We will hold AtCoder Grand Contest 075. This contest counts for GP30 scores.
- Contest URL: https://atcoder.jp/contests/agc075
- Start Time: http://www.timeanddate.com/worldclock/fixedtime.html?iso=20251221T2100&p1=248
- Duration: 180 minutes
- Number of Tasks: 5
- Writer: PCTprobability
- Tester: maspy, HIR180
- Rated range: 2000 -
The point values will be 600-800-1200-1200-1600.
We are looking forward to your participation!
Problem B is similar to https://qoj.ac/problem/8582
People who solved A must have the same brainwave with PCTprobability. It takes me 65:28 to solve the sweet honey A.
Try to write a brute force to generate all valid solution for n=5. That's how I solved it.
why this contest can be an AGC?
I think it's possible to do $$$k=10^6$$$ in D, because you don't actually have to iterate through all maximum pairs $$$(x,y)$$$ as stated in the editorial. Instead, by just knowing the maximum value $$$x$$$, we can calculate the total contribution of all $$$y$$$ with some case analysis. Since we still need matrix multiplication or BM, the complexity is $$$O(k\log n)$$$.
By setting $$$k=10^4$$$, some even more brute force solution might pass, like iterating through the maximum $$$3$$$ values (there are only around $$$10^6$$$ such tuples that are meaningful).
Can anyone prove my example in A? For odd $$$n$$$, pattern is like this (here $$$n = 9$$$):
Everything works, where $$$a[i][j] \neq a[n+1-j][n+1-i]$$$, and the $$$i+j=n+1$$$ diagonal is $$$0101 \cdots 1010$$$.
There is an obvious bijection between the 0 paths and 1 paths without the middle diagonal. And the rest of the proof is the same.
Why B less people solved but less score than D. I solved B but didn't have enough time think D. I lost so many extra rating because of this.
You knew the score distribution, right?
I believe you meant to say: you knew the public standings, right? Pointing out score distribution to a guy complaining about score distribution is... pointless at the very least.
I meant to be a petty little bitch.