jzzhu's blog

Codeforces Round #257 Editorial
By jzzhu, 10 years ago, In English

450A - Jzzhu and Children

You can simply simulate it or find the last maximum ceil(ai / m).

450B - Jzzhu and Sequences

We can easily find that every 6 numbers are the same. It's like {x, y, y - x,  - x,  - y, x - y, x, y, y - x, ...}.

449A - Jzzhu and Chocolate / 450C - Jzzhu and Chocolate

We assume that n ≤ m (if n > m, we can simply swap n and m).

If we finally cut the chocolate into x rows and y columns (1 ≤ x ≤ n, 1 ≤ y ≤ m, x + y = k + 2), we should maximize the narrowest row and maximize the narrowest column, so the answer will be floor(n / x) * floor(m / y).

There are two algorithms to find the optimal (x, y).

  1. Notice that if x * y is smaller, the answer usually will be better. Then we can find that if k < n, the optimal (x, y) can only be {x = 1, y = k + 1} or {x = k + 1, y = 1}. If n ≤ k < m, the optimal (x, y) can only be {x = 1, y = k + 1}. If m ≤ k ≤ n + m - 2, the optimal (x, y) can only be {x = k + 2 - m, y = m}, because let t = m - n, n / (k + 2 - m) ≥ (n + t) / (k + 2 - m + t) ≥ 1.

  2. floor(n / x) has at most values, so we can enum it and choose the maximum x for each value.

449B - Jzzhu and Cities / 450D - Jzzhu and Cities

We consider a train route (1, v) as an undirected deletable edge (1, v).

Let dist(u) be the shortest path between 1 and u. We add all of the edges (u, v) weighted w where dist(u) + w = dist(v) into a new directed graph.

A deletable edge (1, v) can be deleted only if it isn't in the new graph or the in-degree of v in the new graph is more than 1, because the connectivity of the new graph won't be changed after deleting these edges. Notice that you should subtract one from the in-degree of v after you delete an edge (1, v).

449C - Jzzhu and Apples / 450E - Jzzhu and Apples

Firstly, we should notice that 1 and the primes larger than N / 2 can not be matched anyway, so we ignore these numbers.

Let's consider each prime P where 2 < P ≤ N / 2. For each prime P, we find all of the numbers which are unmatched and have a divisor P. Let M be the count of those numbers we found. If M is even, then we can match those numbers perfectly. Otherwise, we throw the number 2P and the remaining numbers can be matched perfectly.

Finally, only even numbers may be unmatched and we can match them in any way.

449D - Jzzhu and Numbers

Firstly, we can use inclusion-exclusion principle in this problem. Let f(x) be the count of number i where Ai&x = x. Let g(x) be the number of 1 in the binary respresentation of x. Then the answer equals to .

Now the task is to calculate f(x) for every integer x between 0 and 220. Let fk(x) be the count of number i where Y0&X0 = X0 and X1 = Y1 (they are defined below).

We divide x and Ai into two parts, the first k binary bits and the other 20 - k binary bits. Let X0 be the first part of x and X1 be the second part of x. Let Y0 be the first part of Ai and Y1 be the second part of Ai.

We can calculate fk(x) in O(1):

The problem can be solved in O(n * 2n) now (n = 20 in this problem).

449E - Jzzhu and Squares

Consider there is only one query.

Let me descripe the picture above.

A grid-square can be exactly contained by a bigger square which coincide with grid lines. Let L be the length of a side of the bigger square. Let i be the minimum distance between a vertice of the grid-square and a vertice of the bigger square. Let f(L, i) be the number of cells which are fully contained by the grid-square.

We can divide a grid-square into four right triangles and a center square. For each right triangle, the number of cells which are crossed by an edge of the triangle is L - gcd(i, L). Then, the number of cells which are fully contained by the triangle is [i(L - i) - L + gcd(i, L)] / 2.

f(L, i) = (L - 2i)2 + 2[i(L - i) - L + gcd(i, L)] = L2 - 2iL + 2i2 - 2L + 2gcd(i, L)

Firstly, we enum L from 1 to min(N, M). Then the task is to calculate . can be calculated by the following steps:

  1. Enum all of the divisor k of L and the task is to calculate the count of i where gcd(i, L) = k.

  2. The count of i where gcd(i, L) = k equals to φ(L / k).

Finally, .

If there are multiple queries, we can calculate the prefix sum of , and , then we can answer each query in O(1).

  • Vote: I like it
  • +25
  • Vote: I do not like it

Comments Comments (81)
| Write comment?
»
10 years ago, # |
  Vote: I like it 0 Vote: I do not like it

The picture of div1E failed...

→ Reply
»
10 years ago, # |
  Vote: I like it 0 Vote: I do not like it

(450 B) We can easily find ... !!!!, lol. It is tricky. I spent all my time trying to find the solution using Matrix Exponentiation.

→ Reply
  • »
    »
    10 years ago, # ^ |
    ← Rev. 2   Vote: I like it +1 Vote: I do not like it

    Actually, you would have to notice, through the recursive definition, that for any n > 2, the number Fi = Fi-1 — Fi-2.

    you start with x | y then the third would be y-x

    x | y | y-x then the fourth is y-x-y=-x

    x | y | y-x | -x then the fifth is -x-y+x=-y

    x | y | y-x | -x | -y then the sixth is x-y

    x | y | y-x | -x | -y | x-y then the seventh is x-y+y=x

    x | y | y-x | -x | -y | x-y | x then the eighth is x-x+y=y

    x | y | y-x | -x | -y | x-y | x | y and we have a cycle

    so you would have to choose an answer depending on n modulo 6.

    I agree, the explanation lacks some insight, I didn't understand the solution for div 2 D.

    → Reply
  • »
    »
    8 years ago, # ^ |
    ← Rev. 2   Vote: I like it +1 Vote: I do not like it

    I put an alternative explanation, in more perspective of Math, on why we have a cycle of 6. And yes, it was bugging me so much, so I decide to review my textbook write something about it :)

    → Reply
»
10 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Do we have some reasoning why Jzzhu and Apples solution is always correct ?

→ Reply
  • »
    »
    10 years ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    By the algorithm, all numbers except 1 and prime P > n / 2 (and maybe one left) get matched. And that's the maximum number of match.

    → Reply
    • »
      »
      »
      10 years ago, # ^ |
      ← Rev. 4   Vote: I like it 0 Vote: I do not like it

      yeah, but how do you ensure that all the remaining numbers of the form 2P can be matched?, i mean why this quantity (the remaining numbers) is always an even number (for match all these), i mean it might be that one number get unmatched from all these

      → Reply
      • »
        »
        »
        »
        10 years ago, # ^ |
        ← Rev. 2   Vote: I like it +5 Vote: I do not like it

        Yes, just let this number unmatched.(The quantity isn't always even!)

        → Reply
        • »
          »
          »
          »
          »
          10 years ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          yes, but then we need a proof to say that this is maximum, because ysymyth told that all numbers are matched (i refer to all numbers but 1 and primes larger than N/2)

          → Reply
          • »
            »
            »
            »
            »
            »
            10 years ago, # ^ |
            ← Rev. 3   Vote: I like it +6 Vote: I do not like it

            Well, if it is even, so all numbers are matched. If it is odd, necessarily at least one number must be discarded and the strategy discards only one number, so it's also maximum.

            → Reply
            • »
              »
              »
              »
              »
              »
              »
              10 years ago, # ^ |
                Vote: I like it 0 Vote: I do not like it

              thanks

              → Reply
              • »
                »
                »
                »
                »
                »
                »
                »
                10 years ago, # ^ |
                ← Rev. 2   Vote: I like it 0 Vote: I do not like it

                Also remember that the discarded number is always even, this way we make sure that at most one number is discarded for every prime and that every discarded number can be paired with every other discarded number.

                So in the end we match all multipliers of a prime not yet matched and if we must discard a number we know that number will not be a waste.

                Among the discarded numbers all can be matched, this leaves at most one unmatched discarded number at the end (in this case, every other discarded number has been matched).

                This strategy is the best because if we must discard a number when looking among the multiples of a certain prime it makes sure that it can be matched in the end and won't be a waste, if it must be a waste it will be the only waste.

                → Reply
»
10 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Can anybody help me on 450D — Jzzhu and Cities, I think my algorithm can work,I use spfa to record which edge update the dist, and finally if it is not this train update the final dist,I can delete the train.But got TLE on 30..Can not understand why.7182221

→ Reply
  • »
    »
    10 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    maybe you can try "slf"

    → Reply
  • »
    »
    10 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    You can also use Dijkstra

    → Reply
    • »
      »
      »
      10 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      I know..But I don't know why I got TLE.

      → Reply
      • »
        »
        »
        »
        10 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        I heard someone hacked by using special data.It makes spfa TLE.You know,when hack succeed the data will be added to test.Sorry for my poor English.XD

        → Reply
        • »
          »
          »
          »
          »
          10 years ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          But I think this algorithm works well..I mean time complexity is alomost O(n).

          → Reply
          • »
            »
            »
            »
            »
            »
            10 years ago, # ^ |
              Vote: I like it 0 Vote: I do not like it

            According to Wikipedia, "For a random graph, the empirical average time complexity is O(|E|)"
            One can make SPFA works very slow with a special graph.

            → Reply
            • »
              »
              »
              »
              »
              »
              »
              10 years ago, # ^ |
                Vote: I like it 0 Vote: I do not like it

              "Empirical complexity" always make me laugh :P.

              → Reply
        • »
          »
          »
          »
          »
          8 years ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          if dijkistra gives TLE.
          if SPFA was hacked.
          if bellman Ford gives TLE.
          How to solve 449B

          → Reply
      • »
        »
        »
        »
        10 years ago, # ^ |
        ← Rev. 2   Vote: I like it 0 Vote: I do not like it

        use ajency list, not matrix, then you do not get timelimit

        → Reply
        • »
          »
          »
          »
          »
          10 years ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          What I use is adjacent list not matrix...

          → Reply
  • »
    »
    10 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    If spfa means shortest path faster algorithm, that isn't linear... is O(k*|E|), which worst case is O(|V|*|E|), if I'm not wrong.

    → Reply
»
10 years ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

For 450E I wrote almost correct code(#1 submission in Div2) except I chose the one number to discard(for each P) between a couple of high numbers instead of 2P and got exactly one pair missing in Test #12.. :P Submission in Py2

→ Reply
»
10 years ago, # |
← Rev. 3   Vote: I like it +1 Vote: I do not like it

The last equation seems wrong. "gcd(i,L)" should be "2*gcd(i,L)".

→ Reply
»
10 years ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

DIV1.B “A deletable edge (1, v) can be deleted only if it isn't in the new graph or the in-degree of v in the new graph is more than 1,”

2 1 2
1 2 10
2 5
2 5

you should output "2", but the right answer should be "1". Did I misunderstand that?

→ Reply
  • »
    »
    10 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    When you delete an edge, you should also delete it in the new graph, so the in-degree of 2 in the new graph is 1 when you are deleting the second deletable edge (1, 2).

    → Reply
»
10 years ago, # |
  Vote: I like it -13 Vote: I do not like it

In this contest when i tried to open codeforces.com I directed to some other site which has same name as codeforces.com but it contain some information about management,health insurance and some other irrelevant thing. It happen with me often and it only happen during contest and waste my valuable time during contest. This problem reside for some time and automatically resolve. Any solution,,,??

→ Reply
  • »
    »
    10 years ago, # ^ |
      Vote: I like it -7 Vote: I do not like it

    Please can you post a more detailed analysis for div2 D ?

    → Reply
»
10 years ago, # |
  Vote: I like it 0 Vote: I do not like it

1)"We add all of the edges (u, v) weighted w where dist(u) + w = dist(v) into a new directed graph." I don't undestand it. It would be nice if there was an example.

2)Can somebody help me optimize my algo or say that it's wrong?

→ Reply
»
10 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Can someone please take his time to explain problem C div2. The editorial is just meaningless. And thank you.

→ Reply
  • »
    »
    10 years ago, # ^ |
    ← Rev. 3   Vote: I like it +5 Vote: I do not like it

    how can you say it's meaningless? it explains everything you need to know...

    But if you still think it's too hard to understand the editorial, focus on the 2nd way of solving it, as it's simpler to understand: Let x be the number of times you'll divide a row or a column. you just have to go through all the values of x: from 1 to and from 1 to to see which gives the highest answer, because there are only different divisors of any number N. So if you check, at each step, the answer for the following cases:

    (m / x) * (n / (k - x + 2)) // do x = m/x and recheck (m / x) * (n / (k - x + 2))

    (m / (k - x + 2)) * (n / x) do y = n/y and recheck (m / (k - x + 2)) * (n / x)

    Just watch out for corner cases that can give you a wrong answer (x > K+1 or x <= 0).

    → Reply
»
10 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Could not understand Div2 D. Someone please explain.

→ Reply
  • »
    »
    10 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Solution is not very difficult. We can solve this problem by using Deikstra's algorithm with heap:

    0) array dist[n] — dist[i] is the value of the shortest path to city i from capital

    1) Dist[1] = 0 — 1 is a capital from statement. Then we should update all distances by using railways, mark "true" for ich city that it was improved by railway in some boolean array ( for example ok[n] ).

    2) Forget about railways and start Deikstra's algorithm with heap. Add all dist[i]!=+inf in a heap and than try to improve our distances by usual roads. But there are some tricks. We should set ok[i]=false in two cases:

    We have an road from v to u with weight s. We try to improve dist[u].

    a) dist[v]+s<dist[u] — this means that using railway is bad as the distance with using railway is bigger.

    b) dist[v]+s==dist[u] — this means that it doesn't matter to use the railway or to use the road. So we don't need in a railway and we can destroy it.

    3) After finishing the Deikstra's algorithm. Calculate the nomber of ok[i]==true. (K — count) will be the answer. 7188851

    → Reply
»
10 years ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

Why the output of the Div-2 D is 2 for the following Input:: Input ::

5 5 3 1 2 1 2 3 2 1 3 3 3 4 4 1 5 5 3 1 4 5 5 5

→ Reply
»
10 years ago, # |
  Vote: I like it 0 Vote: I do not like it

In div 2 problem C editorial, it is mentioned that floor(n/x) can have at most 2*sqrt(n) values can anyone explain it? But according to me x can go from 1 to n, so the total values of floor(n/x) can go from 1 to n .

→ Reply
»
10 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Can somebody help me with problem B of Div 1. I implement the solution of editorial but i get WA on test 5 and i can't debug the code. I think it's absurd. Here is the code: My submission

→ Reply
»
10 years ago, # |
← Rev. 5   Vote: I like it 0 Vote: I do not like it

Sorry, wrong idea

→ Reply
»
10 years ago, # |
  Vote: I like it +3 Vote: I do not like it

I think a simpler way to approach div.2 D/ div.1 B is not by creating a new graph.

One can run Dijkstra's algorithm on the graph single-sourced from node 1, where the heap will store (and sort by in order) : current distance, is train track, and current node. This would guarantee that all the optimum paths without using a train track will be processed first.

After that, we compare the given train tracks to the shortest path and delete off as much as we can.

My practice submission link: http://mirror.codeforces.com/contest/450/submission/7197600

→ Reply
  • »
    »
    9 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Your idea is awesome and the implementation is also beautiful :D

    → Reply
  • »
    »
    5 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    For Div1B my approach (Wrong answer):

    Use Dijkstra. Equal distances are compared on the type of edge. That is, if we can reach somewhere via a road, and some other place via a train route, traversing the same distance, then the road is selected before the train. This is done by assigning a type 0 to roads, and 1 to train routes. Thus, train routes are processed only when necessary.

    Next, I make a set of train edges. We know when a edge is permanent in Dijkstra, when a vertex is popped out of the priority_queue. We can save the edge type, while insertion is done into the priority_queue. Now, if an edge of type 1 is popped out, then we know that this train route is used in the graph, therefore, it is removed from the set.

    In the end, the size of the said set is returned.

    My submission

    Your modified code (to check my Dijkstra, which seems correctly implemented)

    → Reply
  • »
    »
    5 years ago, # ^ |
    ← Rev. 2   Vote: I like it 0 Vote: I do not like it

    Hi, can u explain use of tot[] in your code using an example? Also, is there a case possible where d<=dist[a] but used[a]=1 ? bboy_drain

    → Reply
  • »
    »
    8 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Check out my blog for a well commented and named code: https://mirror.codeforces.com/blog/entry/128562

    → Reply
»
10 years ago, # |
  Vote: I like it 0 Vote: I do not like it

can you offer the offical source code of the question?thanks

→ Reply
»
10 years ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

Can anyone explain the formula in Div1-D solution ?

About its meaning and writing :)

→ Reply
  • »
    »
    10 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    You can look at my code if it helps . I have done exactly according to the editorial.

    → Reply
    • »
      »
      »
      10 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Sorry I wanted to mean "How can i write this formula to a blog or a comment ?". Anyway, actually i want to learn explanation of the formula. Sorry for bad English.

      → Reply
      • »
        »
        »
        »
        10 years ago, # ^ |
        ← Rev. 4   Vote: I like it 0 Vote: I do not like it

        The numbers of i where Ai&x = x is f(x)

        so if you pick from these f(x) numbers

        you will get 2^f(x)-1 groups

        whose result y will satisfy y&x = x

        and it's clear that y has same or more bit to be 1 compare to x

        Its clear that groups whose result have at least 0 bit to be 1

        will contain all groups whose result is 0

        so we add it to our answer

        but it also contains some groups whose result has some bits to be 1

        so we minus those groups which have at least one 1

        but then we will minus too many times for groups have two or more bits to be 1

        so we should add them back

        but for those have three we add them too much ...

        so we just keep add back and minus back

        until the numbers of bit is too many for Ai in this case is 20

        sorry for poor English

        → Reply
»
10 years ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

For 449D-Jzzhu and Numbers

I think in the formula after the sigma should be (-1)^g(x)*(2^f(x)-1)

since you have to choose a number at least

→ Reply
»
10 years ago, # |
  Vote: I like it +5 Vote: I do not like it

My friend make problem E into noip practice.

→ Reply
»
10 years ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

I implemented Jzzhu and Cities like this, but I've got WA on test 5. What's my mistake?

→ Reply
»
10 years ago, # |
  Vote: I like it 0 Vote: I do not like it

The picture of div1E failed...

→ Reply
»
10 years ago, # |
  Vote: I like it 0 Vote: I do not like it

The picture of 449D failed.

Like this:

Then the answer equals to

Unable to parse markup [type=CF_TEX]

→ Reply
»
10 years ago, # |
  Vote: I like it 0 Vote: I do not like it

in Div2 C, how is the answer floor(n/x) * floor(m/y)...

How this floor(n/x) came into picture?

Could someone please explain.

Thanks :)

→ Reply
  • »
    »
    9 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    To maximize dimension of pieces, we must divide choco such that row and col of each piece nearly equal. So we divide n to x choco rows and m to y cols, and take floor because of minimum area

    → Reply
»
8 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Can I use ternary search (on number of horizontal(vertical) cuts for problem C div 2 ?

→ Reply
  • »
    »
    8 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    you need to minimize the denominator not maximize, i blindly coded ternary search without thinking and then regretted it. :/

    → Reply
»
8 years ago, # |
← Rev. 4   Vote: I like it 0 Vote: I do not like it

Sorry, I made a mistake.

You are right.

→ Reply
»
7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

In div1, please anyone explain me why for each right triangle, the number of cells which are crossed by an edge of the triangle is L - gcd(i, L) ?

→ Reply
»
6 years ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

In Div1 D, how to use inclusion exclusion to get the answer as

as given in the editorial.

→ Reply
  • »
    »
    6 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    The total number of possible sets with given n numbers is (2^n — 1). Now, we need to remove sets which have at least one bit set after the & operation of all elements. That is why we subtract 2^f(x) for 1,2,4 etc. But, we have over subtracted the sets which end up with two bits set i.e., subtracted them twice (or more). We need to add them up, hence adding everything with at least 2 bits set. This is regular inclusion exclusion.

    → Reply
  • »
    »
    5 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    raja1999 could you pls explain me as well ,how to get this formulla

    → Reply
»
6 years ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

The idea of Div2 D/ Div1 B given in the editorial is very abstract. Here is how I solved the problem. Firstly add all the roads to the graph, then add all the routes one by one removing the routes which go towards the same city but have a larger weight(longer distance). In this process we may even end up removing some of the roads as well which have a larger weight (this won't affect the answer), so we need to be careful while incrementing the count of removed routes. After this run Dijkstra's algo to find shortest path distances of all cities from the capital. After this for all the routes that still are not removed, iterate over them one by one and for every train route that goes to city v, check to see if the train route has a greater length than the shortest dist to v. If it does, simply remove it and increment count, otherwise if it has an equal length to the shortest dist, try to see if an alternate path can be found to city v that has the same length. For this, iterate over all neighbors of v(not including the capital) and check to see if (shortest dist. to u) + (weight of edge {u, v}) == shortest dist. to v(or the route length). If this condition holds true for some vertex v != the capital, then we can remove the train route and increment count. Since we need to perform deletion operations on the edges, the adjacency list can be implemented as an array of sets of pairs of (neighbor index, weight). Here is the implementation.

→ Reply
»
5 years ago, # |
  Vote: I like it +3 Vote: I do not like it

Please explain 449D. I really can't get my head around the inclusion and exclusion explanation.

→ Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

What is wrong with this solution for Jzzhu and Cities? Please help me.

→ Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Here is my solution for 449B: 117809879

It follows what the editorial says but with two tricks:

  1. You don't need to actually creata a new graph: you can keep p, which we normally use to track the parent in the shortest graph, as the counter for in-degree in the new graph. p[v] will be increased if you find a new shortest path that has the same length as the current one (d[u] + len == d[v]); and if you find a shorter path, reset p[v] to 1.
  2. And do remember to purge obsolete items in the queue (if (d[v] != INF) q.erase({d[v], v});), otherwise you will get TLE (117809032)
→ Reply
»
18 months ago, # |
  Vote: I like it 0 Vote: I do not like it

209956322

Div2 C, Here is the second solution that was mentioned in the Editorial.

→ Reply
»
2 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

Can anyone explain how the formula in Div1-D works?

→ Reply
»
2 hours ago, # |
  Vote: I like it 0 Vote: I do not like it

I didnt understand A at all

→ Reply