### 459A - Pashmak and Garden

Four vertices of a square with side length *a* (and sides parallel to coordinate axis) are in this form: (*x* _{0}, *y* _{0}), (*x* _{0} + *a*, *y* _{0}), (*x* _{0}, *y* _{0} + *a*), (*x* _{0} + *a*, *y* _{0} + *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 *cnt* _{1} * *cnt* _{2}, where *cnt* _{1} is the number of our maximum elements and *cnt* _{2} 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 *k* ^{ d} different sequence at all, so if *n* > *k* ^{ d}, 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* ≤ *k* ^{ d}, 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, 10^{6}]. Let *l* _{ i} be *f*(1, *i*, *a* _{ i}) and let *r* _{ i} be *f*(*i*, *n*, *a* _{ i}), we want to find the number of pairs (*i*, *j*) such that *i* < *j* and *l* _{ i} > *r* _{ j}. For computing *l* _{ i} s, 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, *l* _{ i} would be equal to *cnt*[*a* _{ i}] at position *i* ( *r* _{ i} 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 *l* _{ i} to answer and add *r* _{ i} 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

Thanks for the fast editorial!

We're sorry. Some unexpected events caused that.

Can you please include judge's code that will be great.

Included. ;)

i donot understand the E's solution cannot u explain in more details ??

in problem c it think it will be k^d not d^k

thanks for nice editorial.

How can you get the n * (n - 1) / 2 formula in B (otherwise than by wolfram alpha)?

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 arentotal elements. If we multiply we getn(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.Got it, thanks!

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.

you can't step in programmnig contests without this trivial knowledge :)

The solution of C(Pashmak and Buses) was awesome. Thanks a lot!!

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

Here is it. Look at

`/* That for-loop was modified */`

.I got it, Thanks :)

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..

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]) andsuf[i] =f(i,n,a[i]) for alli. Then you should simultaneously use merge-sort forprefandsuf. And then you can do the same algorithm as for inversions by merging notpref_{ l}andpref_{ r}orsuf_{ l}andsuf_{ r}butpref_{ l}andsuf_{ r}.I'll provide a code for this later.

And here it is: 7516132

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.

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 .

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

in problem C,can someone please explain what the author means by "n d-digits numbers in k-based numbers."??

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=3

first 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

thanks that was awesome

can you please explain your code for generating the sequence, i don't understand it

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

For problem:D..just use fenwick tree to solve it. its fun to solve it with fenwick tree :-) my submission :- 17314867

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?

i can try problem basically translates to finding all pairs such that

1) i < j 2) number of a[i] from 1 to i > number of a[j] from j to n

let count[a[i]] — count of a[i] from 1 to i

for 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 n

now 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 i

https://codeforces.com/contest/459/submission/85599666

Thank you so much for taking out time to answer noobs like me

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..?

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`

do u undertsand this part? i realised this part can explain more than my stupid english

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 ??

fenwik tree gives me sum from 0 to R if i query on R i.e. prefix sum

forwdcnt stores

=>count of elements from 1 to i

=>that have been seen upto i

totcnt will store

=>initially count for all elements in the entire array

=>then as we iterate through the array we decrement the totcnt

e.g. if my array is 1 1 1

intitally forwdcnt[1] = 0 and totcnt[1] = 3

after i = 1: forwdcnt[1] = 1 and totcnt[1] = 2

after i = 2: forwrdcnt[1] = 2 and totcnt[1] = 1

and so on

https://codeforces.com/contest/459/submission/85599666 i think the code is easy and try tracing an example throught this code maybe??

can you tell me the values of

`totcnt && forwdcnt`

for this test case(given in question)before loop totcnt[2] = 3 forwdcnt[2] = 0

totcnt[1] = 4 forwdcnt[1] = 0

after loop begins

and so on

thank you so much.

unable to figure out problem A! Anybody help plz?? I am weak to figure out co-ordinate problems!

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.

each student can take k buses. for d students the number bus arrangements is k*k*k*....*k(till d) so k^d.

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

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.

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,

Merge sort tree is getting MLE??Could someone tell me how to tackle this.My code : http://codeforces.com/contest/459/submission/36654409

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.

How to solve Problem E using BFS/DFS? Can anyone explain me please?

In problem D, what does Fenwick tree store? and how is this tree used to find the answer?

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 ?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

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

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...

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...

Use Fenwick tree instead

i did it by FWT just asking why this didnt work :)

It requires much more of memory to work .. It have multiple points..Memory limit was given for the same

what's the use of fenwick in problem D?

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).

How to solve E when the edges are undirected?