AtCoder Beginner Contest 350 Announcement

7 месяцев назад, # |

wtf this tasks

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7 месяцев назад, # ^ |

-6

Exactly my thoughts. A-F is criminally bad. Idk about G.

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7 месяцев назад, # ^ |

What do you find bad about them ?

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7 месяцев назад, # ^ |

They are trivial and way too standard. I only read G 5 mins ago and it's terrible as well, I can't spoil the solution rn but it's a standard optimization, I regret not reading it earlier.

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7 месяцев назад, # ^ |

← Rev. 2 →

Today, I gave my first atcoder contest and was able to solve 4. How many problems are you able to solve in those contests, generally ? And, how many did you solve today ?

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7 месяцев назад, # ^ |

I know ideas to 6 on average, today all 7 (but didn't code G). I usually don't solve <5.

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7 месяцев назад, # ^ |

Thanks

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hhc0716

7 месяцев назад, # ^ |

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Lotrick

7 месяцев назад, # |

Loved E

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jiya102082004

7 месяцев назад, # |

E — Toward 0 Problem Statement: You are given an integer N. You can perform the following two types of operations:Pay X yen to replace N with ⌊ A N ⌋.Pay Y yen to roll a die (dice) that shows an integer between 1 and 6, inclusive, with equal probability. Let b be the outcome of the die, and replace N with ⌊ b N ⌋.

i didnt understand this question much but the testcases made it worst :(

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7 месяцев назад, # |

Is G just small to large merging?

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celin

7 месяцев назад, # ^ |

Yes. My solution is: For each node I calculated its parent and depth in its component(a tree). Also I used union find to check if two nodes are in the same component. Notice that you can answer each query using calculated values. To merge two nodes, I recalculated the depth and parent of every node in the smaller component out of the two.

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7 месяцев назад, # ^ |

Could you help me find out why this solution gives RTE, TLE and WA? It seems to be what you described.

Submission Link

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DC_H

7 месяцев назад, # ^ |

By a close look I think you should update ancestors inside of dfs on all nodes.

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7 месяцев назад, # ^ |

I don't think so, I'm updating only the smaller child when merging.

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DC_H

7 месяцев назад, # ^ |

What I did in my solution is I updade all parents on smaller tree + all depths and when considering a query 2 u v I take the deeper node by depth let it be v and I check if parent[v] is adjacent to u. I use sets in adjacency list to make that quick.

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7 месяцев назад, # ^ |

← Rev. 2 →

I remember who the parent is for every node + keep a DSU to tell which component is smaller for merging. When merging I update the parents in the smaller tree so that the node from current query becomes its root. The runtime is 24 ms though, so I'm not sure if it's even necessary to merge small to large here?

up: Resubmitted with random swaps and got TLE, so small to large seems necessary.

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7 месяцев назад, # ^ |

← Rev. 2 →

sqrt also passed: split vertices to small an big by current deg. for small store all pairs of neighbours (will be $$$O(n \sqrt n)$$$ total) in set.

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7 месяцев назад, # ^ |

Thanks for the solution! Do you think the small memory limit was designed to kill sqrts, or for something else?

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7 месяцев назад, # ^ |

I guess to kill bitsets

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aryansaprajiit

7 месяцев назад, # |

← Rev. 2 →

For question D, I tried to do it with DSU but failed 16 cases don't know why.

My implementation was like:

long long find_parent(ll u,vector&parent) {

if(u == parent[u]) {
    return parent[u];
}
return parent[u] = find_parent(parent[u],parent);

}

// For union

void union_set(ll u,ll v,ll parentU,ll parentV,vector&parent,vector&ranks,ll &ans) {

ll ru = ranks[parentU];
ll rv = ranks[parentV];
if(ru<rv) {
    swap(parentU,parentV);
    swap(rv,ru);
}
ans = ans + (ru-1)*rv + rv-1;
parent[parentV] = parentU;
ranks[parentU] += ranks[parentV];

}

Any idea what wrong I did?

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alagorithmet

7 месяцев назад, # ^ |

i can't read your code, but there can be more than one connected component, which you may have missed.

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aryansaprajiit

7 месяцев назад, # ^ |

I considered that case also, and I have updated my code now.

That ans = ans + (ru-1)*rv + rv-1; is handling those cases in which when there are multiple components.

1 is connected with 2,3,4. So, 1's rank will be 4.

6 is connected with 7,8. So, 6's rank will be 3.

Now I want to connect 6 with 1. Now, this formula will do its job.

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Thanos_I_am_inevitable

7 месяцев назад, # ^ |

I was not able to understand D. In the last sample test case it is showing 12 as answer. can someone please explain how 12? I am getting more than 12. If i make 1 and 3 as friend and 3 and 4 are already are friend then i can make 1 and 4 friend as well.

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aryansaprajiit

7 месяцев назад, # ^ |

To understand this question, you need to know DSU's working when implemented with path compression and rank optimization.

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Thanos_I_am_inevitable

7 месяцев назад, # ^ |

ok thanks. Let me check dsu.

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aritra2zk9l

7 месяцев назад, # ^ |

← Rev. 2 →

Actually a simple dfs works:- For any connected component find the no of vertices(v) and edges(e) it has. Then the answer is simply incremented by (vC2-e) where vC2=(v*(v-1))/2 for each component.

My Submission https://atcoder.jp/contests/abc350/submissions/52589120

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Slazenger

7 месяцев назад, # ^ |

Your ideal is similar to mine, When there is a pair in so you have to subtract the answer which is 1. My code https://atcoder.jp/contests/abc350/submissions/52618619.

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Actium

7 месяцев назад, # |

How to solve A and B for just 47 seconds? I could not even open the tasks for that time :)

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more-

7 месяцев назад, # ^ |

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brave-kid

7 месяцев назад, # |

+10

how to solve $$$E$$$?

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7 месяцев назад, # ^ |

Just DP/memoization, the only problem is with dice roll possibly giving you a 1, but that just multiplies the expected number of moves when choosing to roll a die by $$$\frac{6}{5}$$$.

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megaminion

7 месяцев назад, # ^ |

I thought that this solution would get TLE so I didn't implement it.

How can we prove that the states number is small enough?

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nu50218

7 месяцев назад, # ^ |

As written in Japanese editorial, any state to consider can be expressed in a form $$$\lfloor N / (2^p 3^q 5^r) \rfloor$$$, where $$$p, q, r$$$ are non-negative integers, since $$$\lfloor \lfloor N / x \rfloor / y \rfloor = \lfloor N / xy \rfloor$$$. Each of $$$p, q, r$$$ have only $$$O(\log N)$$$ possibilities, so in total the number of states is $$$O(\log^3 N)$$$.

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Lotrick

7 месяцев назад, # ^ |

← Rev. 3 →

DP calculating cost to reach 0 from current number. Min of 2 ways:

paying X yen for first way

X + f(num // A)

paying Y yet to throw the dice

(6 * Y + f(x // 2) + f(x // 3) + f(x // 4) + f(x // 5) + f(x // 6)) / 5

You get this from

f(x) = Y + (f(x // 1) + f(x // 2) + f(x // 3) + f(x // 4) + f(x // 5) + f(x // 6)) / 6

To get rid of the infinite recursion, move all f(x)'s to the left

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7 месяцев назад, # ^ |

← Rev. 2 →

The last equation should be —

f(x) = Y + (f(x // 1) + f(x // 2) + f(x // 3) + f(x // 4) + f(x // 5) + f(x // 6)) / 6

Instead of,

f(x) = (Y + f(x // 1) + f(x // 2) + f(x // 3) + f(x // 4) + f(x // 5) + f(x // 6)) / 6

To arrive at the equation —

f(x) = (6 * Y + f(x // 2) + f(x // 3) + f(x // 4) + f(x // 5) + f(x // 6)) / 5

P.S. — You have put Y in bracket and divided by 6 in the last equation.

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Akash_singh11

7 месяцев назад, # ^ |

can you please elaborate more ?

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It seems that someone's brute force passed G.

7 месяцев назад, # ^ |

← Rev. 2 →

Let F(x) be the cost of reducing x to 0. When rolling a dice, we will pay Y cost. The dice on rolling will give either — 1,2,3,4,5 or 6

If it is 1, the cost will be F(x) = Y+F(x/1)
If it is 2, the cost will be F(x) = Y+F(x/2)
If it is 3, the cost will be F(x) = Y+F(x/3)
If it is 4, the cost will be F(x) = Y+F(x/5)
If it is 5, the cost will be F(5) = Y+F(x/5)
If it is 6, the cost will be F(x) = Y+F(x/6)

But, we need to find expectation value of F(x), which is essentially the mean. Hence,

F(x) = (6*Y+F(x/1)+F(x/2)+F(x/3)+F(x/4)+F(x/5)+F(x/6))/6
F(x) = Y+(F(x/1)+F(x/2)+F(x/3)+F(x/4)+F(x/5)+F(x/6))/6

Note, that there is a F(x) term on both LHS and RHS (This is because F(x/1)=F(x),as dividing x by 1, will give x)

Move the F(x) in the RHS to the LHS side, we get —

F(x)-F(x)/6 = Y+(F(x/2)+F(x/3)+F(x/4)+F(x/5)+F(x/6))/6
Thus, 5*F(x)/6 = Y+(F(x/2)+F(x/3)+F(x/4)+F(x/5)+F(x/6))/6
Hence, F(x) = (6*Y+F(x/2)+F(x/3)+F(x/4)+F(x/5)+F(x/6))/5

Now, this can be solved using Recursion+Memoization. My Submission

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Akash_singh11

7 месяцев назад, # ^ |

thanks a lot man. I was really having hard time solving.

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meanup-

7 месяцев назад, # ^ |

← Rev. 2 →

Thank you, Even I'm doubtful initially about cancelling out $$$f(\left\lfloor \frac{N}{1}\right\rfloor)$$$ but later saw your comment and felt good that I'm on track.

In the editorial, it was

$$$f(N) = Y + \frac{1}{5} f(\left\lfloor \frac{N}{2} \right\rfloor) + \frac{1}{5} f(\left\lfloor \frac{N}{3} \right\rfloor) + \frac{1}{5} f(\left\lfloor \frac{N}{4}\right\rfloor ) + \frac{1}{5} f(\left\lfloor \frac{N}{5}\right\rfloor) + \frac{1}{5} f(\left\lfloor\frac{N}{6}\right\rfloor)$$$

$$$ f(N) = Y + \frac{1}{6} f(\left\lfloor \frac{N}{1} \right\rfloor) + \frac{1}{6} f(\left\lfloor \frac{N}{2} \right\rfloor) + \frac{1}{6} f(\left\lfloor \frac{N}{3}\right\rfloor ) + \frac{1}{6} f(\left\lfloor \frac{N}{4}\right\rfloor) + \frac{1}{6} f(\left\lfloor\frac{N}{5}\right\rfloor) + \frac{1}{6} f(\left\lfloor \frac{N}{6} \right \rfloor)$$$

Although it appears recursive due to $$$f(N)$$$ appearing on the right side, by shifting to the left and multiplying the entire equation by $$$\frac{6}{5}$$$, we get:

$$$ f(N) = \frac{6}{5}Y + \frac{1}{5} f(\left\lfloor \frac{N}{2} \right\rfloor) + \frac{1}{5} f(\left\lfloor \frac{N}{3}\right\rfloor ) + \frac{1}{5} f(\left\lfloor \frac{N}{4}\right\rfloor) + \frac{1}{5} f(\left\lfloor\frac{N}{5}\right\rfloor) + \frac{1}{5} f(\left\lfloor \frac{N}{6} \right \rfloor)$$$

Final Problem

$$$f(N) = \min⁡\left(X + f\left(\left\lfloor\frac{N}{A}\right\rfloor\right), \frac{6}{5} Y + \frac{1}{5} f\left(\left\lfloor\frac{N}{2}\right\rfloor\right) + \frac{1}{5} f\left(\left\lfloor\frac{N}{3}\right\rfloor\right) + \frac{1}{5} f\left(\left\lfloor\frac{N}{4}\right\rfloor\right) + \frac{1}{5} f\left(\left\lfloor\frac{N}{5}\right\rfloor\right) +\frac{1}{5} f\left(\left\lfloor\frac{N}{6}\right\rfloor\right)\right)$$$

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Crazy67

7 месяцев назад, # ^ |

but what if die is always roll to 1? Problem seems kinda incorrect to me

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martin0327

7 месяцев назад, # ^ |

The probability of such an event is 0.

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yihuicui

7 месяцев назад, # |

what is problem with my solution in D? Three test case give me RE

N, M = list(map(int, input().split()))
par = [i for i in range(N)]

def find(i):
    if par[i] != i:
        par[i] = find(par[i])
    return par[i]
def union(i, j):
    parI, parJ = find(i), find(j)
    if parI!=parJ:
        par[parI] = parJ
    return 
for _ in range(M):
    a, b = list(map(int, input().split()))
    union(a-1, b-1)
size = [0 for _ in range(N)]
res = 0
seen = set()
for i in range(N):
    pari = find(i)
    seen.add(pari)
    size[pari] +=1
res = 0
for parId in seen:
    res+= (size[parId]-1)*size[parId]//2
    
print(res-M)

Thanks a lot

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Physics212303

7 месяцев назад, # |

← Rev. 2 →

+19

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7 месяцев назад, # ^ |

marvellous

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7 месяцев назад, # ^ |

wow many people think such shit tests preparation is ok?

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7 месяцев назад, # |

+10

Since some people are not impressed with the problems, I'd like to counterbalance and say that I quite enjoyed this round.

In particular I don't think the tasks being doable and their difficulty curve being smooth is a bad thing for a contest like this, what with it having 'Beginner' in the name and all. :)

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dmkz

7 месяцев назад, # |

Solved Problem F by using __gnu_cxx::rope<char> for performing the fast reverse operations (solution) in 1041 ms and Problem G in $$$O(N \times Q \times \log{(N)})$$$ with sorted vectors and binary search in them (solution) in 2764 ms.

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cpp23

7 месяцев назад, # |

Any help with why my code in problem G is wrong?

Code

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Dead_Rank_77

7 месяцев назад, # |

Is it possible to get banned in Atcoder?

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coderdhanraj

7 месяцев назад, # |

← Rev. 2 →

Can anyone tell me why this solution for problem F is giving rte?

Edit: Got AC had some issue in my treap implementation.

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Snapper_001

7 месяцев назад, # ^ |

← Rev. 3 →

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nr0728

7 месяцев назад, # |

← Rev. 2 →

-15

Actually, the brute force solution to problem G is correct. For instance, in order to make the brute force solution run most slowly, we can use the following graph and query $$$2\ 1\ n$$$ every time:

However, the maximum number of queries is $$$10^5$$$, so the best solution (to make the brute force solution run most slowly) is to use $$$50000$$$ queries to build the graph I mentioned, and use the remaining $$$50000$$$ to query $$$2\ 1\ n$$$. We can calculate that the brute force solution will run $$$50000\times 25000=1.25\times 10^9$$$ times at most $$$^{[1]}$$$. With a 3s time limit, brute force can pass easily.

$$$^{[1]}$$$: If we use $$$n$$$ queries to build the graph, and use the remaining $$$100000-n$$$ to query $$$2\ 1\ n$$$, the brute force solution will run for $$$\dfrac{n}{2}\times (100000-n)$$$ times. Obviously, this is a quadratic function with a maximum value of $$$1.25\times 10^9$$$.

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LoliconMe

7 месяцев назад, # |

Why do my answer get TLE if I use bfs, but get AC if I use dfs on Problem G？ DFS dfs: https://atcoder.jp/contests/abc350/submissions/52624937 bfs: https://atcoder.jp/contests/abc350/submissions/52624751

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taogehtsilalos

7 месяцев назад, # |

Hello, I don't understand the solution for Atcoder : is it a small in large merge ?

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Java_Programmer

7 месяцев назад, # |

How to solve F using a stack?

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coderdhanraj

7 месяцев назад, # ^ |

-8

solve it as balanced parenthesis. Reverse the segments with treap or splay tree and for lower-upper cases use scaline to count no of times an index is affected if it's odd change the case.

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7 месяцев назад, # ^ |

I'll just say that your solution is a complete overkill. No need for any of that. Just store all the opening/closing intervals and try to come up with a recursive function that prints the interval [l,r], provided s[l]=='(' and s[r]==')'. It works in O(N).

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coderdhanraj

7 месяцев назад, # ^ |

I didn't get you, how will you reverse the respective segments without them??

It's obvious that we takes intervals [l, r] with given conditions same as we do in balanced parenthesis with the help of stack.

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7 месяцев назад, # ^ |

Here's a link to my submission. Url

The trick is that you don't actually have to perform these operations. You just need to know the "depth" of a parentheses meaning the number of open parentheses that come before it that are not yet closed.

The idea is very similar to the problem of reverals of word order in a string. For example "this is good" reversed os "good is this". This can be achieved by first reversing each word and then the whole string:

"this is good" -> "siht si doog" -> "good is this"

If you can find an efficient way to solve this problem in one pass each depth of parentheses define words of the new "sentence".

Hope this helps. It's a bit hard to explain, but essentially you only care about whether you print the interval from left to right or right to left.

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