We will hold AtCoder Beginner Contest 134.

- Contest URL: https://atcoder.jp/contests/abc134
- Start Time: http://www.timeanddate.com/worldclock/fixedtime.html?iso=20190720T2100&p1=248
- Duration: 100 minutes
- Number of Tasks: 6
- Writer: yuma000, drafear, DEGwer, evima, gazelle, potetisensei
- Rated range: ~ 1999

The point values will be 100-200-300-400-500-600.

We are looking forward to your participation!

What's the approach for E . I gave it an hour but still not able to figure out

https://www.geeksforgeeks.org/minimum-number-of-increasing-subsequences/

i did exactly same link, but it gave me tle , but then later i changed used maps it gave AC. map link Do you know why ?

use container.lower_bound to pass in 50 ms

https://codeforces.com/blog/entry/20553

thanks :)

https://atcoder.jp/contests/abc134/submissions/6476441

I used vector and it gave me TLE. Is multiset faster than vector as both access data in the same way.Can you please help? I'm kind of a noob.

erase takes O(n) time in vector compared to O(logn) in multiset

vector::erase take O(n), but multiset::erase take O(logn) in case you want to erase one element. I used vector and it gave me AC.

Here's my submission.

Good article, learned a new thing. but why answer will be the size of longest decreasing subsequence.

proof!

It is not. It is the number of subsequences, not the length of the longest. Edit: The minimum number of...

but what it mean

"If we focus on the example we can see that the Minimum number of increasing subsequences equals to the length of longest decreasing subsequence where each element from the longest decreasing subsequence represents an increasing subsequence"

it is written in g4g article.

For better understanding, look for Dilworth's Theorem. it says that the size of a maximal antichain equals the size of a minimal chain cover of a partial ordered set

Use multiset in C++ STL

when considering Ai,if ther are such j that Aj<Ai(Aj is the last one in another array),choose the MAX Aj. if you don't you may waste smaller one ,smaller one is more useful than bigger one . to do this you can use multiset

Iterate from a1 to an. For each element a_i color it with the color which has the largest last element smaller than a_i. If there isn't any, color it with a new color.

Greedy works for E. Make another array, X.

In step $$$1$$$, place $$$A_1$$$ in X. In step $$$i$$$, check if $$$X$$$ contains an element that is strictly less than $$$A_i$$$. If there exists such an element in $$$X$$$, replace the first such element in $$$X$$$ with $$$A_i$$$ else append $$$A_i$$$ to $$$X$$$.

The answer will be the final size of $$$X$$$.

Note that the array $$$X$$$ will always remain sorted and hence you can use binary search in each step, giving a final time complexity of $$$O(n \log n)$$$.

You can visualise it as stacking number of discs on top of each other, the array $$$X$$$ contains only the top of each stack, the "representative" element. Each new stack corresponds to a new colour and we can add an element to a stack only if it is larger than every element in that stack. However since the representative element is the largest element of the stack you just need to check if the element to be added is larger than the representative element of the stack.

can u post submission link.

and can u tell why the answer will always be length of longest decreasing subsequence?

Here's my submission.

As for a proof of this, I couldn't come up with one which doesn't require Dilworth's Theorem or replicated a proof of Dilworth's Theorem.

Consider the set

Now for any $(A_j, j), (A_i, i) \in S$, we say $$$(A_i, i) < (A_j, j)$$$ if and only if $$$A_i < A_j$$$ and $$$i < j$$$. Then by Dilworth's Theorem on this partially ordered set, we get our result.

For a longer explanation, our problem now becomes to find the minimal partition $$$S_1, S_2, \dots, S_r$$$ of $$$S$$$ such that for any $$$a, b \in S_i$$$ $$$a < b$$$. Such a partition is called a chain decomposition of $$$S$$$. Our aim is to find the size of the smallest chain decomposition of $$$S$$$. Dilworth's theorem states that the size of the smallest chain decompostion is equal to the size of the largest anti-chain. An antichain of a partially ordered set is a set of elements of the poset such that for any two element $$$a, b$$$ of the antichain neither $$$a < b$$$ or $$$b < a$$$. From our setup, an antichain would be a decreasing sequence and hence the longest antichain will be the longest decreasing sequence. Now it's not hard to prove that the above algorithm does indeed give the longest decreasing subsequence.

You can read more about Dilworth's Theorem and partial orders here or on its wikipedia page.

Thanks islingr greedy proofs are like solving problems of IMO. : )

You can prove it by coloring each integer according to the length of the longest non-increasing subsequence ending at that integer. If $$$i < j$$$ and $$$A_i \geq A_j$$$, the longest non-increasing subsequence ending at $$$A_j$$$ is strictly longer than the longest ending at $$$A_i$$$.

The problem gets reduced to finding longest non increasing sub-sequence of the given sequence. The reason being Dilworth's Theorem

When will the answer to the D problem be -1. i.e when will we not have a good set of choices?

Never

Answer is never -1.

can you explain your approach for D..(Preparing Box) I was thinking that the answer will be the same array which is given to us as input? Please correct me if I am wrong

For indices greater than $$$n/2$$$, whether to put a ball or not is decided by $$$a[i]$$$ itself (because there are no more multiples of it less than $$$n$$$). For the remaining ones, run a loop from $$$n/2$$$ to $$$1$$$. Sum all the box-values of indices which are multiples of the current index. If the parity of the sum and a[i] are same, then we need not put any ball in that box. Else, we need to put a ball in the box.

I did the same thing but I am getting WA for testcase_12,testcase_4 ,testcase_5,testcase_6 .

Can I see your submission, please?

The answer always exist.

There is always a good set of choice.

But how can we work it out? I've tryed 5 times but failed.

Never.

If $$$x_i$$$ is the number of balls in the box $$$i$$$. The problem is equivalent to finding equivalent to solving a system of $$$n$$$ linear equations in $$$x_1, x_2, \dots, x_n$$$ modulo $$$2$$$.

The matrix of this sysetm of linear equation was upper triangular and all the diagonal elements were non-zero and hence it will always have a solution. The proof of this is the algorithm used to solve the problem. You can find $$$x_i$$$ by simply setting

Then simply iterate from bottom to top, first find $$$x_n$$$, then $$$x_{n - 1}$$$, then $$$x_[n - 2}$$$ and so on... This is known as back substitution.

The time complexity will be

For any with box you always have the option to make the total ball in its multiples odd or even depending on whether you fill the with box with ball or not. So answer can never be -1

What's the solution for F?

One solution is to OEIS : http://oeis.org/A062869/b062869.txt

The other solution is to use DP.

What's the recurrence relation for DP?

ABC like 15 minutes...then no idea at all. Gap from C to D much to big.

and D to E ?

idk, thats kind of out of my eysight ;) People solving D must tell.

Well for me D was not that easier but doable as compared to E, I understood what to do in E but due to my shortage of knowledge in STLs, I couldn't implement E but I did D in contest.

What I did was, I ran a first loop from n to 1 and then inside this loop I ran another loop from the last multiple of i nearest to n and then came to i. Something like

`for(int j=last_multiple;j>=i;j-=i)`

and then counted the multiples which are filled with ball or not if the count was of same parity with ith box I needed to do nothing but if the parity was different I needed to fill the ith box with a ball

My solution to D

i found E easier than D, possibly because E's solution came intuitively and i wrote buggy D

D was just a simple brute force.

After like 15 minutes trying to understand the bony language I decided to work on E. After another like 30 minutes I found that I need to count the number of inc subsequences, but had absolutly no idea how to do it in time limit. So I switched back to D... but still was not able to see the simple implementation needs.

How to solve problem F?

F anyone? ?

How to solve D please :)

For D, you can start from N and go backwards till 1, at each stage check the parity of all multiples of the current number which are <=N, and accordingly set or don't set the bit for the current number depending on Ai.

You mean check all the divisors of the current number...then I guess its complexity must be O(n*sqrt(n)) Please correct me if I am wrong

yes, its brute force. start from last element and check for each multiple of i.

N + N/2 + N/3 + N/4 + ... + 1 ..

https://www.geeksforgeeks.org/sum-series-n1-n2-n3-n4-nn/

Actually it's O (N log N), since the amount of iterations you have to do is $$$\sum_{i = 1}^{N} N / i = N \cdot \sum_{i = 1}^{N} 1 / i$$$, which is just N * Harmonic Sequence, which is O (N log N).

Not sure if this is an amateur question, but how is the harmonic series = log N?

I'm not sure if I'll be able to explain this clearly but I'll try.

The Summation of elements of the Harmonic Sequence is: $$$S_1 = 1 / 1 + 1 / 2 + 1 / 3 + 1 / 4 + 1 / 5 + 1 / 6 + 1 / 7 + 1 / 8 \dots + 1 / n$$$

Suppose we generate another sequence bigger than the Harmonic Sequence by rounding the denominators down to their nearest power of 2. The summation of this sequence will be: $$$S_2 = 1 / 1 + 1 / 2 + 1 / 2 + 1 / 4 + 1 / 4 + 1 / 4 + 1 / 4 + 1 / 8 \dots + 1 / n$$$

The value of $$$S_2$$$ is close to $$$\log_2 N$$$ since the amount of elements required to get to the next integer value doubles for every integer. Since $$$S_1 \leq S_2$$$, we conclude that $$$S_1$$$ is also close to $$$\log_2 N$$$.

EDIT: I believe this is explained better on an Algorithms Live! video, but I can't recall which one.

Understood, in fact that's a fairly standard proof, it's just that it's been a while since I've dealt with this kind of math so I had to jog my memory.

Thank you!

How to look at other's solutions at AtCoder?

click submissions and then go to all submissions.

in standings then click on multiplying glass near the name

Thank you..

I tried to solve F using DP but it only passed 30 test cases out of 50 other 20 gave TLE. Is there a solution to the problem which does not involve dp or what were the states of the dp which passed all the test cases?

Could you tell me your DP solution please? I haven't any idea about it.

Basically my dp had 3 states which were the current index, bitmask which stores the values already used and the third state was the sum which had already been created using the already used values. And then the transitions were pretty simple after that.

Thanks.

Here's everything about Problem F: https://arxiv.org/pdf/1202.4765.pdf

My code for problem C returns TLE for the second half of the test cases (probably longer inputs?), any advice to make it more efficient?

I only know the basics of C++, I have looked at a few of others' submissions, but they used vector lists. I know what they are but I haven't learnt how to use them in code yet, so I prefer something I could code & understand.

Thanks.

All you need to do is find the largest and second largest numbers. The second largest number can be equal to the largest number. Thus if the current number is equal to the largest number, then print the second largest number; else, print the largest number.

Here is my code: https://atcoder.jp/contests/abc134/submissions/6467493

That helped. Thanks!

https://pastebin.com/NLHwAzXC for problem d i did the same as the nomral solution but instead of looping trough all the multiples of i it loops through primes i and sum up prime * i and decide to put 1 or no but it givs wa

just bcoz ur approach is wrong. why r u asking the wrong approach to be a correct.

why its wrong

why not to apply dp on seg tree problems. its wrong so its wrong . if u want correct one simply take all multiples of i. tc will be nlogn.

taking primes only should do the same as taking multiplis

I have the same idea. https://atcoder.jp/contests/abc134/submissions/6466295

Get WA too. I still do not know why either. May someone help us?

I figure out why finally. If you loop through primes, you overlap some cases. For example, you multiply 2 and 3, both of which contain the balls in 6. So the solution multiply primes is wrong.

Is my answer clear? Reply me if there is any problem.

Firstly, I taught that F can be solved using dp[i][j] — number of permutations using numbers from 1 to i having the permutation's depth j, but I figured out that is not enough. Can anybody explain the dynamic with 3 states?

I tried problem E by applying LIS twice. But I keep getting wa. Any idea what's the problem?

Submission-wa

Look at this 12 1 1 2 2 3 3 3 3 2 2 1 1 Answer:8 Output:9 Expect 8 find 9

Thanks! Got the issue

Can anybody please explain the state of DP used to solve problem F.

The editorial in japanese says that we should assume the problem as pairing task.

The state of dynamic can be

then we can update the state like dp[i+1][j][k+j] = dp[i][j][k] in O(n^4).

My code here https://atcoder.jp/contests/abc134/submissions/6504932

chokudai please write an editorial explaining F

chokudai Editorial please.

Editorial is out: hereYou can find the editorials on Twitter

Is English editorial available for this contest?

For

ABC, there are no English editorials.Can someone please translate editorial for problem F in English?

You can use google translate anytime! It works.

If you find official editorial and AC submissions for F is so damn painful to look at, you can try to find out the idea in this editorial. Although it's in Japanese but it has an really intuitive example picture and you can use Google Translate to get the idea.

thanks man.

This one helps a lot, Thanks! it's so cool to reshape it into another equivalent solvable one. after the reshape, the transition of DP becomes comparatively clearer.

hi guys i didn't get with C he says WR here is my code in python 3.x l=[int(input()) for _ in range(int(input()))] for i in l: g=[j for j in l if j!= i] if g ==[]: print(*l,sep="\n") break else: print(max(g))