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real會被眼前勝利所迷惑的人<br> 就是會削弱自己力量的人
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會被眼前勝利所迷惑的人<br> 就是會削弱自己力量的人
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會被眼前勝利所迷惑的人<br> 就是會削弱自己力量的人
real: 隨筆 2012-08-21 - 近(?)況
http://chhung6.blogspot.com/2012/08/2012-08-21.html
隨筆 2012-08-21 - 近(? 快封塵了 其實先前寫左好幾篇, 最後冇submit. 過去一年 (2011夏 - 2012夏). 學術 / ACM / 工作方面, 我一直. 主要係 engineering work - tune 結果, 加速, visualize 結果. 同埋寫比較麻煩既 technical report, 仲有 presentation. 近日收到通知:終於有一份 journal paper accept 左! 雖然已經退役, 但有時間, 都會去睇/講下 training. 轉營玩 online contest, 似乎都打得 OK密 (Topcoder, Codeforces, InterviewStreet), 近一兩個月少咗. 最苦惱既問題 - 搵/做咩工好, 最尾搞到年頭先開始搵. 用懶 乜都報下既心態, 報一啲就腳既工 - 嚟中大搞 recruitment talk, 而人工又 相對. Ok 既公司 (u know, 香港始終重視金融). 我冇乜考慮過出國. 但, 喺師兄/朋友/朋輩極力鼓勵下, 試. 然後, 冇諗過. 會有 offer.
real: 九月 2010
http://chhung6.blogspot.com/2010_09_01_archive.html
今年的 Team Formation 總算塵埃落定. 還是在趕 11月中的 Conference Deadline. SRM 483 - 0分 悲劇. 160;期待已久的 Rating 大跌的時機終於來了. 便緊張得錯誤開啟了 500. - -. 看完 250 是一道比較直接的整數除法 (好似係). 稍為冷靜以後, 總算把 250 慢慢的 (202.xx) 搞定. 然後開 500. 想了又想, 想出了算法: DP bit pattern. 中段開了 Division Summary 看. 很多人提交了 900. 有很多甚至時 800 以一的提交. 但自己把心一橫, 堅持做 500. 比賽臨終時, 才發現 Transition 錯了. 不能只 consider 上一格 array element. 再看看 Division Summary 及 Room Summary. 悲劇了 大量 900 的 Submission. 現在的 Challenge Phase, 絕大部份的 900 依然屹立不倒. 2010-09-25 Team Training - Shanghai 2009. 題C 和 題D 的難度.
real: [溫故知新,數論] Prmitive Root modulo n
http://chhung6.blogspot.com/2011/07/prmitive-root.html
溫故知新,數論] Prmitive Root modulo n. 以下定義/定理或者未夠嚴僅. 數學人請見諒.). Theorem 1 (Euler's Totient Theorem). 8801; 1 (mod n. Definition 1 (Multiplicative Order). 的 (multiplicative) order modulo n. 8801; 1 (mod n. 以下為 (hopefully) 較直觀的解說:. 的 order 就是 { x. Definition 2 (Primitive Root). 的 order 是 Φ( n. Corollary 1 (Verifying a primitive root). 8660; 對於每個質因數 p. 65292; 有 g. 8802; 1 (mod n. 如何找出(最小的)一個 primitive root modulo n. 就是「頹試」:. Algorithm 1 (Finding the least primitive root). G EndIf. 12288; EndFor. Mod m) / 利用.
real: 十月 2010
http://chhung6.blogspot.com/2010_10_01_archive.html
SRM 486 - Live. 今次係 300 450 1000. 一打開, 睇完題目 考慮呢題既分數. 令我覺得, google 下會搵到答案. 睇完題意, 有 linearity of expectation DP 感覺). 點知搵黎搵去, 都只係搵到 theoratical expected number of exchanges. 又研究下 Case 3 = 144 / 13 究竟有咩玄機. 又戳下, 答案同 number of inversion 有乜關係. 諗下 google 下. 又諗下. 又 google 下. Member SRM 485 - 被MO屈機. Rush 250 的速度居然比以前了,為甚麼呢? 是 based on 一些 observation. 但因為有 odd/even 的限制,我的算法才正確. 500 全間 904 都沒有人能做出來. 不過顯然對於 W, H = 50. Waihon 則直覺認為暴搜能過,因為 RectangleAvoiding 條件苛刻. 郭 智 亮 說 ﹕ 「 一 個 人 『 叻 』 沒. 合 團 隊 的 運 作 。 」.
real: 三月 2011
http://chhung6.blogspot.com/2011_03_01_archive.html
2011-03-23 Team Training - World Finals 2004. HKG Time - 1900 to 2400. 啱啱尋晚係 research deadline 後. 今次做 World Finals 2004. 標籤: Team Training. SRM 500 - Live. 250 我花了 45 分鐘才通過 sample. 預期會有大量 fail system test 出現. X, x 1) × [y, y 1). 但感覺 500 若能通過 sample. 比起 250,應該有更大機會通過 system test. Petr 等一眾高手完成了全部 3 題. Rng 58 單做 1000 Rank 6th. Rank 4th in room → 沒錢分 sosad. 本人認為,最後討論的 三角形面積方法 的解相當優美. 中學時期,我們都學過用聯立方程組求 線 / 線交點:. 用代入法 (substitution) 或消元法 (elimination) 求出的一般解. 在 x系數 = 0 或 y系數 = 0 的情況需要特別分開處理. 判斷交點 Q 是否在 線段.
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C.H. Hung - Fantasy Author
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real
Codeforces #132 (Div. 2) E - Periodical Numbers. Dynamic Scoring 是 3000 的題目 (所以也不是每個高手在現場解決了). Virtual Participation 時花了 40 分鐘沒能搞定的題. 將一個正整數 X 表示為 n-bit 的 binary-string, S. 若果存在 長度為 k n 的 循環節 (period). Ie, 使得 S[i k] = S[i] for 0 = i n-k). 則稱 X 為 periodical. 給定區間 [L, R], 問當中有多少個 periodical. 求 [1, R] 當中 periodical 正整數的個數. 那麼最後答案 := DOIT(R) - DOIT(L-1). 對於 循環節 長度為 k 的 X,可以將 X 表示成:. X = Y ×. 0001 00.01 00.01 . 00.01) (二進制表示及計算,下同). 0001 00.01 00.01 . 00.01) 是 m-bit 二進位數,其中 m = n. 隨筆 2012-08-21 - 近(?
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