We will hold AtCoder Beginner Contest 350.
- Contest URL: https://atcoder.jp/contests/abc350
- Start Time: http://www.timeanddate.com/worldclock/fixedtime.html?iso=20240420T2100&p1=248
- Duration: 100 minutes
- Writer: physics0523, kyopro_friends, ynymxiaolongbao
- Tester: MMNMM, nok0
- Rated range: ~ 1999
- The point values: 100-200-300-350-450-550-600
We are looking forward to your participation!
GLHF！
Best point values i have ever seen!!!
Why?
How to do D?
a very easy approach other than DSU. first thing to observe is that in every connected component, every node will be friend to every other node. this leads us to the conclusion that every connected component should be a complete graph. in a complete graph, the number of edges is given as:
Now the problem is reduced to something very easy. For each connected component, find the number of nodes it has. this will gives us the required number of edges per connected component. After that you can either subtract m at the end or separately count the number of edges in every connected component and subtract it from the number of edges in a complete graph.
My Submission: Link
I checked evima editorial on yt, it was DFS
i do it with dsu as mentioned by the above user
wtf this tasks
Exactly my thoughts. A-F is criminally bad. Idk about G.
What do you find bad about them ?
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.
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 ?
I know ideas to 6 on average, today all 7 (but didn't code G). I usually don't solve <5.
Thanks
+1
Loved E
Is G just small to large merging?
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.
Could you help me find out why this solution gives RTE, TLE and WA? It seems to be what you described.
Submission Link
By a close look I think you should update ancestors inside of dfs on all nodes.
I don't think so, I'm updating only the smaller child when merging.
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.
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 msthough, 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.
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.
Thanks for the solution! Do you think the small memory limit was designed to kill
sqrts, or for something else?I guess to kill bitsets
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?
i can't read your code, but there can be more than one connected component, which you may have missed.
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.
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
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.
How to solve A and B for just 47 seconds? I could not even open the tasks for that time :)
how to solve $$$E$$$?
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 $$$\frac{6}{5}$$$.
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?
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 $$$O(\log^3 N)$$$.
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
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.
can you please elaborate more ?
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
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
$$$f(N) = \min\left(X + f\left(\left\lfloor\frac{N}{A}\right\rfloor\right), \frac{6}{5} Y + \frac{1}{5} f\left(\left\lfloor\frac{N}{2}\right\rfloor\right) + \frac{1}{5} f\left(\left\lfloor\frac{N}{3}\right\rfloor\right) + \frac{1}{5} f\left(\left\lfloor\frac{N}{4}\right\rfloor\right) + \frac{1}{5} f\left(\left\lfloor\frac{N}{5}\right\rfloor\right) +\frac{1}{5} f\left(\left\lfloor\frac{N}{6}\right\rfloor\right)\right)$$$
but what if die is always roll to 1? Problem seems kinda incorrect to me
The probability of such an event is 0.
It seems that someone's brute force passed G.
marvellous
wow many people think such shit tests preparation is ok?
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. :)
Solved Problem F by using
__gnu_cxx::rope<char>for performing the fast reverse operations (solution) in1041 msand Problem G in $$$O(N \times Q \times \log{(N)})$$$ with sorted vectors and binary search in them (solution) in2764 ms.Any help with why my code in problem G is wrong?
Code
Is it possible to get banned in Atcoder?
Can anyone tell me why this solution for problem F is giving rte?
Edit: Got AC had some issue in my treap implementation.
.
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 $$$1.25\times 10^9$$$.
Hello, I don't understand the solution for Atcoder : is it a small in large merge ?
How to solve F using a stack?
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.
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).
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.
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.
Will the testcases of A update after ABC350?
no
I think it should be that, but the fact is not......
In problem G, the inputs seem to be generated randomly, how come they guarantee that for type 1 queries u, v belong to different connected components
For Sample 3 in C,I think it's OK to print
2
2 3
1 2
But why was it false in judge?(AC*2 & WA*1)
Are you sure that you have no problem with the other two samples?
I solved ABCD,I am very weak...