We will hold AtCoder Beginner Contest 164.

- Contest URL: https://atcoder.jp/contests/abc164
- Start Time: http://www.timeanddate.com/worldclock/fixedtime.html?iso=20200426T2100&p1=248
- Duration: TBD (around 2 hours)
- Number of Tasks: 6
- Writer: kyopro_friends, Kmcode, latte0119, EnumerativeCombinatorics, ynymxiaolongbao, wo01
- Rated range: ~ 1999

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

We are looking forward to your participation!

Addition: About the Unrated ABC163

We are sorry for the inconvenience.

The server down during the last contest was caused by a sudden access to the ALB, and we found that we could solve this problem by doing Pre-Warming.

Also, the bug where the problem statement does not show up has been resolved by changing the caching algorithm. The problem with submissions showing up as IE (Internal Error) was due to the Judge server's scoring algorithm being different than in the past. This too has already been fixed.

Just out curiosity, rating 1999 in Atcoder is roughly how much in CF?

2200-2300

Damm 2200 is a beginner

I think it's around 1800-1900 on cf

I'm 1400~1500 on Atcoder.

If 1999 on Atcoder means 1800~1900 on CF.

Than I'm 1200~1300 on CF.

Hello everyone I'm a beginner.

Common, it might not be your saturation level (might not have attended enough contest)

In AtCoder, when you participate in only a few contests your rating is much lower than your actual strength. You need to participate in at least 10 contests in order to get accurate rating.

I just participated in a few contests and now have around 1550. Are you sure what you're telling is true?

It will need a little more contest for your rating to be converged enough to your "actual" rating, because the rating shown is actually the lower bound of ninety-something-percent confidence interval of rating. Generally it needs about 14 times of participation to rated contests until the rating shown becomes high enough. (The "provisional" hint shown on the left of your rating also suggests this)

hope everything will be fine this time!

hope everyone will get full marks this round！

Hope it will never have problems(like 163) this time...

why, what was problem with problems of 163?

Just read this blog

I hate Matrix Construction.

I have given up.

F is too hard to solve for me.

Am I the only one who did a randomized solution for F?

My method was like this:

Code

Unfortunately your solution fails against the following test case. Your solution outputs

`-1`

, but a valid answer exists.Test CaseSimilar input with larger input leads to TLE too.

Ah, I see — this is one of the test cases with exactly one solution? Thanks for hacking my wrong method!

I initially expected that there would either be a lot of possible outputs or none at all. Turns out I was wrong :)

I like problem F last time. But I have to give up this time.

How did you think of D?

E is much difficult

Don't you realize you calculate the same pair twice

1-indexed, for

$$$1 \leq i \leq j \leq |S|$$$

instead of

$$$0 \leq i \leq j \le |S|$$$

Usually the discussion starts after the contest finished.

Haven't you seen the restrictions?

oh by mistake i have get the result with bruteforce with array starting indexing convention . so sorry for that

My first Atcoder contest...wow, this is beginner? I'm stuck on D. Feel like an idiot.

How to solve E?

it's easy to show that $$$5000$$$ silver coins is enough to go to every city,because you can go through every edge with them.so just break one city into $$$5000$$$ points $$$(city,money)$$$ and use dijkstra.

isnt the maximum 2500

$$$2450$$$

actually $$$2450$$$ is enough because the longest shorest path is not longer than $$$49$$$.

but i was silly so i used $$$5000$$$ XD

Just to be safe, I used up to 10000 silver coins, then did a state dijkstra.

10000 got AC? At first I also wrote

2 * sum of edges(10000 at worst) to be safe but got TLE, then changed tosum of edgesand got accepted.The edges will be of the order N*5000*5000 when all c[i] are 1. Will dijksta run in time for this number of edges?

You don't need that many edges. Just connect (node, coins) -> (node, coins + c[i])

No, the problem specifically says that the maximum number of edges is 100. Therefore, in the state graph the maximum number of edges is 100*5000 = 500000.

why so many edges?only $$$5000\times(n+2m)$$$ edges needed

yes, my mistake.

But can $$$O(n^4 logn)$$$ pass???

Shortest path -> state [N][K] -> minimum time to reach node N with silver coin K. we can take Maximum K as 2500. since the max cost of an edge is 50. and we will at most take N-1 edges to reach target.

I think my E is $$$O(A_{max}^2 * N^3)$$$. link to submission

could someone spot the mistake in my code for E? been losing sleep over it submission

nvm, fixed it.

how did you fix code? i met the same WA.

my INF was too small

How to solve D ?

The English Commentary for Problem D: Multiples of 2019 is here

Thanks a lot! That explanation helped me.

There must be simpler way, this is not level of D.

This may help. https://atcoder.jp/contests/abc164/submissions/12378162

for a string 2114 pref[0] would give us 4%2019 pref[1] would give 14%2019 pref[2] would give 114%2019 and pref[3] would give 2114%2019. right? then you afterwards what you implemented using map is unclear to me. please explain this.

It is just prefix sum logic.

Can anyone help with problem D?

I somehow arrived at if the range (i,j) should be divisible by 2019 then

$$${(1,j)}/{10^j} \equiv {(1,i-1)}/{10^{i-1}} (\mod 2019)$$$

where (i,j) is the base 10 representation of the number $$$S_i S_{i+1}....S_j$$$

Idk how to solve it further. Please help. I have also read the editorial and saw various solutions but I don't understand how are they reaching the final form?

thanks a lot

Bro i had a time complexity doubt since 10^8 takes 1 sec ,so 2019*2e5 =403800000 which should take 4 sec but soln is accepted.Can u explain ?

Hi, I think for simple operations, the order 9 per second will execute instead of order 8. So,I think that works, https://codeforces.com/blog/entry/17799?#comment-226617 have a look at their discussion!

thanks!!

How you solve D? I used brute force but it gave me TLE? Do you have any ideas?

Just construct number from start contiguously and store frequency of remainder. If $$$n$$$ is length of string at stage $$$i$$$ number will be $$$\,\,\,N\,= \,d_0*10^{n-1}+d_1*10^{n-2}+...+d_{i-1}*10^{n-i-1}$$$

let $$$R_i= N\mod2019$$$

So at any stage if remainder is $$$R_i$$$ see how many time this remainder has occurred previously excluding current one and add this to final result. For this you ca just use an array or Map whatever.

That is nice explanation, thanks!

I can't understand the logic that if for an index i,if the remainder is equal to the remainder at any other index j,then how the number formed by the elements between i& j would be divisible by 2019? Can someone please explain.

Can you please provide me the code of this approach, I'm getting WA.

Sure.this.

And to explain his doubt. let's say remainder $$$r$$$ is same for two indices $$$i\,and\,j\,(i<j)$$$ so number formed at index $$$i\,and\,j$$$ will be of the form $$$N_i=2019*k_1+r\,\, and\, N_j=2019*K_2+r$$$ so number between segment $$$[i,j]$$$ will be $$$(N_j-N_i)$$$ whose remainder got canceled out and we got $$$2019*(K_2-K_1)$$$ which is clearly divisible by 2019.

thanks man,that helped a lot

Can you please explain why your code to this input

`12019`

gives this output`2`

, but 12019 mod 2019 != 0?The number formed from i=2&j=2 is divisible by 2019,so there are 2 pairs:(1,5) &(2,2). https://atcoder.jp/contests/abc164/tasks/abc164_d In the constraints ,you can see i can be equal to j.

Thanks.

By the way all the digits are guaranteed to be between

`1`

and`9`

in the original problem.Thanks, my bad :(

My explanation + code

Take for example

`divisor = 13`

and`S = 39262`

.`2 / 13`

is`2`

, save it.`62 / 13`

is`10`

, save it.`262 / 13`

is`2`

, but we have already seen this reminder. What does it mean?`(262 - 2) % 13 = 260 % 13 = 0`

, so`260`

is a multiple of`13`

. But we are interested only in`26`

. Turned out if`x * (any power of 10) mod divisor = 0`

, then`x % divisor = 0`

if divisor isn't`2`

or`5`

.`39262 / 13`

reminder is again 2 and we have seen it two times:`(39262 - 262) % 13 == 0`

and`(39262 - 2) % 13 == 0`

, so add them both.Hey, thanks for the detailed explanation. This problem is similar to ABC(atcoder beginner contest) 158E but I'm getting WA in 158E problem which is a general case. Here's my code Can you help in that?

1st test is for P = 2 (you can see all test cases here). And for 2 and 5 "if x * 10^k % divisor = 0$, then x % divisor = 0" doesn't work. For example, 7 * 10 % 2 = 0, but 7 % 2 = 1.

So what how should I check divisibility for p=2 and p=5?

A number is divisible by 2 if and only if its lowest digit is even.

A number is divisible by 5 if and only if its lowest digit is 0 or 5.

Storing a prefix counter of the appearances of $$$S_i$$$ such that $$$S_i == 0 \; (mod \; P)$$$ for $$$P = 2$$$ or $$$P = 5$$$ is enough.

What is the name of this kind of alg?

It has a brilliant application of prefix sum algorithm https://www.geeksforgeeks.org/find-if-there-is-a-subarray-with-0-sum/

this helped me a lot. hope it helps

I got the "sum", but not the "multiple".

Am I mistaken in saying that today's ABC D has an identical solution to ABC 158E? The two problems seem to be asking for exactly the same thing.

Actually today's problem is easier, for the modulo is given as 2019, which is a relatively prime of the base, 10.

I remembered that I had seen it, couldnt solve it neither then nor now.

Same XD !!

We just need to take the extra case of p=2 and p=5 in problem 158E. Rest is all same:)

Are standings hidden ?

No

This is what I see during the whole contest and still it is showing the same.

Click on the dropdown to the right of "Customize" and see if you have overly restrictive filters.

You are right, Thanks.

What's the approach for D ? I used the stoi and substr function , got WA.

Hey! You can refer the commentary here

The problem F can be easily solved by maximum-flow with lower bounds. See this submission.

Nice one :)

can you please elaborate more i don't understand is this graph for the first sample? i understand that you solve for each bit independently but how do you add the edges?

I thought of this but I didn't have enough time to implement it :( Should it pass considering there can be around $$$64N^2$$$ vertices/edges?

There are only $$$\mathrm O(n)$$$ edges not weighted $$$1$$$. I think the Dinic algorithm still runs in $$$\mathrm O(m+\sqrt mn)=\mathrm O(n^2)$$$ time although I can't proof it.

My approach

This is my approach which got me TLE I pre-computed all the modulo values but to find (L, R) pairs I had to use a nested for loop. Can someone please tell me how do I find the number of (L,R) pairs in O(n) time ?

Hi, read problem 158E from this. https://img.atcoder.jp/abc158/editorial.pdf

It is almost similar but instead of

2019you have to solve it for generic numberP.My D approach

k starts from 0

Iterate from right and calculate (10^k%2019+digit at i-th index)%2019

store this in map and for every element in map calculate NC2 of that frequency

This problem is like calculating number of subarrays having sum%m as 0

can you please explain why this problem is like calculating number of subarrays having sum%m as 0 ??

You are given a array you calculate prefix sum at every i-th index mod m

Now if prefix sum at two index(suppose i & j) is same then there is a subarray from (i+1,j) whose sum will be zero. So we can store each sum in map and choose any two index having same sum in NC2 ways. In this question instead of prefix we calculate suffix sum + power of 10. And we can choose indexs in NC2 ways.

Video Tutorial for D

You could use vector.swap(v) instead of memcpy, which basically copies nothing but a pointer.

indeed, the runtime would have been even smaller, but at this point, it's still very fast.

Why $$$0 <= a_{i, j} < 2^{64}$$$ in problem F? For some weird unsigned long long practice? :)

I made me nervous that my codes were always CE. Actually,it's the "time" keyword that made this problem. Hope you all learn from my lesson.(How silly I was!) GL and HF.

Can someone explain easy to implement solution for E?

$$$dp[i][silver]$$$ = shortest path from $$$1$$$ to $$$i$$$ such that you're left with $$$j$$$ silver coins.

$$$silver$$$ is at most $$$2500$$$ — $$$(N * 50)$$$

I am not able to understand how 2500*50 states can be minimized by n*m*2500 iterations (n is vertex). If we see bellman ford , n*n iteration is required to minimize all n vertexes.

HELP!!!

For problem E:

I used dp.

$$$dp[i][j][k]$$$ means now standing at the $$$i-th$$$ node ,and have $$$Min(600,j)$$$ silver coins the $$$minimal$$$ time to go to k.

We can use $$$dp[i][j][k]$$$ to update other situation as follow.

And I used shortest path to update it,but I fail in the second test which is line(pass all of the others!),what's wrong?

https://atcoder.jp/contests/abc164/submissions/12395345

I am sorry I can't help you with the issue but why

600?I see

I am not sure of what you see but is good to see that you can see something.

I'm wrong , 600 is too small.Thank you very much.

Is 600 sliver coins enough?

No ,I fix it ,but I get TLE now...

Can you tell me if $$$O(n^4logn)$$$ can pass?

You can try the following :

dp[i][j] denotes minimum time to go to the ith node with j silver coins from 1.

Now repeatedly update this dp (n+1) times, as in bellman — ford. ( + 1 because initially I don't initialize it )

UPD : There is a case where this gets Wrong Answer. Sorry. Theoretically it requires O(m*m) iterations.

Can I see your code, thank you

Here you go

thanks bro

why iter n+1 times? for bellman-ford, it repeats at most n-1, plus 1 init should be n.

Thanks for pointing this. I tried to figure out what went wrong and I found that my solution is wrong as it requires O(n*n) iterations.

It passed due to weak testcase.

Testcases where I get WA : (An extension to 3rd sample case)

thanks bro. so do you mean the iteration should be num of edges? i'm curious how you figure it out? i have tried the whole day but no clues found.

It is O(n*n) because,

1) To reach node 1 to node t, You will buy coins O(n) times.

2) Among above O(n) cities, it takes O(n) edges to move to node where we buy the coins next.

Your submission can pass with given time complexity, but you are using much memory so it increases the time.

Thanks!

After I change it into $$$dp[i][j]$$$ ,it got $$$AC$$$

For problem F, without noticing $$$U,V<2^{64}$$$, I used

`long long`

instead of`unsigned long long`

and got WA for 2 times.Can I solve D recursively ?

Thanks for the quick English editorial !

my first contest at atcoder. the first three questions felt pretty easy.But got stuck on the D th problem if anyone can make a video tutorial on that.it would be a great help

This is another person's comment to the blog. He/She made one.

Not terribly important, but I'm just going to point out that there's a typo on English B where Takahashi's name is misspelled as Takashi once.

If we do not consider time efficiency, is it possible to use the differential constraint system to solve the F problem？ I used it but got wa

I wander why this submission on F failed in only one test

For question E, I have a test data that let my code that Accepted in Atcoder now to output wrong answers.

Data is: 3 2 1 1 2 1 2 1 3 2 4 1000000000 1 1000000000 1 1000000000 1

my code that Accepted in Atcoder now : https://atcoder.jp/contests/abc164/submissions/12437105

If row 111~115 is uncommented, you can use this data

Does E allow SPFA to pass

If anyone need explanation ,code and example for problem D here

Thanks!

I made video tutorial on first four problems . It may help you . Link

I accidentally found a "hack" for F; this solution is wrong but passes: submission

Error detailsLines 100 and 101 should have

`last`

, not`first`

.This test case produces an incorrect solution: