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Replying to @ConsciousRide
Last time I built the same for a streaming app where it uses global content IDs regional rights index offline candidate generation. Keep user preference vectors..lists cached per region...fetch cached candidates → apply country/licensing bitset → lightweight rerank→ return
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Replying to @bengunnnnn
konnte übrigens nicht dein bild mit dem werkzeug verwenden deshalb bitset vom bauhaus
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Replying to @andy19959
bin einfach seine timeline durchgegangen lol konnte aber torx bitset nicht finden deshalb musste ich das bitset von bauhaus nehmen
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Anfang 20: Geil, meine neue Konsole kommt heute Ende 20: Geil, mein neue Akku Stielstaubsauger und meine neuer Akkuschlagbohrschrauber mit Bohr - und Bitset kommen heute
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誤った記法であることは理解していつつ、動的bitsetとかはO(N/64)って書きたくなっちゃう
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Replying to @sapi_kawahara
う~ん?論理演算は難しくないですか?nを変更するたびにカルノー図作る必要ある気がします。 まぁ拡張性考えるにしてもboolを整数に変換するより良い方法があるのは確かですね。flgCountみたいな変数を別途作れば良いし、bitset使えるならそれで一発ですし。
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Replying to @PR0GRAMMERHUM0R
context Bit containers are core data structures used in modern software. They are widely used in the Linux kernel priority queues, bloom filters, and huffman coding data compression. They are included in many standard libraries, like the BitSet class in java.
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bitset 並列化に持っていくのが難しい問題って珍しいね〜
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[K] FA 同じ文字へのワープだと思うと,ワープ用に頂点を 2 倍して Dijkstra [D] みなさん: 最も近い 2 べきに行く? 3 べきも使うかも ta: 2^x * 3^y の形だけ全列挙して,最大の素因数で割るとして DP → 証明: AC [F] no → 実装 k=0 の式が間違っていた (1 敗, k 側なので関係なかった) ans=0 で初期化してしまっていた (1 敗) [J] 連結成分をマージしておく 1 つに全部入れるのが最適で,増加量を尺取りしたいがだるすぎる! (毒の合計, 隙間の合計, 隙間を全部後ろに動かしたときのダメージ差分) をセグ木に載せて二分探索すると楽に [G] no: 色が全部違っても bitset しかない がんばって詰めたが実装する時間なし 1. (色, その色が塗られている頂点集合が変化しない時間帯) が N 2Q 個あるので,これらをソートして並べる 2. これを B = 2048 個くらいずつ取って bitset<B> で処理 i. クエリ時刻から bitset の mask を得る関数を用意 (mask が同じになる時間帯をまとめ,時間でソートしておけば,二分探索で得られる) ii. 各頂点が塗られているかの bit 列を作成 (各頂点について,何 bit 目から何 bit 目まで立つというのは前計算できる) iii. DP しながら,各クエリに対して mask を取ってきて popcount を取る だいぶ速いと思う
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Q3: Longest substring without repeating character: 2 pointers Bitset We have to keep track of the character seen using bitset and the max of the longest length of substring that doesn't have any character that repeats
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Just completed 50% of this Bit Manipulation series. After XOR series, I will cover: - AND Series - OR Series - Gray Code (in-depth) - BitSet Data Structure - Advanced Tricks & Patterns Might take 30 days, might be less. Depends on how deep I go. Even after the series ends, I'll keep adding valuable Bit Manipulation content whenever I find interesting patterns, techniques, or problems worth sharing. Bit Manipulation is one of the hardest DSA topics to master, so I did grind hard on it. My goal is simple: If you complete this series, you shouldn't need another Bit Manipulation resource for interviews, competitive programming, or DSA. Though prerequisites are you should know basics of Bit manipulation.
30 Days. 30 Bit Manipulation Tips & Tricks. For the next 30 days, I’ll share powerful bit hacks, clever tricks, uncommon patterns, and practical optimizations used in CP and real-world problem solving. No boring theory. Only useful, practical, and important concepts.
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Some nice and easy optimization in Go's bitset library. A few operations got a really nice boost (almost twice as fast). Low hanging fruits that I found while working on something entirely different. bits-and-blooms.github.io/bi…
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Replying to @possdiff @ChShersh
Because the std allows vector<bool> to be implemented as a bitset. Meaning 8 entries could need 1 byte in one implementation, but in another platform or compiler or std library it could require 8 bytes. That can break things if you access it as a raw ptr.
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Replying to @Uwu123231634
I didn't know Rust devs need to implement their own bitset when it's available in C out of the box en.cppreference.com/cpp/util…

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Replying to @ChShersh
Try std bitset maybe, always cool
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A 実装バグらせて終わり B ucupかなんかで似たの見たことある気がしたし、DAGの数え上げみたいになって解けた気もするが、なんか実装してたらよくわからんくなった D bitsetでdpとかビームサーチで答え見つかりそうと思って書いたけど、バグらせてて終わり
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Counting Bloom Filters are surprisingly easy to implement and support full CRUD operations, but at the cost of much larger size compared to bitset implementations. Rotating Bloom Filters handle this tradeoff pretty well, though the implementation is more complex.
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C : clunky bridging byte and bitset with to_integer std::byte myByte{0xFF}; std::bitset<8> bits(std::to_integer<unsigned char>(myByte));
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any data structure choice must be driven by access pattern, not category: a custom bitset with small-buffer optimization and algebraically updateable hashing is required for the hot-path DFS that repeatedly clones, hashes, and probes without committing, while the standard-library DynamicBitSet is the right tool for a simple one-shot debug assertion. a new blog post that discusses this theme in the context of an optimized BitSet implementation in Zig .. debasishg.github.io/blog/why…

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