**Problem Authors**: GT_18, karansiwach360

**Author`s Code**: 42406878

**Code Complexity**: *O*(*LogN*)

**Problem Authors**: csgocsgo, karansiwach360

**Author`s Code**: 42406891

**Code Complexity**: *O*(*NLogN*)

**Problem Authors**: csgocsgo, Ezio07

**Author`s Code**: 42406848

**Code Complexity**: *O*(*N*)

**Problem Author**: Ezio07

**Author`s Code**: 42406857

**Code Complexity**: *O*(*NLogN*)

**Problem Authors**: dhirajfx3, GT_18, TooDumbToWin

**Author`s Code**: 42406908

**Code Complexity**: *O*(*NLogN*)

**Problem Authors**: karansiwach360, hitman623

**Author`s Code**: 42406921

**Code Complexity**: *O*(*N*)

**Problem Authors**: 300iq

**Author`s Code**: 42406934

**Code Complexity**: *O*(26^{2} * *N* + 26 * *Q*)

**Problem Authors**: dhirajfx3, DeshiBasara

**Author`s Code**: 42406942

**Code Complexity**: *O*(|*x*| * *LogN* * 26) per request.

We hope you enjoyed the problems!

Hope to see you all in next edition of Manthan.

Can someone tell me why i am getting this error in my code it is working fine in my compiler 42407188

your array should be initialized with a constant value. In case of your code:

int a[n+1]andint si[l]are wrong!Ezio07 It would be very helpful if you could add Time Complexity for all the proposed solutions. Thanks!

Yes, added!

but the time complexity for problem D is $$$O(n^2 log n)$$$:(

another solution to problem A:

As was written, the answer it's number x for series of numbers: 1, 2, 4, 8,..2^x. This series of numbers it's geometric progression. The sum of the geometric progression is given by the formula: S(x) = (b(x)*q-b(1))/(q-1). For 1, 2, 4,..2^x q=2, b(1) = 1 and b(x) = 2^(x-1) => S(x) = (2^(x-1)*2-1)/(2-1) => S(x) = 2^x-1. The following condition must be met for the answer to be correct: S(x)<=n => 2^x-1<=n => n+1-2^x>=0. The code this solution: 42408562. (sorry for my not so good english)

Another solution :))

hey...can you explain me... how did u get this logic? thanku in advance @ Dragonol...

The logic is based from the binary form of n:

for exemple: n = 10. The binary form of 10 is

1010, you will need1000,0100,0010,0001to form any number x (1<=x<=10)The answer is actually the length of binary form of n. In fact,

log2(n)=length_of_binary_form_of_n—1so the answer shuould belog2(n)+1.Thanks. Now I got it

Thank you for Explaination. I got it now

This is a beautiful solution. :)

Hi,

I used cout<<(ll)(log(n)/log(2) + 1)<<'\n'; I got wrong answer. http://codeforces.com/contest/1037/standings/participant/19485663#p19485663 Can you please tell why?

Your code seems to be correct. Even ideone is giving the expected answer on 8 that is 4. https://ideone.com/tgBeiN

Yes, but codeforces is giving wrong answer.

GT_18 karansiwach360, can you please provide the reason?

It's clear case of precision error.

If you try

`cout<<(log(n)/log(2)+1-4);`

, (for n=8) you'll get`-1.60245e-016`

. On casting to (ll), the result you get is 3.Yes. I think Something may have gone wrong on their side.

///Deleted

loge(n)/loge(2) == log2(n), not loge(n-2).

ops!! My fault :)) Sorry

Your solution sucks. If

ncould be up to 10^{18}, it wouldn't work.http://codeforces.com/contest/1037/submission/42365860

Yours fail too when n is up to 10^18, you suck.

That's because the problem condition for n is 1 ≤ n ≤ 10^9. If you want n up to 10^18, then here you are:

try to check

n= 2^{59}andn= 2^{59}- 1I don't understand . They both come out with 60. So i do the subtraction of 2^59 and 2^59-1, the answer is 0.Why?

most probably it's wshift-count-overflow try this

It also writes 60 for

n= 2^{59}- 1but not for n=2^(59) do you have correct solution?

correct solution is to do all in integers. just go from 0 to 60 and check whether it is a right answer

`32 - __builtin_clz(n)`

can anybody tell me why iam getting wa in test case 11 in D problem https://ideone.com/r2c4K5

A condition

if(n==200000) { cout<<arr[i]<<" "; } why this condition ??? In the end you just have to print Yes or No...

2 2 1 1 2

output: Yes

this might be the condition

Got accepted

I was declaring array size smaller than input

Can someone tell me D's solution? I read the D's tutorial and read other source codes, but I can't understand. What does D's tutorial say and what is the solution?

For understanding what the tutorial is trying to say, first you need to understand how graph is stored in an adjacency list and BFS Order Traversal. If you understand these two things, then you can read ahead.

Basically, the way you store nodes in adjacency list, determine how your BFS Order Traversal is going to look like.

So, for checking the

givenBFS Order, you store the neighbours of a node in the list according to the order they appear in thegivenBFS Order Traversal. Basically, "we can sort the adjacency lists of all the nodes in the order in which they are present in the input BFS Sequence." (as mentioned in editorial)Once you get the adjacency list according to

givenBFS Order, Do BFS on the list and then check whether it matches the given order or not.Here is my editorial to D. Let me know if it helps :)

42423685 oh Great minds think alike .

Can somebody tell me what is wrong with my code (42403836) for 4th problem.Actually I am getting WA on test case no.11 which is a large test case that is why I am not able to debug it.

maybe this condition

My answer is yes in this case what should be its answer?

Yes, you are right,answer should we yes

Thank you that helped me a lot !

see this comment

I submitted again, with little bit of moification -> (42416659) now it is passing the case you are pointing to,but still it is not passing the test case no.11

for problem D, this approach worked for me.

But this didn't work. I don't know why. if someone can explain please do.

Link for AC submission: https://codeforces.com/contest/1037/submission/121338039 for wrong submission: https://codeforces.com/contest/1037/submission/121337127

Can we do Problem D using Dijkstra's algorithm with edge weights set to 1 ?

Input order should be in increasing order of distances from the source.

If you ever use Dijkstra with all edge weights as 1, use BFS instead. But I don't think it would work. You have to add all nodes connected to the last node in some order, but you can't mix them with other nodes (even if they are on same distance from 1).

My submission for B is in queue for too long. 42412545 What should i do ?

everyone get stuck in queue. maybe they're repairing the system

Can someone tell me why i am getting this error in my code it is working fine in my compiler 42399951

You don't initialize local visited array with zeros

Can Someone Explain Problem B? Why are other elements modified when only the median needs to be modified?

Median is middle element in a sorted array.

Now, modifying Median may break the sorted order of the array elements. So we make sure elements before the median are still smaller than new median and all the elements after median are larger than new median too.

Got It.Thanks For Help.

Still wondering: optimality proof for A, anyone?

Can anyone explain G more clearly?

For each character

c, you need to precalculate grundy numbers of strings between consecutive occurrences ofc, their prefixes and suffixes. When a stringsbreaks into multiple strings, the first part is a calculated suffix, last part is a calculated prefix. For other parts, you can keep prefix xor of the calculated xors. So suppose you have a string`adbecbdabfr`

and query on`dbecbdabf`

and you choose character`b`

(as your first move), the precalculated grundy numbers will be of`ad`

,`ec`

,`da`

and`fr`

and their prefixes and suffixes. To answer the query, you take`d`

suffix from`ad`

and`f`

prefix from`fr`

.`ec`

and`da`

can be retrieved from prefix xor.There are

n+ 26 such strings. You can consider them as queries while precalculating and order them properly such that the required values are calculated before the query.Why actually "The best possible way to distribute the coins is to create packets with coins 1,2,4,…2k with maximum possible k such that 2⋅2k−1≤n. " is there any proof about that ? I have solved the problem and I still don't know why this happens!

To be able to make 1 you defnitely need the coin of value 1 so it has to be included.d now to make 2 you can include another coin of value one to sum to 2 or you can use a coin with value 2 so that you can make 3 as well so using coin with value 2 is optimal. You can go on inductively to show 2

^{k}is optimal.Lets assume that we can take

`<= log2(n)`

coins. We can choose some of them in`2^log2(n) = n`

different ways. So there are at most`n`

different sums. One of them is`0`

. So we have at most`n-1`

non-zero sums. It is not enough to cover`n`

numbers(`1, ..., n`

). Thus answer isnot`<= log2(n)`

, answer is`>= log2(n)+1`

. So,`log2(n)+1`

is optimal answer.Now we have to find

`log2(n)+1`

coins. These are`1, 2, 4, ..., 2^(log2(n)+1)`

. We can prove its optimality by induction. First`i`

coins cover numbers`[1 ... 2^i-1]`

. Next coin is`2^i`

. With that coin we can obtain also sums`2^i, 2^i+1, ..., (2^i+2^i-1 = 2^(i+1)-1)`

. So with first`i+1`

numbers we can cover`2^(i+1)-1`

coins.I see that many people make arrays/vectors of size 2e5 + 5 if max number of elements is 2e5. Why is it not 2e5?

Maybe for work on boundary of array and taking this size to sure don't go out of size of array ... actually for more security

i see you have never got WA/RE because of wrong bounds

Could you explain what is it and when it happens?Because in my view if an array goes out of bounds, there is a mistake in the code.

Can anybody tell me what is wrong with my solution for D. It is getting wrong answer on case 12 My code

can anyone tell to me in task A why we can not do this with smaller number of packets?

in problem B why we need to traverse the whole array and find the answer but we simply know the required value of the median(s) from the question, we can perform the operations from the middle element to s.

Let me clarify problem E a bit, for current explanation goes more into details and less into the idea.

First, we build final graph of friendships. Then we delete from it all nodes with less than k friends — those buddies weren't going anywhere anyway.

The interesting observation is that all the remaining people can go to the trip on the last day. Now we can just play the time backwards. After we removed edge which appeared on day n and also all the nodes with friends number below k, all remaining nodes can go to the trip on (n-1)-th day. Then do the same thing for n-1, n-2, et cetera. Need to print results in reverse order though :)

Thanks for the explaination.

https://codeforces.com/contest/1037/submission/42430271

I am doing different approach can anyone figure out where I am going wrong!!!

Can you please describe how the third sample test case works??????? I couldn't figure it out.

Auto comment: topic has been updated by Ezio07 (previous revision, new revision, compare).solved F with Divide & Conquer 42425616. Actually it's quite similar to the one in the editorial.

Why the code of E is so long......

Just have a look at the solve() function rest is just template.

SolutionNice solution for E...

Will definitely give it a try

Problem F

Can someone push me in the right direction? I can't figure out how to get

landrfor each element in linear time.https://www.geeksforgeeks.org/the-stock-span-problem/

Thanks for help.

For problem F, Can anyone give an explain about how to calculate number of segment in [l, r]? Also, why the segment is not k, 2 * k — 1, 3 * k — 1 ... ?

Did you get the answer? I'm also confused why is it 2*k and not 2*k-1 in the explanation. EDIT: Got the answer. Here is a great explanation.

thanks, it's a good solution.

Can somebody tell me what is wrong with my code(42398601) for valid BFS problem?

Could you explain, how did you solve the query part in O(|x|* LogN *26) per request? I solved it in O(|x| * LogN * LogN * 26) . I first build the suffix tree of given string. Then to find whether there is an element in range [l,r] in given subtree, I use binary search on each node of segment tree which are visited during query. Hence, there would be O(logN * logN) for each character (from 0 to 26) for each index of x in the worst case.

Consider of example the

ithposition of the array of leaves has valuel_{i}, now we will be maintaining a set of persistent segment trees from left to right, that is persistent segment tree forithleaf is obtained by adding value 1 to thel_{i}position of persistent segment tree of leafi- 1. To do this, maintain three arrays, seg_tree, L_ptr, R_ptr. The node represented by integer id c will represent-seg_tree[c]: The sum of values in the segment represented by current node.

L_ptr[c]: The left child of node represented by c.

R_ptr[c]: The right child of node represented by c.

Now to add the value

l_{i}to the persistent segment tree of leafi- 1 and get the root node for the persistent segment tree for leafi, we can adopt the following algorithm.If we have persistent segment of node

i- 1 as c, we can obtain persistent segment tree for nodeiby calling add(c,0,|S|,l_{i}).Code snippetNow to find whether the current subtree contain any value in the range

L≤R,Let

a+ 1 be index of left-most leaf in the current subtree, andbbe the right-most leaf in the current subtree andp_{j}represent the persistent segment tree of leafj.Now we need to only take two segment trees

p_{a}andp_{b}and query if sum of values in the segmentL...Ris non zero.To do this we can apply the following method or say call find_sum(

p_{a},p_{b}, 0, |S|, L, R)MethodIf returned value of the above function is non-zero then there is some value in the range L ... R. You may also like to see the find_first method in the code given in the tutorial to find the smallest value which is greater than L but less than R.

So this is how we can check in O(log|S|) time whether there is some leaf in the current subtree with value in range L ... R.

Let us know if something is still not clear.

Thanks :)

Problem D. Test Case 39. Given Test Case - 2 1 2 2 1

My opinion —

`Yes`

. But the checker says`No`

. Can someone Explain to me why the answer is`No`

?You always start from root which is 1. So it's impossible that you visit any other node before 1. Therefore, sequence must always start with 1. Here it starts with 2.

If I have started at 1. Then because first no is 2. I would have printed no. But here the case is opposite. Checker is saying No, and I'm saying Yes.

In the given input, tree has two nodes. Root is 1, and it has single child which is 2. In this case, there's actually single BFS possible which is "1 2", not "2 1". You always start with 1. If first is not 1, then just print "No", no need to check anything else.

Then what does this mean ??

`The BFS algorithm in this problem always starts with 1, however, it is not guaranteed that a[1] = 1.`

It means that correct BFS algorithm should start with 1, but we may give a sequence with a[1] != 1 which is necessarily an incorrect BFS sequence and you have to print "No" for such sequence.

Thanks. Got it. Was confused by the announcement.

lol, I can't believe that. Added a && v[0] == 1 and code passed. So stupid.

Can you guys write the last part of the Problem F properly, more clearly?

hitman623

This is not clarification of the method from the tutorial, but slightly different method of calculating total count for each number.

On the figure above you can see progression of

`b`

arrays for given array`a`

andk= 3. To calculate total count for number 8 you want to calculate count of all numbers in the triangle (their count is proportional tor-l) and then subtract blue ones (their count is proportional toi-l) and green ones (their count is proportional tor-i).Turns out, you don't really need all that case work. Here.

Yeah this is what we expect from the editorial that HOW it turns out that you don't need it.

Just do cout<<(long long)log2(n)+1 in Problem A

don't do that, please

About problem G. Editorial says we treat precalclulating like queries. So we iterate over character c and use two prefix/suffix dps and one range query. But yet we dont know grundy for each gap between two equal characters, we cant precalculate prefix xors. We can use segment tree, but than complexity becomes n*A*A*log(n) which is too long. So how to accomplish precalculation?

TooDumbToWin

Here is an O(n) solution to D :

42464470

Maybe this one is simpler :)

42377485

In problem D we need to check if the BFS sequence is valid or not. The editorial solution is really easy to understand. I guess same approach can be used if the problem was to check for valid DFS sequence.Only change would be that a DFS will be used instead of BFS. Can anyone confirm this, I am not 100% sure about it!

Yes, the editorial approach will work in that case also.

can anybody help me understand 1037E(Trips) problem??

See this explanation from saluev

thanks:))

Can anyone explain the E question clearly? I dont get the author's logic

Can anyone please tell me why my code gives wrong answer for problem B ? Here's the link:

https://codeforces.com/problemset/submission/1037/48669584

For D, I made an array

`par`

, for which,`par[i]=j`

means that`j`

is the parent of the node numbered`i`

. I constructed a queue,`Q`

that is used to imitate the BFS. Now, I add the current node (first node initially) to the queue, and check whether the next`k`

nodes (where`k`

is the size of the adjancency list for the current node) have the current node as their parent, while adding these nodes to the queue`Q`

. After these`k`

nodes are checked, I remove the current node from the queue.This will work for any order of neighbor selection for the BFS. The submission can be found here. It doesn't work for the test case 12. Any ideas? I think the implementation is correct, but there is some problem with the concept.

here is my submission:69550598. I ran bfs once and stored the parents and the no.of direct children for every node. Then,i made a queue and a counter j. the no. extracted from the queue is the current parent. so i check the next children no. of nodes in the input from j. if any of them does not have p as its parent,it would not have been added by any way of dfs traversal and thus we answer No. If it is a child then add it to the queue. If the loop runs successfully,we increment j by the no. of children so that we can check for the children of the next parent. I hope this explains my code.

Can someone explain Dynamic Programming approach for problem C?

Can anyone please tell me whats wrong with my code for the question B

Can somebody point out what is wrong with my solution?

https://codeforces.com/contest/1037/submission/59094440

Can anyone pls explain why for F, the answer is number of subsegments of length k, 2k, 3k ... ?

Shouldn't it be k, 2k-1, 3k-2...? Like for example in sample input 1, there are 1 subsegments of length 2 and one of length 3. when k=2.

Thanks.

We can perform Problem D in $$$O(n)$$$ using two-pointers.

80832291

Aren't you using a vector<set<>> ? So time complexity can't be O(n). Shouldn't it be O(nlogn)? I'm not sure of my answer though

Actually I never end up using

`vector<set<int>>>`

after initialising... so you can just remove that part of code and it will be $$$O(n)$$$

Can anyone help me with my submission for problem D? 83125545 I dont understand why the output of the following test case is "NO"

6 1 2 1 5 2 3 2 4 5 6 1 5 2 3 4 6

Please help!

I've another solution for D : 116992209 1. If 1st node is not a 1 then "No" 2. Take a map and put the index of all the array elements of given sequence. Take another map and in it put the locations of their parent. Now check if our map 2 is sorted or not? If it's sorted it is "Yes" else "No"

Plz. can anyone tell me in BUGABOO D why we are sort the adjacency lists of all the nodes in the order in which they are present in the input BFS Sequence?