We will hold Toyota Programming Contest 2024#2（AtCoder Beginner Contest 341）.
- Contest URL: https://atcoder.jp/contests/abc341
- Start Time: http://www.timeanddate.com/worldclock/fixedtime.html?iso=20240217T2100&p1=248
- Duration: 100 minutes
- Writer: math957963, leaf1415, PCTprobability
- Tester: yuto1115, cn449
- Rated range: ~ 1999
- The point values: 100-150-250-400-450-475-575
We are looking forward to your participation!
lovely score distribution :)
Hope I can AK!
Me too.
Me too.But it is impossible for me.
475 F & 575 G ?
They don't want you to get tired before think-cell Round 1. btw that dog is pure evil
Wish I could pass A, B, C, D, change the rating from 23 to 100.
But if you pass ABCD,you will up to over 100,even 200
If you pass ABCDEFG in half an hour,you will up to over 2000,even 2300.
Hope I can AK!
Hope ABCDEF!
Me,too.
Me,too.But I've only passed ABCDE and it's still impossible for me now.
I have only passed ABCD :(
I only hope to solve ABCDE, but today I solved ABCDEF :)
If I can pass ABCDEF,maybe I can up to 1000+. But if F is 525+, maybe I can up to 1100.
Hope I can pass ABCDE!!
Hope I can get 1Dan in this contes!
I hope to AC A-E. Good luck!
Atcoder down?
trash round,atcoder worse and worse
Can we please talk about our solutions?
is the editorial posted?
noo why do they remove one piece in problem F :(
I wonder why they like segment tree so much.
Can someone explain why my problem E failed: My Solution
Why this code for D is getting WA?
LCM is n*m/gcd(n, m), not just n*m.
Thanks.
Can you explain to me the solution to your problem D? I can't come up with an effective solution
Binary Search for answer, number of values that are multiples of X for an integer N = $$$N / X$$$. In general here only one of the two holds for Kth smallest value $$$value \vert N$$$ or $$$value \vert M$$$. But notice how for all multiples of $$$lcm(N, M)$$$ they are counted twice (once for N, once for M) so remove them. Hence finally for any integer X you get the formula $$$ X / N + X / M - 2 * lcm(X, Y)$$$. Now this range can be $$$[1, max(N, M) * K]$$$. To check fast user binary search.
could elaborate why the upper bound is
Assuming $$$m = max(n, m)$$$ $$$m \times k$$$ gives the $$$kth$$$ number that is divisble by m. Till this range there are $$$(m \times k) / n$$$ numbers divisble by n and k numbers divisble by m. You can easily see $$$(m \times k) / n + k - (m \times k) / lcm(m, n) >= k$$$.
My Problem G failed, could anyone find why I'm wrong?
I divided the sequence to the $$$O(\sqrt{n})$$$ blocks, and I calculate the maximum average perfix of each block.
For a query $$$i$$$. I calculate the answer in the block of $$$i$$$ first. Then, for each block next the block which $$$i$$$ belongs, I compare the maximum average perfix with the now answer, if it's bigger, I will infer the position which get the maximum is good, and calculate it into answer.
My proof is below.
First, the last element of the maximum average perfix of a block must bigger(or equal) than the number of maximum average perfix. If the maximum average perfix if bigger than now answer, we could infer that last element if bigger than the now answer. So we choose the last element is better.
Thanks.
Was there a typo in the post you guys made or is it actually being hold?
The first line of the post was this:
We will hold Toyota Programming Contest 2024#2（AtCoder Beginner Contest 341）.
I've written smth interesting for problem E here.
Why is the rating update today so slow?
For E , I am storing indexes where the element is equal to the previous element. For a range l to r , I am finding largest index from the stored indexes which is smaller than or equal to r , if it is greater than l the answer is NO otherwise the substring is good . I am getting WA. Do I miss something . Submission
Loved Problem B&G . Best Atcoder beginner round ever!
G blows my mind.