Hello Codeforces!

On May/15/2019 17:35 (Moscow time) Educational Codeforces Round 65 (Rated for Div. 2) will start.

Series of Educational Rounds continue being held as Harbour.Space University initiative! You can read the details about the cooperation between Harbour.Space University and Codeforces in the blog post.

This round will be **rated for the participants with rating lower than 2100**. It will be held on extended ICPC rules. The penalty for each incorrect submission until the submission with a full solution is 10 minutes. After the end of the contest you will have 12 hours to hack any solution you want. You will have access to copy any solution and test it locally.

You will be given **7 problems** and **2 hours** to solve them.

The problems were invented and prepared by Roman Roms Glazov, Adilbek adedalic Dalabaev, Vladimir vovuh Petrov, Ivan BledDest Androsov, Maksim Neon Mescheryakov and me. Also huge thanks to Mike MikeMirzayanov Mirzayanov for great systems Polygon and Codeforces.

Good luck to all participants!

Our friends at Harbour.Space also have a message for you:

Hello Codeforces!

We are collaborating with an external sponsor who is offering **a unique scholarship opportunity for women who want to study Cybersecurity or Fintech**, both for Bachelors and Masters level.

**The scholarship includes:**

- An up to $10,000, one-time financial award, paid to the University
- An all-expenses-paid trip to the EMEA Summit 28-30 October, 2019 in Berlin
- A Cybersecurity professional to serve as an industry mentor

**Apply before May 30th** and Harbour.Space will also consider you for a scholarship from our own funds.

Don't miss this chance to study with us in Barcelona!

If you have questions about the scholarship, please contact us at hello@harbour.space!

Congratulations to the winners:

Rank | Competitor | Problems Solved | Penalty |
---|---|---|---|

1 | Radewoosh | 7 | 174 |

2 | mnbvmar | 7 | 183 |

3 | I_love_Tanya_Romanova | 6 | 119 |

4 | Farhod_Farmon | 6 | 124 |

5 | xiaowuc1 | 6 | 142 |

Congratulations to the best hackers:

Rank | Competitor | Hack Count |
---|---|---|

1 | algmyr | 57:-1 |

2 | mnbvmar | 16:-2 |

3 | xavier_cai | 11:-1 |

4 | halyavin | 10:-3 |

5 | avm | 7 |

And finally people who were the first to solve each problem:

Problem | Competitor | Penalty |
---|---|---|

A | Dalgerok | 0:01 |

B | Farhod_Farmon | 0:05 |

C | sillysilly | 0:04 |

D | nuip | 0:10 |

E | Farhod_Farmon | 0:23 |

F | ugly2333 | 0:13 |

G | Dukkha | 0:57 |

**UPD:** Editorial is out

Will there be a scholarship for men also?

No, but if you tell them you identify as a female, you'll get in.

too edgy

everything is too edgy for you because you are geniosity

wtmoo

This is now possible if you're using the new Snapchat filter

Hey SoulAdor ! We have scholarships for both men and women and we post about them regularly on Codeforces! If you are interested in one, you can apply on our website! Everyone applying to the university is taken into consideration for a scholarship and no separate application is required.

Is he considered as a woman ?

too funny

Educational round logic :D

good luck everyone

Seems this round will be started before publishing division 3 rating change xDRmind:Educational round correspond the third category in System problem something and i hope Reconciliation to All Participants

wait this round is dedicated to women? Then I guess its a div 4 guys shit.

This is why you're blue xD

?xDxD what you mean??xDxDxDxD=))

Good luck to everyone!

I have no experience of rating adding in edu round :(

hope to become an expert = =

good luck

thx, but i may have lost in speed.. TAT

What's the difference between Div 3 Rounds and Educational Rounds besides the rating limit?

Problem difficulty. Div. 3 problems are generally easier than Educational rounds.

I do not agree with you my friend.

I do not agree with you my friend.

Div 3 are full of greedys!

That's accurate, no doubt about it, but it doesn't mean dificulty in Div 3 problems is lower, at least the way I perceive it.

You have compete in so much CF rounds . If we take into account what I see is you have solve max (3) problems in Educational rounds sometimes (1) and if we look into Div 3 most of the time you solve (4) problems.I am not saying that div3 rounds are so easy but it is easier than Educational rounds and you prove most of the time by solving more problems in Div3 if we compare with educational.Rest is up to you.:P

Typo perhaps? Paid

by theUniversity notto the.I have submitted for B around 10 times it says idleness limit exceeded please help

flush the print. You can read the linked page, given in problem.

Why were you guys here during the contest, lol?

I stuck at C so was taking a break. Still trying to understand why it is timing out. Perhaps it would be better if I jumped to D instead of coming here :)

Are you using Union Find?

I was using DSU with weighted Union Find. Got WA though :(

Why would you need weighted?

Kinda, I created disjoint sets and then calculated connected components. Worked fast with my random N=500k test. Here is the solution.

Hmm. Python :/

It looks like my solution has a problem with cycles. I created a big sample, when I set M to 20000 it returns result in 0.2 secs but when set to 30000 it loops forever. let me convert it to a proper Union Find.

Can you tell what i did wrong in this code 54197918

You're using flush after second cin, not sure it effects something but delete it. In you first query you're sending "? 1 1", that returning a[1]*a[1], you need to add sqrt of that value to v (first v.pb(x)). One more thing, your solution is too crowded for that problem. After finding first number you can easily find others with a simple approach.

Any hint for E and F? I think F can be solved by divide and conquer, but I don't know if my idea is correct..

I think E is related to finding the inversions in the array but couldn't improve the time complexity from O(n^2)

You can use 2 pointers because it is monotonous. I used binary search and got WA on test 57 but not sure if the method is wrong or i have a bug.

For E: Fix l, starting from 1 to x. The idea is similar to 2 pointers and you will need some precomputated arrays.

Precalculate p[x] = are numbers <= x sorted, s[x] = are numbers >= x sorted, l[x] = max number <= x, f[x] = min number >= x, you can check if (l, r) is good in O(1), just do binary search.

F is just counting the coefficient of each $$$b_i$$$. You can process the $$$b_i$$$'s in sorted order and use a BIT to efficiently count how many terms smaller than your current $$$b_i$$$ are in a range, and their contribution to the coefficient.

Actually there are $$$O(n^2)$$$ range pairs, and I got stuck on this ranges. I try to consider left range and right range separately, but I cannot figure out how to count that for each i efficiently.

Say you're processing the ranges containing $$$a_i$$$. These are $$$[l, r]$$$ with $$$l \le i \le r$$$. If $$$a_j < a_i$$$ and $$$j < i$$$ then you add $$$1$$$ to the coefficient for each of the ranges that contain $$$a_j$$$. The right endpoint doesn't matter, so for each right endpoint $$$a_j$$$ is contained in $$$j$$$ ranges, and the total contribution to the coefficient is $$$j \cdot (n - r + 1)$$$, the latter being the number of right endpoints. You can get a similar expression for $$$a_j < a_i$$$ and $$$j > i$$$. To query efficiently you can build a BIT that allows you to get the sum of $$$j$$$ for $$$a_j < a_i$$$ and $$$j < i$$$, and another one for the right side. Then when you finish processing an element, you add it to the BIT.

What 'BIT' refers to?

Binary Indexed Tree.

please anyone show his code for problem B in c++14 PLEASE...

Here you are: 54185189

code for B

The queries are a1*a2 a3*a4 a1*a3 a5*a5，It is possible to deduce a1 a2 a3 a4 from the first three queries and we got a5 by asking a5*a5, Then the array can be determined uniquely.

Has someone's nlog^2n in E passed? I kept getting TLE on test 6. I binary searched the r for each l. Edit: Passed after I dropped a logn.

I was thinking on similar lines but wasn't that n^2 logn? How fast do you test if a particular l,r is valid?

No that is not n^2logn. What you are thinking is to check all possible (l,r) pairs whether it gives non-ascending array or not. Correct idea would be to get minimum r (using bs and some preprocessing) for each possible l (1..n) and adding n-r to the ans.

No, what I did was to fix a particular l. Then calculate the first 'r' such that [l, r] is valid. Then added n — r + 1 to the answer. Check my code here. I maintained the max prefix and suffix till the array is sorted for calculation.

your code is better than editorial.

Thank you!

How to solve B?

Check answer for $$$a_{1}$$$ and next 4 elements, $$$a_{1}$$$ is equal to GCD of query answers hacked

Why?

Ask four queries: a1*a2, a2*a3, a3*a4, a4*a5

brute force all 6! = 720 permutations, and check if the it matches with the above.

https://codeforces.com/contest/1167/submission/54187357

Ask (a1, a2), (a2, a3), (a3, a4), (a4, a5). Since ai*aj are unique you can get first 5 values easily, and the remaining value is a6.

Any test cases which can be used as hacks?

Hello, I tried solving problem C and I was getting runtime error on test 3. I could not find the error, can someone help me please? Here is my code: http://codeforces.com/contest/1167/submission/54204878

You are taking input in O(n^2) fashion. That won't work I guess.

I guess so, thanks

What is test 57 on problem E?

Test cases visible now

10 1000000 404504 367531 741030 998953 180343 781888 161191 855804 689526 976695

What's with the super tight bounds in F? My O(n log n) solution takes 1996/2000ms. 54195181

Happy hacking :P

My code WA test 1???? 54208002

check the answer on CodeBlacks why this code Run on Codeforces get "WA" and this test example is test one !!!

The problem D inputs by "cin" will get "WA" ,replace "scanf" get "AC" ,why??

among 8 div2 participants eho is better than me 7 of them are fake accounts. it is unfair lol i should be the 2nd, not 9th

`[submission:54208002]`

This code "WA" test 1???Is in problem E wrong to do binary search? Isn't it the same as 2 pointers?

What a pity! For E, cin/cout got a TLE while scanf/printf passed.

Input/Output speed is not a problem because I also used cin/cout in E and pased. You just need to add ios_base::sync_with_stdio(0) to make cin/cout as fast as scanf/printf.

Thanks very much. Indeed, with sync_with_stdio(0) my solution passed. But it still has a gap with scanf/printf.

For cin/cout with sync_with_stdio, it takes 1606 ms. While for scanf/printf, it takes 1278 ms.

Try using cin.tie(0); cout.tie(0) It is usually faster than scanf/printf

Thanks!

But for this problem, it still takes 1578 ms, not as fast as scanf/printf.

You can use "\n" instead endl

C and D were easier than B.

Why did my solution to problem D gave TLE: code

Removing stack and converting string to character array passed the pretest. Don't know why..

Because adding two strings is linear to their length. If you just push the char in the end is

`O(1)`

. Changing it with += that equals to push_back it gives acc 54211273 (your sub with this change)Oh I didn't knew abt it. Thanks :)

Authors should have announced that there will be interactive problems in the contest beforehand.

Why?

Many of Division2 participants didn't have any idea about interactive problems (I am also one of them). Because usually div2 contests doesn't contain this type of problems. To know and learn about them all on a sudden in a running contest is taught and time consuming.

There is nothing to learn, interactive problems don't differ much from usual problems. My first interactive problem was on NEERC 2010, I had green level then but I solved it without any difficulties. I don't understand why so many people are afraid of interactive problems.

Any hacks available for problem (B) ?

Thank you creators for such cool problems. I really enjoyed it.

why this solution(54181282) is getting wrong answer on testcase 2?

Wrong output

The line producing the error is probably this:

`for(i=s.size()-11;i>=0;i--)`

`s.size()`

returns an unsigned int and if s.size()<11, s.size()-11 will underflow and take a value of the order 2^32.Change

`s.size()`

to`(int)s.size()`

thank you.

For s with length less than 11

Is it possible to solve C with a method other than Union Find?

dfs

use map.

https://codeforces.com/contest/1167/submission/54196574 Why Do I get TL ?

It is Memory limit exceeded

Why was problem b hacked?

Because according to your method, the fourth and fifth Numbers can switch places

因为你第四个数和第五个数可以换位置拉。

why this https://codeforces.com/contest/1167/submission/54206553 is getting runtime error ?

I did it like the blog example :(

using same code but with cout<<flush is accepted

Seems like you don't separate the numbers in your answer (with spaces).

what numbers ? the code can't even run after the first query

You forgot the

`&`

inside`scanf`

XD XD

is D the simple stack implementation ?

You can also solve it recursively, by using the fact, that all valid sequences S = (A)B, or empty.

54212912

Yes. Actually, you just have to invert the color before processing a

`'('`

and after processing a`')'`

.@MMXIX can you explain your approach in detail,and how does reverting colors gives us required answer

You have to minimize depth, so as you encounter two same brackets in sequence invert your color . This will minimize the depth.Simple greedy approach.

Thanks for your comment. I did this in the same way and got accepted but I was in doubt as it 'D' problem.

Was O(N logN) supposed to pass on E? I got TLE on test 50 with such an approach. If it was intended to pass, why set the bound as high as 1e6?

There are O(N) solutions using 2 pointers. However, I think only simple O(NlogN) solutions passed (binary search).

I used bs and got WA on test 57. Still can't understand what's wrong.

Finally found the bug. It is not related to binary search.

I don't know what is the hacking format of interactive problems.

Somebody help me with an example,please.

The problem B could be more interesting if the number of elements would be greater.

Yeah, it would have been better if the number of permutations were too large for all of them to be checked.

xx

memset gives wrong answer and loop initialization got Accepted.

`sizeof(vs)`

is a constant value, you have to multiply it by the size of the array.Correct command would be

`memset(vs, false, sizeof(bool) * (n+1))`

because type of elements in the array is bool.`sizeof (vs)`

will typically be larger (8 bytes) than`sizeof(bool)`

.`vs`

is a pointer,`sizeof`

a pointer is generally 4 bytes (8 bytes for x64 compilation).Can anyone explain their approach of solving problem D?? I am not able to make up any thoughts about it

let r0 be the nesting of string with color 0, similarly define r1.

Initially, r0 = 0 and r1 = 0.

now ,each time you encounter an opening bracket, if r0<r1, color this bracket 0 (also increment r0)else color this bracket 1 (also increment r1).

Similarly, each time you encounter a closing bracket, if r0>r1, color this bracket 0 (also decrement r0)else color this bracket 1 (also decrement r1).

This guarantees that both strings will be balanced bracket sequences because r0+r1=nesting of original string =0 and r0 or r1 can't be negative (because we ensured it in the loop above) so r0=0,r1=0.

This also guarantees that max nesting is minimised because only the min is incremented when encountering an opening bracket (always better than incrementing max) and the max is decremented when encountering a closing bracket (always better than decrementing min).

Can you explain the proof a little briefly? , it is a nice approach.

can someone please explain approach to question D??

The judge doesn't judge.

@Radwoosh can you give approach and idea to solve problem D

MikeMirzayanov this is happening again :/

MikeMirzayanov Judge out of order!!

Why is Codeforces so slow after every contest? My past 3 submissions in the virtual contest have not been judged.

What is the approach for F?

For fixed i, find all j's with (a_i >= a_j), then answer will be the sum of min(i + 1, j + 1) * max(n — i, n — j). Go through i's in increasing order and maintain several fenwick trees.

Thanks, a really cool technique I have to say!

I'm very confused. Can someone please explain how to solve E? I have been reading multiple AC solutions but still can't understand the idea behind that!!

Save the maximum index and the minimum index of each different elements in a. Then think of how to judge a f(l,r) is valid or not in O(1)

In problem C, I'm using DFS and getting a TLE in 7th test case. Can anyone help?

Maybe you are not building the graph in an efficient way. Suppose you have 2 5 7 9 Then you don't have to do

2->5->7->9

5->2->7->9

7->9->5->2

9->2->5->7

Instead do something like

2->5

5->7->2

7->9->5

You don't need to make exact graph as you only have to count connected components.

you can skip visited groups

It would be nice, if someone please explain me the problem C using sample Case. I didnt figure out it yet for a while xD. Thanks in advance <3

Lets take first case there are 5 groups and 7 people. Each of 5 rows starts with some number that means how many people in Ith group. For example in first group 3 people with number 2, 5, 4 and so on. And problem says if person sends news, their friend can also send info to their friends. After this if person in this group he can share news to this friends and if this friends in other groupes they can share news to other friends in that groupes. At the end we can say that we need to find every component and people in this component can share info with this size of component(including himself). First case if we take person with number 1 he is in two groupes {1, 2} and his friend 2 is in group {2, 5, 4} and if we add this two groups its {1, 2, 5, 4} everyone can share new with themselves that means 4 is the component size. You can solve this problem with dfs or dsu)

You said: "First case if we take person with number 1 he is in two groupes {1, 2} and his friend 2 is in group {2, 5, 4} and if we add this two groups its {1, 2, 5, 4} everyone can share new with themselves that means 4 is the component size."

But My point is it should be.. if we take person with number 1 he is in two groupes {1, 2} and his friend 2 is in group {2, 5, 4} and{2, 6, 7} and if we add this three groups its {1, 2, 5, 4, 6, 7} everyone can share new with themselves that means 6 is the component size." Can you please clear my confusion. Thanks xD <3.

2 6 7 doesnt mean fifth group contains {2, 6, 7} first number means how many people in this group, in this example 2 people in fifth group with numbers {6,7}

Oh hoooo. Thats clearly show how bad I am in problem statement reading. But Thanks a lot xD. Take love <3

could you help me check where is wrong? ID:54220798

In E, is it okay to find the max l value and min r value such that l<=r , and f(l,r) is valid. Then we know that f(l-k1,r+k2) is also valid for all possible values of k1 and k2.

That is true but not enough. For example, leaving one element only inside leaves array sorted, if l != r, then we can leave l + 1

couldnt understand

I mean, what if l = n and r = 1. Which max l value and min r value do you choose? If you iterate all, then you count some segments several times.

but l<=r

What about this case. 1 4 2 5. f(4, 4) = 1 2 5. it is valid and we cannot make l bigger or r smaller, because then l > r. So you are saying that all f (4 — k1, 4 + k2) are true for positive k1 and k2, as I can tell. Well that is right. But we also have f(1,2), for example -> 4 5. And it is not of this form, 4 + k2 = 2 so k2 is negative.

Oh, I understood. So you mean that 4-k1 and 4+k2 is correct but it leaves out other cases... because l is unable to become greater than 4 and r unable to become smaller than 4.... Is this what you meant?

Yes

Why using Dfs to get number of nodes in each connected components gives wrong https://codeforces.com/contest/1167/submission/54223559

You are assigning the value of c to val[i] while running dfs, c being the number of connected components, this doesnot ensure that the parent of all these components is 'i' and therefore the answer is coming out to be less than than the actual number of components. Just a little bit of change and there you go..!!! 54225792

PS-- This might give you TLE in system tests as you have not used weighted dsu.

Hope that I can reach 1600, it is only 12 ratings away but it seems that I am not performing well this time... Btw I really appreciate problem B as an interactive task, as we don't get to meet them often and this is a great opportunity to train up skills related to interactive tasks.

You're right. This is my first interaction problem.Wish you can be blue name

Thanks!

I think you will :)

congratulations！

Thanks!

I'm new here. Are these contests are Python friendly? Most people solved C with Union Find in C++ and passed the tests, but the same algorithm in Python gets TLE

In div2 and div3 (and most div1) problems can be solved with Python. With medium plus problems, if there is too much list reading or query you will need Pypy3. There are lots of difficult problems looks impossible with Python but we're far away from them (I'm trying to solve some of them on Codechef and even with 5x time limit it's still a big challenge, my current problem). And here Python solution for C: link

Thanks, the key was to change input() to sys.stdin.readline() https://codeforces.com/contest/1167/submission/54233244

Anyway it's strange that the time limit for python and c++ is the same number.

And why do you need work with stdin? In real life in 99% you need to write class of function, so why not just give a prototype of the function or class that you need to implement?

Sorry for my questions, I'm new here.

Actually you're right about writing a class of function, but here I most time coding what I need for current problem. I'll use libraries if I go up in rank :) For stdin, I only use it when there is lots of reading like in that problem.

Don't worry about time limit, it's something classic CP thing but Codeforces gives extra time for problems. You'll start to need C/C++ solutions if you become a Div1.

See solution of @algmyr he passed the problem probably with union find

I wonder when I can see the ratings changing...

Easy busy contest

Why Radewoosh's submit 54183325 didnt get TLE? In his solution (sorry for my poor English) he didnt use rank's heuristic and got AC. I tried to submit solution without rank's heuristic and got TLE 54233682 . Could someone help handle it, please?

And im about problem C.

Because he used a technique called 'Path Compression'.

In his code

not

So after a fin(v) operation, the nodes on the path from v to the root will all directly connect to the root.Therefore, when you perform fin(i) (i is a node on the path from v to root) will just return oj[i]

Wait for tutorial

dont tell me what to do

I have passed problem G, but I have no idea whether my solution has a right complexity.

My solution is 54235792. It uses a method similar to two-pointer to find the nearest pair that is the same far from point x.

Could someone please tell why it's correct or construct a data that will lead to a TLE?

I got compilation error on problem D during system test after it passed during the contest? How is that? :D

54189857

Here is my $$$O(N)$$$ solution for problem

E: 54241182The general idea is to maintain, first of all, a global rightmost $$$L$$$ and a global leftmost $$$R$$$ such that ranges $$$[1,L]$$$ and $$$[R,X]$$$ are valid, then for each $$$r$$$ the rightmost $$$l$$$ such that the union $$$[1,l] \cup [r,X]$$$ is valid too.

Used DSU for problem C, got WA test 3 and I cant find the mistake, can anyone help? 54184178 Edit: Found the error, I was not checking if nodes are already connected before connecting them.

same for me.

then I used dfs and it got ac.

I don't know what is the mistake.

I also got WA test 3. You should check in your connect function that you don't connect the same element.

Hello, I tried solving problem C and I was getting TLE error on test 3. Somebody please help in optimizing my code. I didn't get the editorial properly. Here is my code: https://codeforces.com/contest/1167/submission/54269390

The answer is the same for vertices in same connectivity component. You dont need to clean the array.

How to solve problem G?

can any one explain E problem please ?

problem D, solution1 got accepted whereas solution2 gave TLE. The only difference between the two is in line number 21. In solution1 we append 0 in string by res+='0'. In solution2 we append by res=res+'0'. I thought these two operated the same way. Can someone give insight to the reason?

the first solution, you modify the res string, fill the data in string if possible, or create new string if not possible, just like vector push_back

the second solution, you create new string

the first solution is possible because c++ string overload that operator http://www.cplusplus.com/reference/string/string/operator+=/

Someone please help me with problem C. I am getting MLE. Any help would be appreciated. https://codeforces.com/contest/1167/submission/54526141

78428168 78426655 Why did this same code got accepted with C++17 while it got TLE on C++17(64)? All i did was change the array size.. That should've given a RE instead of TLE

Can anybody explain?

Good round