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Codeforces (c) Copyright 2010-2024 Михаил Мирзаянов
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As for D, there is a simple O(N) solution. First of all, let's forget about a and b less than 0 We can observe that score is increased on each integer as many times, as this integer appears a divisor of some other integer. This happens floor(n / i) — 1 times for number i. Then we can see, that we have a mirror situation with a and b less than 0. And with a < 0, b > 0 and a > 0, b < 0 works the same. That's why answer is sum of i * (floor(n / i) — 1) for i from 2 to n multiplied by 4.
Here is my solution 45732442
Why every "i * (n / i — 1)" can be choosed?
eg: 6->2, -6->2, 6->-2, -6->-2 in all the above case 3 is the divisor of the integer 6 so suppose n=18 then 3 appears as divisor for 6,9,12,15,18 so the score becomes 3(|x|:divisor) * 5(floor(n/i)-1) * 4(for a,b +ve 0r -ve cases).
Thank you!给你花花！
A can be done by an O(n) algorithm by just traversing once along the array
In problem D.
Could someone elaborate more on the proof that all the nodes are in the same component?
Set of numbers that can transform into a number x is always even (if a can transform to x, -a can either), so the graph built by connections that based on two numbers which is transformable to each other or not will have vertices with even degree, so the graph is Eulerian, which means you can travel through all the vertices without coming back to previous vertices. And so all the vertices must be in a same component.
I guess.
Suppose you transform a into b and |a| < |b| then x = b / a right? It's clear that a, b and x are in the same component. You can see that |b| ≤ n and |a| ≥ 2 so |b / a| ≤ n / 2, which means you are able to transform x into 2x as well, that makes x, 2 and 2x belong to the same component. So for every node that has at least an edge attached to it, we know that node is in the same component as 2.
O(sqrt(N)) code for D: https://codeforces.net/contest/1062/submission/45742622
Cool solution, that uses the same idea as my, but faster
fascinating solution :)
can you please explain it? especially the line below:
https://codeforces.net/contest/1062/submission/45724128 It's this solution but optimized. The answer is: for each i, for each multiple of i, sum j / i as can be seen from this code. N / i is the number of multiples and you can take that sum in O(1). Now, just group the i's by the value of N / i.
Can someone share their code for problem E with same solution as the editorial?
My implementation is almost the same as the editorial.
One difference is that my range queries return the two smallest and two largest dfs-order labels in [l, r]. We will either delete the smallest and use the LCA of the second smallest and the largest, or we will delete the largest and use the LCA of the smallest and second largest.
The editorial instead suggests to make additional queries on [l, u) and (u, r] to simulate deletion of node u.
You can find my solution here. It's the same approach as the author's, provided a quick explanation to the thought process commented above the code.
If you have any questions, please let me know :)
Thank you very much! Your clean code helped me learn and implement lca and solve this problem.
You're welcome! Glad i could help :)
Can someone share their code for problem D with same solution as the editorial?
following.
Codefores is the best!!!
Am i the only one who solved D in less than 10 lines?
Me too, and it was a O(n) solution.
There is an O(n) complexity algorithm for A. Since we need to check for the longest set of increasing difference-by-1 consecutive elements in the array, we don't need to use all pairs i,j. It can be done by for i in size of array, if a[i]==a[i-1]+1 and a[i]==a[i+1]-1 then we increase the delable_currentsize_var by 1; else (means we find the barrier of the increasing set) we update max_delable_size = max(max_delable_size, delable_current_size) and reset the delable_currentsize_var to 0.
https://codeforces.net/contest/1062/submission/45727163
An alternative solution to problem F:
First, find any topological order of the DAG.
Observation 1: A city is important if and only if all the cities before it can reach it and all the cities after it can be reached from it.
Observation 2: In a DAG, vertex v can be reached from any other vertex if and only if v is the only vertex to have no outedges.
From 1 and 2, we can find which cities are important(by adding the vertices in the topological order and maintain a set of the cities with an outdegree equal to zero). This is O((m + n) log n) and with some tricks it can be done in O(m + n).
Now, let's focus on semi-important vertices:
Observation 3: Assume that v is a vertex without outedges in a DAG. Then v can't be reached from exactly 1 city if and only if:
So through restoring the vertices with exactly one outedge and how many times each vertex is pointed to by these vertices, we can easily deal with semi-important vertices in O(m + n).
Submission: 45770465
Correct me if i'm wrong. The topological ordering is not so important with two node that can't reach each other, right?
So with test case 3,
5 4 1 2 3 2 2 4 4 5
if i change the topo order before we apply solve(tsq, G), particularly node 3 and node 1 ( swap(tsq[3], tsq[4] ) ), your code will give answer 4, instead of 5, which is WA.
Swapping node 3 and node 1 should be swap(tsq[0], tsq[1]).
Nice solution. I have found a similar one, but mine is a little bit more complex.
What is the intuition behind C's solution? Someone explain please.
Solve the problem by hand for some cases, and you should realize that each time the numbers keep doubling. If the whole array is ones, then a number at step i will contribute (2^(i — 1)).
However that doesn't hold for zeros, so you assume everything is ones, then subtract the excess.
it is like when you transfer from binary to decimal and you should take attention that you should take all ones first so if the number is 101001 you can say the number is 111000 * take care that if the number is 111111 the decimal will be 2^6 — 1 and if if the number is 111000 the number is 2^6 — 2^3 so the solution is 2^(all segment) — 2^(number of zero's) hope that is clear for you
Can you help me? What is wrong in this code 45786145 ? I changed all things but even is getting the same result.
45787128
Thank you very much!!!
There is a dynamic programming approach that independs of pots of 2.
Let n be the number of 1's in the interval. Now you can have two processes: 1) the total deliciousness of the n pieces will be the deliciousness of the n-th piece plus the total deliciousness of n - 1 pieces. 2) the deliciousness of the n-th piece will be the total deliciousness of n - 1 pieces plus 1.
Let x be the total deliciousness for the n pieces, and m be the number of 0's in the interval. Then the answer for these guys will be x * total deliciousness of m pieces (since, instead of starting with 1, these guys are starting with x)
if anyone solved problem D by editorial approach, please provide the code.
is the tutorial for E has a error? it should be out[u]=max(out[v1],out[v2],...),or my understanding of in[] is wrong?
"a child of u" includes all the direct and undirect childs of u. Apparently, inu = max(inv1, inv2, ..., invk) (v is a child of u, both direct and undirect) and inu = max(outv1, outv2, ..., outvk) (v is a DIRECT child of u) are basically the same.
Thanks!
In second sentence in E we should have the reverse implication i.e inu ≤ inv ≤ outu implies u is an ancestor of v, instead of u is an ancestor of v implies inu ≤ inv ≤ outu which is written, since this is the direction used later in the proof.
You're right. It should have written as u is an ancestor of v if and only if inu ≤ inv ≤ outu
Can someone explain me why in the 7 pretest of A problem answer is 2 but not 4?
Okay, I just read the task one more time ;D
The statement says "the maximum_ number of consecutive elements" not "the total number of consecutive elements".
Can someone explains why the complexity of the final step in tutorial F is O(N+M),please? (......After that we pop those nodes from the stack (or whatever), mark them as not visited and continue to iterate(迭代) to si−1.) (......To calculate in we reverse the directs of the edges and do the same. Because each node is visited 1 time by nodes on (P) and at most 2 times by candidate nodes ) in such a way why some nodes won't be visited more than 1 time?
Assume we have: 1->2->3->4 1->5->3 2->6->4 6-> a set of nodes S1 5-> a set of nodes S2
Let go backward from 4 to 1. First, at 4, because R_6 points to 4 so we start a search at 6, each node in S1 is iterated one time. After that, we reach 3, because R_5 points to 3 so we also start a search at 5 and each node in S2 is iterated one time. There maybe a chance that nodes in S1 are revisited (there is an arc from 5 to 6). Okay so let's deal with the worst case here: nodes in S1 are visited the second time. Then we reach 2, we start a search that goes from 2 to 5 and to all nodes in S1. So that's the third time we visit nodes in S1. The important thing is, this time we mark all nodes in S1 as visited and we will never visit them again. Therefore, each node is visited at most three times, one time by the nodes lie on the longest path and two times by the candidate nodes.
thank you bro ;D
Can Anybody explain why I am getting WA on Test Case 15 ?
My Code