| # | User | Rating |
|---|---|---|
| 1 | jiangly | 3810 |
| 2 | Benq | 3676 |
| 3 | Kevin114514 | 3655 |
| 4 | maroonrk | 3463 |
| 5 | strapple | 3390 |
| 6 | Um_nik | 3387 |
| 7 | tourist | 3384 |
| 8 | heuristica | 3322 |
| 9 | turmax | 3319 |
| 10 | jiangbowen | 3291 |
| # | User | Contrib. |
|---|---|---|
| 1 | Qingyu | 156 |
| 2 | nik_exists | 150 |
| 2 | maspy | 150 |
| 4 | Um_nik | 144 |
| 5 | Errichto | 139 |
| 6 | adamant | 137 |
| 6 | AmShZ | 137 |
| 8 | BledDest | 132 |
| 8 | maroonrk | 132 |
| 10 | qwexd | 129 |
|
-8
The AI Era The single largest catalyst for the recent shift is the explosive rise of Large Language Models (LLMs) used for cheating Thousands of new or low-rated accounts use AI to instantaneously solve problems A through D in Div. 2/Div. 3 rounds. Because they solve them virtually instantly, they achieve mathematically impossible performance deltas for their actual skill bracket. The Deflationary Effect: These AI-driven accounts perform like 1900-rated users while carrying a baseline 1500 rating. They effectively "suck" rating points out of the active user pool. When honest users lose to these fast-solving bot accounts, their ratings drop heavily. This creates a massive deflationary pressure that ripples all the way up to Masters, Grandmasters, and ultimately LGMs Burnout, Inactivity, and Rating Hording As ratings become harder to defend, many true GMs and LGMs choose to sit on their ratings or only compete in high-stakes global events where AI noise is less prevalent. When the highest-rated active nodes in the Elo network stop regularly distributing points back into the ecosystem through standard rounds, the ceiling naturally sinks Rating Abandonment Issue As top ratings become increasingly brutal to maintain, legendary grandmasters (LGMs) and International Grandmasters (IGMs) have shifted their behavior. Many top-tier users choose to play on anonymous secondary accounts ("alts") rather than risking their primary handle’s 3000+ rating on an unpredictable round filled with LLM noise When a 3200-level genius plays on a 2200-rated alt account, they completely disrupt the pool. They obliterate their rank bracket, taking massive points away from legitimate Master and Grandmaster players, without ever pushing those points back into the "LGM bank account" on their main handle. Conclusion Ultimately, this has triggered a shift in community mindset: many top veterans (like me, I just shifted from AtCoder to CF like yesterday) now view rating as a highly volatile, secondary metric and focus purely on absolute rank, ICPC preparation, and the joy of problem-solving rather than chasing an artificially deflated number |
|
0
Isn't the feedback on rejected submissions a pretty significant bug? I just tested it and a WA submission gave me a specific mathematical hint identifying the flaw in my proof. Since wrong submissions have no extra penalty and there are 15 attempts, this effectively lets contestants use the judge as an interactive proof debugger: submit → get a hint → fix → resubmit This seems quite different from normal contest judging and could potentially make some problems much easier
Rejected submission Take a spanning tree (T) of (G). It has (n-1) edges, so there are [ m-(n-1)=m-n+1 ] edges outside (T) Choose any subset of the edges outside (T) For each chosen edge (e=(u,v)), add the unique path from (u) to (v) in (T) This gives a subgraph where every vertex has even degree: the added edge and the path together form a cycle Thus, the set of edges with even degree is completely determined by which edges outside (T) we choose Once those are fixed, the tree edges are forced: starting from the leaves of (T) their inclusion is uniquely determined by the requirement that every vertex has even degree So every subset of the (m-n+1) non-tree edges gives exactly one valid (S) and every valid (S) is obtained this way Therefore the number of such subsets is [ \boxed{2^{m-n+1}} ] Accepted submission Take a spanning tree (T) of (G). It has (n-1) edges, so there are [ m-(n-1)=m-n+1 ] edges outside (T) Now choose any subset of the edges outside (T). The edges of (T) are then uniquely determined by the requirement that every vertex has even degree: start from the leaves of (T), and for each leaf, its edge to its parent is forced by the parity of its other incident edges. Remove the leaf and continue Thus every subset of the (m-n+1) non-tree edges gives exactly one valid (S), and every valid (S) is obtained this way. Therefore the number of such subsets is [ \boxed{2^{m-n+1}} ] With 15 attempts and no WA penalty one could deliberately submit progressively incomplete proofs and use the judge to locate the missing pieces. |
|
On
melechosher111 →
Identical code, different verdict — TLE vs AC on Round 1124 (Div. 2), 9 hours ago
-7
"I asked Friend #1 to send me his code directly, and I copy-pasted and resubmitted it from my own account" MOSS didn't flag yo ass? |
| Name |
|---|


