # |
  Vote: I like it +15 Vote: I do not like it

When will the editorial for F be added?

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16 months ago, # |
  Vote: I like it +5 Vote: I do not like it

how to solve 702F — T-Shirts?

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15 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Tutorial for F?