AtCoder Beginner Contest 164 Announcement

5 years ago, # ^ |

← Rev. 2 →

+34

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.

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Trgt_2021_explosion

5 years ago, # ^ |

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

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aramun_haesulla

5 years ago, # ^ |

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.

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okwedook

5 years ago, # ^ |

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

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WaterColor2037

5 years ago, # ^ |

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)

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towhid1zaman

5 years ago, # |

+16

hope everything will be fine this time!

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ShineEternal

5 years ago, # |

hope everyone will get full marks this round！

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mycontest

5 years ago, # |

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

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aramun_haesulla

5 years ago, # ^ |

+10

why, what was problem with problems of 163?

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solvemproblr

5 years ago, # ^ |

Just read this blog

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5 years ago, # |

+59

I hate Matrix Construction.

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5 years ago, # ^ |

+13

I have given up.

F is too hard to solve for me.

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box

5 years ago, # ^ |

← Rev. 2 →

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

My method was like this:

First, decompose the matrix into 64 layers, with one layer for each bit, since bits on different layers don't affect each other.
Then, for each enforcement that's "AND = 1" or "OR = 0", set the corresponding row/column to 1 or 0.
If there are collisions (one cell set to both 1 and 0) then immediately return impossible,
Otherwise, for each of the cells that were neither assigned 0 nor assigned 1, randomly assign it 0 or 1.
Check if the resulting matrix fulfills requirements.
If not, randomly assign them again. If you have assigned them enough times (100000) and each time it failed, return impossible,

Code

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WaterColor2037

5 years ago, # ^ |

Unfortunately your solution fails against the following test case. Your solution outputs -1, but a valid answer exists.

Test Case

50
0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

Similar input with larger input leads to TLE too.

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box

5 years ago, # ^ |

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 :)

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5 years ago, # ^ |

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

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Lmntrix11

5 years ago, # |

How did you think of D?

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Abhinandan_Arya

5 years ago, # |

← Rev. 2 →

-94

E is much difficult

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5 years ago, # ^ |

+22

Don't you realize you calculate the same pair twice

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5 years ago, # ^ |

1-indexed, for

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

instead of

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

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5 years ago, # ^ |

+11

Usually the discussion starts after the contest finished.

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5 years ago, # ^ |

Haven't you seen the restrictions?

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Abhinandan_Arya

5 years ago, # ^ |

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

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HBM88

5 years ago, # |

+14

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

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towhid1zaman

5 years ago, # |

+17

How to solve E?

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5 years ago, # ^ |

+18

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.

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errorgorn

5 years ago, # ^ |

isnt the maximum 2500

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5 years ago, # ^ |

+19

$$$2450$$$

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5 years ago, # ^ |

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

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5 years ago, # ^ |

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

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Agnimandur

5 years ago, # ^ |

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

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ExplodingFreeze

5 years ago, # ^ |

10000 got AC? At first I also wrote 2 * sum of edges (10000 at worst) to be safe but got TLE, then changed to sum of edges and got accepted.

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raja1999

5 years ago, # ^ |

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?

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cheissmart

5 years ago, # ^ |

← Rev. 4 →

+32

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

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Agnimandur

5 years ago, # ^ |

-19

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.

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5 years ago, # ^ |

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

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raja1999

5 years ago, # ^ |

yes, my mistake.

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5 years ago, # ^ |

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

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ILoveBitches

5 years ago, # ^ |

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.

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cheissmart

5 years ago, # ^ |

← Rev. 2 →

+19

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

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hitch

5 years ago, # ^ |

← Rev. 2 →

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

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hitch

5 years ago, # ^ |

nvm, fixed it.

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haishand

5 years ago, # ^ |

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

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hitch

5 years ago, # ^ |

my INF was too small

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Ugnwd

5 years ago, # |

How to solve D ?

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aj54

5 years ago, # |

+18

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

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ritik_patel05

5 years ago, # ^ |

Thanks a lot! That explanation helped me.

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5 years ago, # ^ |

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

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dalip_patra

5 years ago, # ^ |

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

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jhjfbsbfkbjfnfnfjfj

5 years ago, # ^ |

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.

map<ll,ll>mp;
ll ans = 0;
for(ll i=0;i<n;i++)
{
    if(pref[i]==0)ans++;
    ans+=mp[pref[i]];
    mp[pref[i]]++;
}

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dalip_patra

5 years ago, # ^ |

It is just prefix sum logic.

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aKILLL

3 years ago, # ^ |

← Rev. 2 →

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?

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youtsuha

5 years ago, # ^ |

thanks a lot

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Chodermal1

5 years ago, # ^ |

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 ?

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aj54

5 years ago, # ^ |

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

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Chodermal1

5 years ago, # ^ |

thanks!!

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aziz17_01

5 years ago, # |

-8

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

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Trgt_2021_explosion

5 years ago, # ^ |

← Rev. 3 →

+10

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.

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5 years ago, # ^ |

That is nice explanation, thanks!

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shahil_005

5 years ago, # ^ |

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.

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Matuagkeetarp

5 years ago, # ^ |

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

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Trgt_2021_explosion

5 years ago, # ^ |

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.

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gaurav1903

5 years ago, # ^ |

thanks man,that helped a lot

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akulenok

5 years ago, # ^ |

Can you please explain why your code to this input 12019 gives this output 2, but 12019 mod 2019 != 0?

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shahil_005

5 years ago, # ^ |

← Rev. 2 →

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.

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akulenok

5 years ago, # ^ |

Thanks.

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WaterColor2037

5 years ago, # ^ |

By the way all the digits are guaranteed to be between 1 and 9 in the original problem.

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akulenok

5 years ago, # ^ |

Thanks, my bad :(

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alex223

5 years ago, # ^ |

← Rev. 6 →

My explanation + code

Take for example divisor = 13 and S = 39262.

Reminder of 2 / 13 is 2, save it.
Reminder of 62 / 13 is 10, save it.
Reminder of 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.

For 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.

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Matuagkeetarp

5 years ago, # ^ |

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?

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alex223

5 years ago, # ^ |

← Rev. 2 →

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.

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Matuagkeetarp

5 years ago, # ^ |

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

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Hepic_Antony_Skarlatos

5 years ago, # ^ |

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.

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a14789654

5 years ago, # ^ |

← Rev. 4 →

What is the name of this kind of alg?

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gaurav1903

5 years ago, # ^ |

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

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a14789654

5 years ago, # ^ |

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

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shenxy13

5 years ago, # |

+54

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.

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JKLover

5 years ago, # ^ |

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

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lemongrab

5 years ago, # ^ |

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

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dunjen_master

5 years ago, # ^ |

Same XD !!

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PRESS_F_2_PAY_RESPECT

5 years ago, # ^ |

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

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5 years ago, # |

← Rev. 2 →

-18

Are standings hidden ?

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towhid1zaman

5 years ago, # ^ |

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5 years ago, # ^ |

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

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Norrius

5 years ago, # ^ |

+20

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

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5 years ago, # ^ |

You are right, Thanks.

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kashita

5 years ago, # |

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

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aj54

5 years ago, # ^ |

Hey! You can refer the commentary here

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amane-ame

5 years ago, # |

+68

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

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amane-ame

5 years ago, # ^ |

← Rev. 2 →

+55

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pavel.guskov

5 years ago, # ^ |

Nice one :)

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O_E

5 years ago, # ^ |

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?

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spar5h

5 years ago, # ^ |

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?

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amane-ame

5 years ago, # ^ |

← Rev. 2 →

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.

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wingman__7

5 years ago, # |

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 ?

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lohit_97

5 years ago, # ^ |

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

It is almost similar but instead of 2019 you have to solve it for generic number P.

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saksham9575

5 years ago, # |

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

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Ahmed_Allawati

5 years ago, # ^ |

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

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saksham9575

5 years ago, # ^ |

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.

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stefdasca

5 years ago, # |

Video Tutorial for D

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5 years ago, # ^ |

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

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stefdasca

5 years ago, # ^ |

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

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ADJA

5 years ago, # |

+62

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

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2018LZY

5 years ago, # |

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.

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praktice

5 years ago, # |

Can someone explain easy to implement solution for E?

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stefdasca

5 years ago, # ^ |

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

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

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siddharthabhi30

5 years ago, # ^ |

← Rev. 2 →

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.

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

5 years ago, # |

← Rev. 3 →

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.

$$$dp[i][j][k]+d[i]$$$ to update $$$dp[i][max(j-c[i],0)][k] $$$
$$$dp[i][j][k]+(the—time—from—it—to—i)$$$ to update $$$dp[it][Min(600,j+(the—number—of—silver—coins—required—from—it—to—i))][k] $$$

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?

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lohit_97

5 years ago, # ^ |

I am sorry I can't help you with the issue but why 600 ?

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5 years ago, # ^ |

I see

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lohit_97

5 years ago, # ^ |

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

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5 years ago, # ^ |

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

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JKLover

5 years ago, # ^ |

Is 600 sliver coins enough?

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5 years ago, # ^ |

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

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

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5 years ago, # ^ |

← Rev. 2 →

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.

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ccsu_deer

5 years ago, # ^ |

Can I see your code, thank you

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5 years ago, # ^ |

Here you go

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ccsu_deer

5 years ago, # ^ |

thanks bro

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haishand

5 years ago, # ^ |

← Rev. 2 →

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

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5 years ago, # ^ |

← Rev. 3 →

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)

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haishand

5 years ago, # ^ |

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.

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5 years ago, # ^ |

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.

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5 years ago, # ^ |

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

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