fcspartakm's blog

Educational Codeforces Round 15 Editorial
By fcspartakm, history, 8 years ago, translation, In English
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8 years ago, # |
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Has anyone tried solving D using a binary search?

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    8 years ago, # ^ |
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    WA#15. Also there are some strange things in tests. My answer is OK, even is not same with right answer.

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    8 years ago, # ^ |
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    Yes, http://codeforces.net/contest/702/submission/19497839.

    Seems to work, so far

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    8 years ago, # ^ |
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    I don't have WA on case 15: http://codeforces.net/contest/702/submission/19508679

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    8 years ago, # ^ |
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    Try the function t(x) = time to reach the destination with x repairs. You'll find this function is linear and actually there are two points were an extremum can occur — at x = 0 and near x = d / k. Thus binary search won't work very well.

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      8 years ago, # ^ |
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      It is possible with binary search because we aren't looking for extremums but the single minimum. In the binary search compare two consecutive function values (in the middle of range). If they're decreasing set lower bound to right middle; otherwise set upper bound to left middle. My solution is similar but with ternary search not bs.

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8 years 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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    8 years ago, # ^ |
    ← Rev. 2   Vote: I like it +3 Vote: I do not like it

    Yes) http://codeforces.net/contest/702/submission/19487949 No( Hacked...

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    8 years 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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      8 years ago, # ^ |
      ← Rev. 2   Vote: I like it +6 Vote: I do not like it

      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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        8 years ago, # ^ |
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        You are right%%%

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        8 years 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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          8 years 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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            8 years 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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              8 years 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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                8 years 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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                  8 years ago, # ^ |
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                  So there is no difference between the two kinds of checking above :)

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    8 years ago, # ^ |
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    Still not hacked: 19492314

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      8 years 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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        8 years 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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      8 years 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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    8 years ago, # ^ |
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    Yes ! Not hacked till now :D http://codeforces.net/contest/702/submission/19494286

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    8 years 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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      8 years ago, # ^ |
      ← Rev. 2   Vote: I like it +5 Vote: I do not like it

      The only thing that it "cuts back" is one simple if statement: if(b <= a) cout << b * d;

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        8 years 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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        8 years ago, # ^ |
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        a < b by task statement.

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8 years 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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    8 years 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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    8 years ago, # ^ |
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    errr... I think it is pure math... you can see my solution below 19500454

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    8 years ago, # ^ |
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    thanks a lot.

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8 years 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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8 years ago, # |
← Rev. 5   Vote: I like it +6 Vote: I do not like it

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

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      8 years 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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    8 years ago, # ^ |
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    The in your function is it double or integer division please ?

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      8 years ago, # ^ |
      ← Rev. 4   Vote: I like it +3 Vote: I do not like it

      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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8 years ago, # |
← Rev. 4   Vote: I like it +24 Vote: I do not like it

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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    8 years ago, # ^ |
    ← Rev. 5   Vote: I like it +21 Vote: I do not like it

    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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8 years 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.net/contest/702/submission/19495043 here's the submission code

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    8 years 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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8 years 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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8 years ago, # |
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Problem C using the Two-pointers approach: 19503041

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    8 years ago, # ^ |
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    can you please explain your algorithm.

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      8 years 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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8 years 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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    8 years ago, # ^ |
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    when the 1st element is 1e9 , it's the worst case

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8 years 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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    8 years ago, # ^ |
    ← Rev. 3   Vote: I like it +74 Vote: I do not like it

    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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      8 years 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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        8 years 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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          8 years ago, # ^ |
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          Thank you

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          8 years 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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    8 years ago, # ^ |
    ← Rev. 2   Vote: I like it +116 Vote: I do not like it

    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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      8 years ago, # ^ |
      ← Rev. 2   Vote: I like it +23 Vote: I do not like it

      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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        8 years 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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          8 years 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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            7 years ago, # ^ |
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            Because random_shuffle() uses rand().

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    3 years ago, # ^ |
    ← Rev. 2   Vote: I like it +8 Vote: I do not like it

    I have another solution that doesn't require treap

    Consider we process all the queries together, and for every item, we try to make the customers buy it.

    for every customer, we maintain $$$S_i$$$(the money the ith person currently has).

    let's maintain $$$\log{C}$$$ buckets. for person i having $$$S_i$$$ money, we put him in the $$$\lfloor \log{S_i}\rfloor$$$th bucket later on, I will use the term "person $$$i$$$ goes one bucket down" meaning that $$$\lfloor\log{S_i}\rfloor$$$ will be decreased by one after buying the following item, and that he should be put into the lower bucket.

    now when we see process an item costs $$$V$$$ (let $$$k:=\lfloor\log{V}\rfloor$$$)

    we know that

    1. everyone in bucket $$$j < k$$$ won't buy this item

    2. Some people in bucket $$$k$$$ buy this item, and everyone that buys this item goes some buckets down

    3. all people in bucket $$$j > k$$$ will buy this item, and some of them will go one bucket down

    more specifically, for case 2, for everyone in this bucket that can afford $$$V$$$, will buy this item and goes some buckets down.

    for case 3, we can maintain a tag on each bucket $$$x$$$ which means how much everyone's $$$S_i$$$ is decreased. Then, for every $$$S_i - V < 2^x$$$ should go one bucket down.

    to maintain it, we can simply use a set to simulate a process, since everyone goes down a bucket at most $$$\log{C}$$$ times.

    let $$$Q$$$ be the total number of item, $$$N$$$ be the number of customers

    Total Complexity $$$O(N\log{N}\log{C} + Q)$$$ amortized

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8 years 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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8 years ago, # |
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What test hacks a ternary search in problem D?

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8 years 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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    8 years ago, # ^ |
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    what about min ?

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      8 years ago, # ^ |
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      You can use sparse table for it I guess.

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        8 years 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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8 years ago, # |
← Rev. 2   Vote: I like it +5 Vote: I do not like it

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

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8 years ago, # |
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how to slove E?

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    8 years 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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    8 years ago, # ^ |
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    19516878

    implemented editorial!

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8 years ago, # |
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(E):amazing question and tutorial!

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8 years ago, # |
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How to solve problem D using ternary search?

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

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

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8 years ago, # |
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When will the editorial for F be added?

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8 years ago, # |
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how to solve 702F — T-Shirts?

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8 years ago, # |
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Tutorial for F?

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4 years ago, # |
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Can anyone tell how to solve problem F?

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4 years ago, # |
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Tutorial is not available

......

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