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Автор macaquedev, история, 19 месяцев назад, По-английски

Prior to yesterday, there were some cheaters who managed to get to red, who were smart enough to evade detection, and prompted LLMs in a very careful way to get maximal performance out of them, for example lequoctran181.

However some of them just don't even try to hide it. It's actually pathetic, and with yesterday's release of o4-mini and full o3, we'll have hundreds if not thousands of them reaching red... ugh.

arnabmanna 309859410 309858141

Just look at this. It's actually just sad at this point...

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19 месяцев назад, скрыть # |
 
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I was asked to ping MikeMirzayanov about this as well.

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u have no proof

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    19 месяцев назад, скрыть # ^ |
     
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    Trust me, no one is gonna write descriptive comments while contest. There is too much time pressure.

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      19 месяцев назад, скрыть # ^ |
       
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      it may appear to be rather suspicious that comments are written but that will not suffice as evidence of cheating

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      19 месяцев назад, скрыть # ^ |
      ← Rev. 2  
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      If you want to, it's not actually that hard or time consuming to add comments and change variable names.

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      19 месяцев назад, скрыть # ^ |
       
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      Sometimes I do that because I don't want to get up from my chair to get a piece of paper and I need to organize what I'm thinking better.

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        19 месяцев назад, скрыть # ^ |
         
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        Yes, I do that sometimes as well. But that results in comments like this:


        // ok so we need to basically find the largest set of nodes, which are ancestors of each other, such that none of them is an ancestor of the other in another tree // ok so can we convert this to a useful format using tin and tout? // take path from root to leaf, and we need to find largest set of ranges such that none of them completely contains the other // but we need to also support incremental add and remove queries // how to do that?

        not something like // Take input int n, k; cin >> n >> k; vector<int> a(n); for (int i = 0; i < n; i++) { cin >> a[i]; } // Binary search check function auto Check = [&](int m) { long long ans = 0; for (int i = 0; i < n; i++) { ans += max(a[i] - m, 0LL); } return (ans <= k); }; // Binary search int low = 0, high = 1e9, idx = -1; while (low <= high) { int mid = (low + high) >> 1; if (Check(mid)) { idx = mid; high = mid - 1; } else { low = mid + 1; } } // Print answer cout << idx << '\n';

        (yes I know this is slightly exaggarated, the original code is not nearly as bad, but it's the same spirit)

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    19 месяцев назад, скрыть # ^ |
     
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    Found the alt