### Erfan.aa's blog

By Erfan.aa, 6 years ago,

### 459A - Pashmak and Garden

Four vertices of a square with side length a (and sides parallel to coordinate axis) are in this form: (x0, y0), (x0 + a, y0), (x0, y0 + a), (x0 + a, y0 + a).

Two vertices are given, calculate the two others (and check the ranges).

Total complexity : O(1)

Sample solution: 7495194

### 459B - Pashmak and Flowers

If all numbers are equal then answer will be n * (n - 1) / 2, otherwise the answer will be cnt1 * cnt2, where cnt1 is the number of our maximum elements and cnt2 is the number of our minimum elements.

Total complexity : O(n)

Sample solution: 7495202

### 459C - Pashmak and Buses

For each student consider a sequence of d elements from 1 to k that shows the bus number which is taken by this student on each day. Obviously, there are kd different sequence at all, so if n > kd, pigeonhole principle indicates that at least two of this sequences will be equal, so that two students will become close friends and no solutions exist. But if n ≤ kd, then we can assign a unique sequence to each student and compute the answer. For computing that, we can find the first n d-digits numbers in k-based numbers.

Total complexity : O(n * d)

Sample solutions: 7495236

### 459D - Pashmak and Parmida's problem

First of all, we can map the given numbers to integers of range [1, 106]. Let li be f(1, i, ai) and let ri be f(i, n, ai), we want to find the number of pairs (i, j) such that i < j and li > rj. For computing lis, we can store an array named cnt to show the number of occurence of any i with cnt[i]. To do this, we can iterate from left to right and update cnt[i]s; also, li would be equal to cnt[ai] at position i (ri s can be computed in a similar way).

Beside that, we get help from binary-indexed trees. We use a Fenwick tree and iterate from right to left. In each state, we add the number of elements less than li to answer and add ri to the Fenwick tree.

Total complexity : O(n * logn)

Also we can solve this problem using divide and conquer method. You can see the second sample solution to find out how to do this exactly.

Sample solutions: 7495225 7495225

### 459E - Pashmak and Graph

In this problem, a directed graph is given and we have to find the length of a longest strictly-increasing trail in it.

First of all consider a graph with n vertices and no edges, then just sort the given edges by their weights (non-decreasingly) and add them to the graph one by one.

Let dp[v] be the length of a longest increasing trail which ends in the vertex v. In the mentioned method, when you're adding a directed edge xy to the graph, set dp[y] value to max(dp[y], dp[x] + 1) (because of trails which ends in y and use this edge). You need to take care of the situation of being some edges with equal weights; for this job we can add all edges of the same weights simultaneously.

Total complexity : O(n + m * logm)

Sample solution: 7495216

• +46

 » 6 years ago, # |   +29 Thanks for the fast editorial!
•  » » 6 years ago, # ^ |   +8 We're sorry. Some unexpected events caused that.
 » 6 years ago, # |   +9 Can you please include judge's code that will be great.
•  » » 6 years ago, # ^ |   +3 Included. ;)
 » 6 years ago, # |   0 i donot understand the E's solution cannot u explain in more details ??
 » 6 years ago, # | ← Rev. 2 →   0 in problem c it think it will be k^d not d^k
 » 6 years ago, # |   0 thanks for nice editorial.
 » 6 years ago, # |   0 How can you get the n * (n - 1) / 2 formula in B (otherwise than by wolfram alpha)?
•  » » 6 years ago, # ^ |   +9 This formula is the number of unique pairs you can form with n given elements.Each element can be paired with n - 1 elements, and there are n total elements. If we multiply we get n(n - 1), but we must note that we've counted all pairs twice, so we must divide this number by 2 in order to get the right amount.
•  » » » 6 years ago, # ^ |   0 Got it, thanks!
•  » » 6 years ago, # ^ |   0 If all numbers are equal then we have to choose 2 numbers from total n numbers. Then there are total nc2 = n!/((n-2)!*2!) = n*(n-1)/2 ways.
•  » » 6 years ago, # ^ |   +5 you can't step in programmnig contests without this trivial knowledge :)
 » 6 years ago, # |   +3 The solution of C(Pashmak and Buses) was awesome. Thanks a lot!!
 » 6 years ago, # |   0 I solved E with the same algorithm I get wrong answer in test 5 and don't know what's wrong ! this is my submission http://codeforces.com/contest/459/submission/7507479
•  » » 6 years ago, # ^ |   0 Here is it. Look at /* That for-loop was modified */.
•  » » » 6 years ago, # ^ |   0 I got it, Thanks :)
 » 6 years ago, # |   +5 Can anyone explain how to solve problem D using divide and conquer. I think the idea used in this problem is similar to finding the number of inversions in an array using divide and conquer.But, I am not getting how to solve it exactly..
•  » » 6 years ago, # ^ |   0 I can.In order to avoid using of data structures firstly compress all the numbers. Then you should calulate pref[i] = f(1, i, a[i]) and suf[i] = f(i, n, a[i]) for all i. Then you should simultaneously use merge-sort for pref and suf. And then you can do the same algorithm as for inversions by merging not prefl and prefr or sufl and sufr but prefl and sufr.I'll provide a code for this later.
•  » » » 6 years ago, # ^ |   +3 And here it is: 7516132
•  » » » » 8 months ago, # ^ |   0 I was able to get an AC using ordered set but on using merge sort tree i am getting tle. Is it becuase the time complexity for using it is O(n*logn*logn) ??Thanks in advance.
•  » » » 4 years ago, # ^ |   0 Wherever you don't need to do the entire merge sort you have already two components you just need to sort them apart and focus only on the step of merging them .
 » 6 years ago, # |   0 can anyone please tell me where my solution is lagging to the problem 5. Help is appreciated. Thanks. http://codeforces.com/contest/459/submission/7556964
 » 6 years ago, # |   0 in problem C,can someone please explain what the author means by "n d-digits numbers in k-based numbers."??
•  » » 6 years ago, # ^ |   0 n d-digits numbers basically means first n decimal based numbers and you need to represent those n numbers in the k-based numbers.For example, n=6, k=2, d=3first 6 numbers are 0,1,2,3,4,5. And in the 2-based number system(binary) 000, 001, 010, 011, 100, 101. And we get 6 different sequences as described in the editorial. As the bus number starts with 1 we need to add 1 to each digit of the number. So the numbers in k-based or sequences now become 111,112,121,122,211,212.Hope this helps. My solution
•  » » » 22 months ago, # ^ |   0 thanks that was awesome
•  » » » 15 months ago, # ^ |   0 can you please explain your code for generating the sequence, i don't understand it
•  » » » 5 months ago, # ^ |   0 can you please explain the code??
 » 6 years ago, # | ← Rev. 2 →   0 http://codeforces.com/contest/459/submission/11762173 can someone help me with optimizations.used the divide and conquer technique. my code is getting tle. thanks :) ShayanH
 » 5 years ago, # |   +3 For problem:D..just use fenwick tree to solve it. its fun to solve it with fenwick tree :-) my submission :- 17314867
•  » » 9 months ago, # ^ |   0 Its fun but I am not able to understand why were doing what were doing. I mean I understand why this works but I cant still explain to myself. Can you explain me please?
•  » » » 7 months ago, # ^ |   0 i can try problem basically translates to finding all pairs such that1) i < j 2) number of a[i] from 1 to i > number of a[j] from j to nlet count[a[i]] — count of a[i] from 1 to ifor all i we will add to fenwick tree count[a[i]]now we will again iterate from beginning and keep on removing count[a[i]]what does my fenwick tree have at any i?for each "value" number of times that "value" can be obtained "value" is the count of some element from i + 1 to nnow a simple query of count[a[i]] — 1 shall give me how many js are there such that count[a[j]] from i + 1 to n is lesser than count[a[i]] from 1 to ihttps://codeforces.com/contest/459/submission/85599666
•  » » » » 7 months ago, # ^ |   0 Thank you so much for taking out time to answer noobs like me
•  » » » » 6 months ago, # ^ |   0 i am still not getting i have read your explanation 3 times...and i also know fenwik tree but still not getting...can you give some more explanation..?
•  » » » » 6 months ago, # ^ |   0 i didn't understand your last two liness for each "value" number of times that "value" can be obtained "value" is the count of some element from i + 1 to n
•  » » » » » 6 months ago, # ^ |   0  for (int i = 1; i <= n; ++i) { f.update(totcnt[a[i]], -1); totcnt[a[i]]--; forwdcnt[a[i]]++; ans += f.query(forwdcnt[a[i]] - 1); } do u undertsand this part? i realised this part can explain more than my stupid english
•  » » » » » » 6 months ago, # ^ |   0 But fenwik tree is used to answer subarray sum queries ?? why are we using it here and please can you can just tell what is totcnt and forwdcnt ??
•  » » » » » » » 6 months ago, # ^ |   0 fenwik tree gives me sum from 0 to R if i query on R i.e. prefix sumforwdcnt stores=>count of elements from 1 to i=>that have been seen upto itotcnt will store=>initially count for all elements in the entire array=>then as we iterate through the array we decrement the totcnte.g. if my array is 1 1 1intitally forwdcnt[1] = 0 and totcnt[1] = 3after i = 1: forwdcnt[1] = 1 and totcnt[1] = 2after i = 2: forwrdcnt[1] = 2 and totcnt[1] = 1and so onhttps://codeforces.com/contest/459/submission/85599666 i think the code is easy and try tracing an example throught this code maybe??
•  » » » » » » 6 months ago, # ^ |   0 can you tell me the values of totcnt && forwdcntfor this test case(given in question) 7 1 2 1 1 2 2 1 
•  » » » » » » » 6 months ago, # ^ | ← Rev. 2 →   0 before loop totcnt[2] = 3 forwdcnt[2] = 0totcnt[1] = 4 forwdcnt[1] = 0after loop begins i totcnt forwdcnt 1 1: 3 1: 1 2: 3 2: 0 2 1: 3 1: 1 2: 2 2: 1 and so on
•  » » » » » » » » 6 months ago, # ^ |   0 thank you so much.
 » 5 years ago, # |   0 unable to figure out problem A! Anybody help plz?? I am weak to figure out co-ordinate problems!
 » 5 years ago, # | ← Rev. 2 →   0 In problem C , Can anyone explain what does 'sequence of d elements from 1 to k' means in the editorial and how did we arrive to k^d , also how the pigeonhole principle is helping us here? Thanks. EDIT :- I solved the problem. Nice Editorial and Nice problem, I got it now. Thanks anyways.
•  » » 3 years ago, # ^ |   0 each student can take k buses. for d students the number bus arrangements is k*k*k*....*k(till d) so k^d.
•  » » 2 years ago, # ^ |   0 Each student can choose from k buses each day, so for each student there are k^d options to choose from for d days. Now there are total n students and we have to distribute these k^d options among them, if n>k^d, then at least two students will get exactly the same sequence for d days
•  » » 8 months ago, # ^ |   0 The distinct options for choosing k buses in d days is k^d. One student will choose one option. If the number of student is greater then the number of option then there is minimum 2 student who choose same option.
 » 3 years ago, # |   0 Could someone please illustrate this part:For each student consider a sequence of d elements from 1 to k that shows the bus number which is taken by this student on each day. Obviously, there are kd different sequence at all,
 » 3 years ago, # |   0 Merge sort tree is getting MLE??Could someone tell me how to tackle this.My code : http://codeforces.com/contest/459/submission/36654409
 » 3 years ago, # |   0 Really good contest, with innovative problems. Here are all my solutions to the problems of this contest. I liked C, especially the Pigeonhole Principle and I liked D for the data structures. I wrote an editorial on it here.
 » 20 months ago, # |   0 How to solve Problem E using BFS/DFS? Can anyone explain me please?
 » 16 months ago, # |   0 In problem D, what does Fenwick tree store? and how is this tree used to find the answer?
 » 13 months ago, # |   0 For problem D, i compressed the input and calculate cnt1[i] and cnt2[i] as l_i and r_i in the editorial, but instead of using a fenwick tree i used a multiset and iterate j from 2 -> n. For each j, i inserted cnt1[j-1] to the multiset and used multiset::upper_bound to count how many numbers larger than cnt2[j]. Cplusplus.com said that both multiset::insert and multiset::upper_bound takes O(log n) time, so overall it will take O(n log n). However, my solution: 68404208 got TLE in test 10. Can someone please help me figure out what is wrong ? Maybe i miscalculated the complexity ?
•  » » 12 months ago, # ^ |   0 Just in case anyone having the same problem, using multiset is ok, but using std::distance is O(n) for non-random-access iterator => TLE
 » 12 months ago, # |   0 My D solution using BIT with important comments line which can help you to make understand this easily https://github.com/joy-mollick/Segment-Tree-And-Divide-And-Conquer/blob/master/Codeforces-D%20-%20Pashmak%20and%20Parmida's%20problem(BIT).cpp
 » 8 months ago, # |   0 Why cant we use PBDS was problem D? it is giving me runtime error.. Can anyone explain me why ? Link to my solution- Your text to link here...
•  » » 7 months ago, # ^ | ← Rev. 2 →   0 could you identify the error?i had encountered a similar problem on using PBDS but there memory constraints were given less clearly there it was 28 MB https://codeforces.com/contest/1354/problem/D here though that isnt the problem...
•  » » » 7 months ago, # ^ |   0 Use Fenwick tree instead
•  » » » » 7 months ago, # ^ |   0 i did it by FWT just asking why this didnt work :)
•  » » » » » 7 months ago, # ^ |   0 It requires much more of memory to work .. It have multiple points..Memory limit was given for the same
•  » » 7 weeks ago, # ^ |   0 I thought of a simple solution using multiset first but it got TLE. Then I learnt about PBDS from gfg and used unordered_map in my code and it got AC.However, the same program with map got TLE. I am really not sure what's happening inside as I need to read how the data structure works but anyways, you can see my solution if it helps. My solution: 100429402Idea: I created the ordered_set of pairs by iterating backwards (from last index) in the array taking count of i-th element and the element as pair. Now I start iterating from 0th index and remove the current element and its count from the set and find the number of elements less than current count of i-th element using order_of_key.
 » 7 months ago, # |   0 what's the use of fenwick in problem D?
 » 7 months ago, # |   0 In E, what if we're asked unique vertices (No, I am not curious, I interpreted the question wrongly, thought about it around 3 hours, then realized I had read the question wrongly).
 » 5 months ago, # |   0 How to solve E when the edges are undirected?
 » 4 months ago, # |   0 I tried D with seg tree, got a TLE at tc 9, any help would be appreciatedMy submission
 » 2 months ago, # |   0 D could be done even by persistent segment tree
 » 4 weeks ago, # |   0 Great work....