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Codeforces (c) Copyright 2010-2018 Mike Mirzayanov

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Will the tutorial for Div 1-E be available in the future? Edit: it wasn't available just for a few minutes.

can someone who can give proof of C ????

Div 2 C:

It doesn't depend at all if there exists any letter other than a or b in the given string. You can for sure ignore those letters, so the editorial says to erase them. Now, what you have is a string consisting only of a and b's. Also two consecutive b's can be merged as one. So your final string will look something like (a...a)b(a...a)b(a...)...

You can now consider this problem as sum of all possible product of subsets of a given set, where each element in the set is the number of a's delimited by b.

For example: In the string "aaabaabaaab", set formed will be {3,2,3,0} (0 can be ignored). Now if you have a set {a1,a2,...,aN}, then sum of all possible products of this set is equal to (1+a1)*(1+a2)*...*(1+aN)-1.

Proof:

Write the required answer as follows:

S = Sum of products of subset with (size=1)+(size=2)+...(size=N)

S = (a1+a2+...aN)+(a1.a2+a1.a3......+aN-1.aN)+...+(a1.a2.....aN)

After factorization,

S = (1+a1)(1+a2)...(1+aN)-1

Hope it helps..:)

Thanx got it

Hey, can anyone explain why do we need to increase the blocks size by 1? :p (Div2 — C)

Because if we have x 'a' symbols in the block we can pick the first 'a' from this block or the second 'a' or the third or ... or the n-th or we can just not pick any of 'a' from the block.

For a given block you need to decide the number of ways to pick an "a". As it is possible that you do not choose any "a" from that segment we need to add 1 to each segment. Addidtionally you'll be required to subtract 1 from your thus obtained answer as it includes the case where no "a" is selected in all the segments and thus an empty subsequence.

Thanks, it helped a lot!

a + b + ab = (1+a)(1+b) -1; a + b + c + ab + bc + ac + abc = (1+a)(1+b)(1+c) -1 in this fashion we can calculate our ans i.e sum of product of every possible combination

As a comment on problem Div1-B : consider the trie of all strings that made of only 'a' and 'b' chars, It's a complete binary tree of height n. actually we should choose k leave of this tree such that the strings of these leaves be in the range [s,t] and the minimal sub-tree that consists these leaves and the root of the trie is as large as possible the final answer is the size of this sub-tree!

Ok! now we greedily add leaves and compute the answer! (counting,bfs or...) Now we can solve the more general case of this problem! :)) (for example: each string consists lowercase English letters!)

Can someone explain C with dp approach ?

I did using 3 parameters:

i: the current position in the string;

n: meaning if i have or not a 'b' in the current subsegment;

p: meaning if i already formed a valid subsegment.

Basicaly,

if the current caracter is 'a' and i already have a 'b' i can start a new subsegment or continuos a previous;

or, if the current caracter is 'a' and i don't have a 'b', i can just continuos the subsegment;

if is 'b' i set that i have a 'b';

else, i just go.

submission

Thanx

I made it bottom up this way: The state will be the current index, with the guide question: "how many ways to make valid sequences 'til here?".

the answer for dp[i], will be all the ways to dp[i-1], plus, if the current char is 'a', you can start one new sequence here (so, plus one), or extend all the previous valid sequences before the last 'b' (because you must have an 'b' between 'a's), which means that you can extend all the valid ways until the index of the last 'b', that turn out to be dp[index_of_last_b].

solution: https://codeforces.com/contest/1084/submission/46886918

Thanx

Anyone please help me to find the fault in my 46870177 of 1084B - Kvass and the Fair Nut.

Guy, I read your code and your error is make the division integer, if you variables is unsigned long long all your operation with your variables will give a result with this type,backing to the problem when you user (int) will give error when the division gives a large result. My submission 46889415 of your code.

Yes that was a very silly mistake. Thank you.

In Div2 D or Div1 A

Test Case :

3

5 24 24

1 2 16

1 3 14

Although, the answer would be correct by approach described above. But I am interested in calculating answer when the turning vertex is 1. By this approach it will be (8+10+5) where 8 will come from 24-16 and 10 will come from (24-14). But in actual that path is not at all possible. Because if we start from 2 and reach to 1 gas is = 13 but to go to 3 we need 14, cann't go to 3. same will be the situation when we start from 3 and want to reach to 2.

Could anyone explain me why this is not affecting final answer to problem?

If at any moment the sum becomes negative then the path is not optimal since you can remove the negative part, so even if you incorrectly consider it, the answer wont change since there will be a better solution anyways.

Thanks!!

But how can you guarantee that there will always be a better solution ?

Because removing the negative prefix leads you to a bigger sum

Why

`⌊v−(s+n−1)/n⌋`

not`⌊v − s/n⌋ + (s % n) ? 1 : 0`

?That are the same thing.

Can anyone explain in detail how to solve Div. 2 D?(I will be very grateful.)

Google Kadane's algorithm (for finding maximum sub-array sum) and then compare that technique here also. It helps me!

After struggling for some time on this problem, I finally have a proof of the algorithm that the author is proposing. Here it goes (along with the detailed explanation on how to solve the problem):

First, some terminology, that I'll use (I assume that we initially root our tree at the vertex 1): Let's define

the sum on a pathstarting at a vertex u andending at a vertex vas the sum of weights of all the vertices on the path minus the length of all the roads on the path, andthe sum on a pathstarting at a vertex u andleading to a vertex vas the sum of weights of all the vertices on a path (excluding v) minus the length of all the roads on the path.Note:In this, I consider that v is always an ancestor of v.We propose that the answer is simply the maximum of the (sum of the 2 longest paths leading to a vertex v + w[v]) over all the vertices in the tree. In case there are lesser than 2 such paths having a positive sum, we can take the other path(s) as having a sum of 0, they simply include no other vertex except v.

To elaborate more on the paragraph above, we can perform a dfs starting from the root node and since we know that there are only 2 possibilities: either the optimal path passes through the vertex v that we are currently at or it does not. I can compute the answer as the max. of the 2 possibilities. The thing proposed above is the max. sum of a path that passes through the vertex v, the other possibility can be considered by recursively computing the answer over all the children of the vertex v.

This is almost a standard kind of an approach that can be solved by a dfs. Here is my code for reference: 46893082 .

Now, onto the proof of this approach, mainly the thing about how can it guarantee that at no instant the fuel becomes negative. We prove this by contradiction.

Let us assume that our algorithm finds an optimal path in which the fuel becomes negative somewhere in the middle. Let's assume that the path goes from a vertex a to a vertex b and the answer was found at a vertex x. In this case, I can decompose the path (a -> b) as (a -> x -> b). Now, there are 2 cases:

1) The fuel becomes negative on some edge on the path (a -> x): Let's assume that the corresponding edge is from u to v. Now, consider that the total fuel at u was x and the edge weight is y. Now, in this case, the max. sum on the path that gives us our optimal answer is the max. sum on the path (

u->b) + (x-y), and (x-y) < 0. So, following the path from (u -> b) gives us a greater sum than the optimal path, which is a contradiction to our choice of optimal path, and since (x — y) < 0 , our algorithm would not consider this as the longest path ending at v, but it would rather consider some other path having a positive sum or simply the path starting off at this vertex whose sum is 0. This shows that the algorithm would never fall into such a situation.2) The fuel becomes negative on some edge on the path (x -> b): Let's assume that the corresponding edge is from u to v. The same reasoning as above can be applied and we can get a better path (having a greater sum) than the optimal path by starting off at vertex v, and since our final answer is the maximum over all the vertices, it would rather consider that path as the answer as the optimal path. Again a situation that is not possible.

This shows that on the path chosen by our algorithm, we would never run out of a fuel. Further, our algorithm considers for all the vertices the optimal path passing through them and takes their maximum. Thus, our algorithm gives us the correct answer.

Hope it helps. :)

great! it helps a lot!

Nice post! Can you help me out in this? I have used the in-out dp approach to solve this problem. I don't know why it is failing on test case 34. https://codeforces.com/contest/1084/submission/46897634

I don't actually know what's the in-out dp approach that you are referring to. But anyways, I don't think any other person would understand your code better than you do. So I'd suggest you to find out the reason for why it fails on your own. You might end up learning a new thing from that.

Will you please go through my test case just 6-7 comments above. I think that don't come in both of these categories.

Consider case 2 of my proof. Here, x = 1 and b = 3. The algorithm at vertex 1 considers the optimal path to be of value 13 + 10 = 23. This is the only possible path. But on this path, the fuel becomes negative. This happens at the edge (u, v) where u = 1, v = 3. Now, in this case, it is optimal to start at the vertex v = 3, where the answer = 24. I agree that it's an invalid path and the algorithm considers it. But that's exactly what case 2 states, if this would have been a path that our algorithm considered as the answer, the fuel would have become negative on the edge 1 -> 3 and we would have been better off starting off at 3. But yes you're right. Here the algorithm at the vertex 1 chooses a path that is not possible. So, we can't say that the algorithm actually finds a valid path passing through every vertex, but we can guarantee that it finds an optimal path based on the reasoning given by the above 2 cases.

Ok got it. Thanks for going through my test case.

Can someone please explain problem number D of Div. 2 ?

Look at my reply to the above comment.

Why these downvotes ? I just wanted to ensure that he gets a notification in case he doesn't look at this blog post again thinking that he got no reply.

Thank You..!! appreciate your help :)

Thanks for your fast Editorials. <3

Prob div1E: Will a greedy solution work instead?

For each rectangle I define net exclusive area to be: "exclusive" area of each rectangle (which is area not shared by any other rectangle) — a[i].

The greedy solution is to find the rectangle with the lowest exclusive next area and delete it(only if the value is negative) and update the neighbor rectangle values and continue the process.

If we have 3 rects and lowest negative value at middle. There is can be situation when optimal solution is only middle. But your algo will exclude it. For instance: 3 3 7 1 4 3 4 1 3

If deleted rect will be always utmost than it will give correct solution.

"If deleted rect will be always utmost than it will give correct solution." Can you explain this?

On samples where you algo will always delete rect that is utmost among left ones, you will get correct answer. Otherwise result correctness is not guaranteed.

Why do the rectangles in problem E have to be non-overlapping for the solution to work?

They are overlapping, but not nested. If they were, finding area of intersection would not be as easy as

x_{j}·y_{i}.Actually most of solutions will work without any changes. Obviously, optimal answer will not contain nested rectangles, so we can't get worse answer. Also if we left the same formula for intersection, pair of nested rectangles will give less score, than it should give actually. So we will not get the answer, that is bigger than optimal.

Thanks, I hadn't thought of that. The only thing is that

y_{i}will no longer be in decreasing order. The nested rectangles can also be removed in linear time for convenience before doing dp, if I'm not wrong.Oh, I doubt that they can be removed, because they have different

a_{i}.The test for example:

I didn’t say to remove them, just keep the solution unchanged. My argument is that in the solution itself there shouldn’t be any nested rectangles, and the dp formula somehow compensates that by giving a negative score to having nested rectangles. Am I missing something?

EDIT: Never mind, I misinterpreted your last comment.

As I see only reason is that with this assumption simpler to build envelope.

Even as sad in editorial total time "O(n) if rectangles are already sorted".

can anyone explain me this part in Problem C (DIV 2): For each i (1≤i≤k), spi= 'a'. For each i (1≤i<k), there is such j that pi<j<pi+1 and sj= 'b'.

We must create an array 'p' of indices that correspond to the letters 'a'. Between all adjacent elements of this array should be an index corresponding to 'b'

thanks !!!

For Div1E I've wondered that my O(nlgn) solution get TLE for 2sec with n=1e6. Even after using vector instead map and releasing tie between cin and cout.

There is an easier way to solve problem Div2 E.

Let dp[i] represent the number of all distinct prefixes that end at i and are not less than S's prefix or larger than T's prefix, initially dp[0] is 1, and we treat strings as one based.

Initially, dp[i] = dp[i — 1] * 2, because you can expand all previous options by an 'a' or 'b'.

However, this recurrence counts prefixes larger than t and smaller than s, but the over counted strings can be at most 2, which happens if s[i] == 'b', or t[i] == 'a'. but in these cases you can remove them individually. (They can happen when you use S[1..i], and extend it by an 'a', or use T[1..i], and extended it by a 'b')

Once all prefixes are calculated, the answer is the sum of the dp array, but any dp[i] can't contribute larger than K prefixes ( since we need at most K distinct strings), so instead dp[i] contributes by min(dp[i],k)

This approach could also be extended if the question did not limit to using only 'a' and 'b' but had rather used the entire lowercase letters.

Solution: https://codeforces.com/contest/1084/submission/46911068

In 1083E, shouldn't the formula be

dp_{i}=dp_{j}+x_{i}.y_{i}-x_{j}.y_{i}-a_{i}. Otherwise it will just give the area covered by all the rectangles.And what is the convex hull trick? Can someone explain/provide a link.

Never mind found it: Convex Hull Trick

Could someone explain what sequences do I need to count in Div2-C? Are they "aaa..." or" aaa...baaa..."?

i have counted the number of a's in the 'aaa..a' sequences which has 1/more 'b' between them. see my code, you will get it. my code: https://codeforces.com/contest/1084/submission/46881140

Div 2 C.

Scan the string from left to right and maintain two numbers — the number of sequences found so far (

r) and the number of those sequences that have a 'b' above them (t). When encountering 'b' assignrtot, and when encountering 'a' addt+ 1 tor.https://codeforces.com/contest/1084/submission/46883657

Can u explain or proof why this approach works?

When you encounter 'b' all the sequences found so far will have a 'b' above them. When you encounter 'a' you get more sequences with the index of that 'a' as the last element. Those are all the sequences found so far having a 'b' above them with the index of the new 'a' added at their ends, plus a one-element sequence containing the index of the new 'a'.

Can anybody explain div 2 B fair nut's else answer? How did the author reach it?

Can anyone explain div 2E ? In the editorial, I understood till the part where problem is reduced to (k-2) strings and the common prefix of S and T are removed. But after that (from second paragraph), I can't understand much. I have understanding of both tries as well as binary tree. Can anyone help?

Each string is a path in trie. We want to choose

kpathes betweensandtand maximizec, which is equal to number of vertexes in union of them. For every "a" in stringsand every "b" in stringtwe have complete binary tree in trie with some height.In complete binary tree with height

hwe can consider multiset {h,h- 1,h- 2,h- 2,h- 3 (4 times), ...}. It can be proved that maximal union oflvertical ways in complete binary tree is sum of firstlelements of this array, and if we want to choose some number of ways in two trees, we can combine their multisets. Let's store this huge set as arrayp, wherep_{i}is count of elementh-i. For complete binary treeplooks like {1, 1, 2, 4, ...}. We can sum arrays to combine multisets.It's not difficult to sum arrays for

O(n) complete binary subtrees fast.Could somebody help me find out my mistake in my solution?

It fails on the 14th test case

https://codeforces.com/contest/1084/submission/46908137

So basically i have three 1-D arrays up,down and cross. Initially i root the graph at node "1" to form a tree

up[i] holds the value for the maximum gasoline possible ending for an upwards path ending at i

down[i] holds the value for the maximum possible gasoline on a downwards path starting at i

cross[i] holds the value for maximum possible gasoline on a path with i as the turning vertex

Initially up[i]=down[i]=cross[i]=gasoline[i] for all i

up[i] is nothing but maximum over up[i] and up[j]-(distance between i and j)+gasoline[i] for some child j of i

similarly down[i] is maximum over down[i] and gasoline[i]-(distance between i and j)+down[j] for some child j of i

cross[i] is (for all pairs of children j and k): maximum over

1) cross[i]

2) (going from j->k)

up[j]-(distance between i and j)+gasoline[i]-(distance between i and k)+ down[k]

3) similarly go from k->j

What is wrong with my logic? :( I have read a solution below but i would also like to know what is wrong with my solution.I'll be very grateful if you could help me out cause i've been struggling for a while. Thanks in advance

Is the runtime for problem C missing a

Qlog^{2}Nterm? On each query swapping two values, each of thelogNupdates to the segment tree's nodes may involve an LCA lookup inlogN.You can do it in

O(1) using Sparse-Table.Didn't know about that, thanks!

Actually if you store middle node for non-vertical path too, you'll need to compute lca not more than once per query, because it happens only when you merge to vertical paths and in result you get a non-vertical one. So you can still compute lca with binary jumps.

Interesting. How can you perform the case where two vertical paths add to a vertical path in O(1) if you are doing LCA with binary jumps?

If two paths together form a vertical path too, you can say, that it’s endpoints are nodes with the maximum and the minimum depth. After that you just check that all other nodes lie on a path between them.

How can you check that the other nodes lie between them without doing LCA lookups?

It is enough to have

`isParent(u, v)`

function, which uses only in and out times.Div 2 C: Can anyone explain what's wrong with my code? Maybe I divide on ‘a’ — bloks incorrectly? 46914444

Your mod constant is wrong. It's 1e9 + 7, not 10e9 + 7.

No, it is correct.

It's wrong mate. In C++, 1e9 === 1,000,000,000. You have used 10e9 == 10,000,000,000.

Whatever you say, boss. :P

Besides, you access beyond the vector's range — the condition 'ptr != n' in the two inner while loops should be the first.

Guys, what is needed to solve Div2 Problem D? I know DFS but I am not able to understand the approach to be used for this problem. For me, the editorial above is providing no clue/ help. Can someone help me with some explanation on this problem?

Thank you.

For a node

i,dp[i] denote the answer when you start at nodeiand move down. you calculatedp[i] asd[leaf] =w[leaf] anddp[i] =max{dp[j] +w[i] -c(i,j) :jis direct child ofi}. Consider a path which has maximum value, this path has a nodeNwhich is at minimum depth, either the path starts at this node and goes straight down or passes throughNand divides into two paths both of which moves down. So to calculate the answer you take maximum ofdp[i] — path moves straight down or,dp[j1] +dp[j2] +w[i] -c(i,j1) -c(i,j2) — path passes through nodeiand moves down toj1 andj2 withj1,j2 as direct child of i.and update the answer. here is the code https://codeforces.com/contest/1084/submission/46876865

Thanks a lot. Will give it a thorough read.

If I understood correctly the explanation for div-1 B, it will be better to change "Let's add the first 1 at the end." to "Let's add the first 1 in the end.".

So the idea is like this, let's name our array

awhich has lengthm, we have to add [1, 1, 2, 4, 8, ...] (length isH) to our arrays lastHelements. To do this we forget about first 1 and add second 1 toa_{m - H + 1}. We do this for each group. Then start from elementa_{0}and add 2 *a_{i}toa_{i + 1}. And finally add every 1, which we intentionally forgot, for each group, at the needed index.Thanks, you are right. I'll fix it.

Does this Div. 1 F require many constant optimalizations? I came up with the intended solution but got TLE on test 10 or 11 plenty of times.

Try submitting your solution with GNU C++ compiler without diagnostics.

Got Accepted. Thanks!

In the solution of div2 B you wrote that if sum of all a[i] is <=s the answer is -1, but actually if sum==s the answer is 0.

Can somebody explain what is going on in problem : https://codeforces.com/problemset/problem/1084/B ?

What does least mean in

`kvass level in the least keg is as much as possible.`

?we can get 1 liter of kavss from any not empty kegs, and each time we pour 1 liter from a keg, the level of it will decrease by 1. After we get

sliters, the levels in some kegs will be decreased. So our task is to maximize the lowest level.How to do the Div1 C question, the editorial is unclear to me. I am not able to understand how the queries to swap the values of the two nodes will work with the segment tree. Please help

|p−x|+|x−1|+|p−1|=2(max(p,x)-1) CAn somebody explain this ??.. I am a newbie

If we expand modules we will get this:

So we know from problem situation that it is always true.

When you get the formula

`|p−x|+|x−1|+|p−1|`

, you know that x >= 1 && p >= 1, so then you will convert it to`|p-x| + x + p - 2`

. Then just expand the module and you will get the final formula.Thanks bro

I am is a beginner and i was solve to proble A, but i have a problem with this cide:

## include <bits/stdc++.h>

using namespace std; int main() { int n; cin >> n; vector a(n); for (int i = 0; i < n; i++) cin >> a.at(i); vector sorted(n); copy(a.begin(), a.end(), sorted.begin()); sort(sorted.begin(), sorted.end()); int x; for (int i = 0; i <= n; i++) if (sorted.back() == a.at(i)) { x = i; break; } int onX, nX; for (int i = 0; i < n; i++) { if (a.at(i) == a.at(x)) onX += x * 4 * a.at(i); nX = (i — x) * 2; if (nX < 0) i *= -1; nX += i * 2; nX += x * 2; nX *= a.at(i); } int sum = onX + nX; cout << sum << endl; return 0; }

When i was compiling that code using my compiler that was has result 16 (correct). But when i submitting this code on pretest 1 that not has result 16 (that result is 28, and sometimes resulting random number) What's wrong??

You are using values of uninitialized variables

OK thanks, i understand.. :)

In DIV 2 problem B , is floor(v-ceil(s/n)) equivalent to floor(v-(s+n-1)/n) ?

I have the exact same question. Actually I don't get why "n-1" is necessary in "s+v-1" — I thought that "s/n" was enough.

Can anyone explain Div2 B problem?