A1
A2
B
C
D
E
Some notes/backstory about the round in general:
- Sorry about the mistake in C! I hope it didn't affect you too much, if at all. You can blame this on my bad organization — a tester did actually catch this mistake, but somehow I forgot to correct it.
- The only tester feedback I received besides corrections was to swap the problems C and D. Based on the solve counts, I guess it worked out.
- A lot of people received WA17 or WA20 on D because of not making the default values in the array small enough. With $$$0$$$ as the default, you'd get WA17. Others, who used $$$-10^9$$$ or $$$-10^{10}$$$, got WA20. I actually made this mistake myself, and my brute-force didn't catch it because it only seemed to happen with large enough $$$n, k$$$.
- I was skeptical about placing E at the end because of how simple the solution was, but that seems like it was a good choice now.
- I understand criticism about the problems being too "standard" — in fact, I'm not really proud of having a problem like D, but I thought it was necessary for balance (turns out it wasn't anyway). At the very least, they could serve to teach people new algorithms.
There is another solution for 102646A2 - Product of Triples (Hard Version)
Iterate $$$x$$$ over all numbers from $$$[1,n]$$$
For each number $$$x$$$, fix $$$i$$$ to be one of the divisors of the number $$$x$$$.
Calculate $$$y$$$ = $$$x/i$$$, now iterate $$$z$$$ over all divisors of $$$y$$$. The three numbers would be $$$i$$$, $$$z$$$, $$$y/z$$$. Just check for the $$$a≤b≤c$$$ and count them !
galen_colin
can any one tell me why WA in this problem https://mirror.codeforces.com/gym/102646/problem/E my code ---> https://ideone.com/yB1W29 my idea is to search about sccs inside the graph then remove all edges from this scc and so on till remains k edges
The entire input graph is already strongly connected. So removing edges from this isn't guaranteed to work, and is essentially equivalent to removing $$$m - k$$$ random edges, which has no guarantee of producing the maximum number of SCCs.