台灣樂天全站4.8折起

平時我自己上網搜尋資料就還蠻喜歡看 QNAP 威聯通 TS-251A-4G 2Bay NAS 網路儲存伺服器

的

因為可以一網打盡真的是太方便!!!!!

就算沒買過肯定逛過聽過看過 QNAP 威聯通 TS-251A-4G 2Bay NAS 網路儲存伺服器

吧!!!

QNAP 威聯通 TS-251A-4G 2Bay NAS 網路儲存伺服器

功能:

QNAP 威聯通 TS-251A-4G 2Bay NAS 網路儲存伺服器

描述:

★搭載 Intel? Celeron? N3060 雙核心 1.6GHz 處理器?

★硬體加密引擎?，AES-NI?

★4K 影片硬體解碼，影片線上即時轉檔與離線背景轉檔?

★內建 USB QuickAccess 接口隨插即用，不用網路也可存取資料?

★OceanKTV 應用程式，隨點隨唱，打造專屬 KTV 環境?

★Linux Station 支援 Ubuntu? 正式授權作業系統，可搭配 HDMI 輸出桌面?

★支援網路虛擬儲存擴充櫃 (VJBOD)
















?

標題:

演算法題目求詳解

發問:

V. Pan has discovered a way of multiplying 68X 68 matrices using 132,464 multiplications, a way of multiplying 70 X70 matrices using 143,640 multiplications, and a way of multiplying 72 X72 matrices using 155,424 multiplications. Which method yields the best asymptotic running time when used in a divide-and-conquer... 顯示更多 V. Pan has discovered a way of multiplying 68X 68 matrices using 132,464 multiplications, a way of multiplying 70 X70 matrices using 143,640 multiplications, and a way of multiplying 72 X72 matrices using 155,424 multiplications. Which method yields the best asymptotic running time when used in a divide-and-conquer matrix-multiplication algorithm? Compare it with the running time for Strassen's algorithm.

最佳解答:

SOLUTION: The strassen's divide-and-conquer algorithm for matrix multiplication has the recurrence for its time complexity (considering only the counts of scalar multiplications): T(n) = 7 T(n/2), with the solution T(n) = O(n^log(2,7)) = O(n^2.807355...) If, just like Strassen's algorithm, we divide an nxn matrix into smaller matrices (recursively), but each with size (n/68)x(n/68), then when two 68x68 matrices can be multiplied by 132464 scalar multiplications, we'll have the recurrence T(n) = 132464 T(n/68). Similarly, the other two ways will have their recurrence: T(n) = 143640 T(n/70) and T(n) = 155424 T(n/72), respectively. The solutions are (respectively) (with big O sign omitted) n^log(68, 132464) = n^2.795128..., n^log(70, 143640) = n^2.795122..., and n^log(72, 155424) = n^2.795147.... So, the 2nd method is the fastest (in theory), and all of the three Victor Y. Pan's algorithms are better than Strassen's. Notes: (1) I used log(b, v) to mean "log v with base b". (2) The solution to the recurrence T(n) = a T(n/c) is n^(log(c, a)) in our case, which is covered by the "Master Theorem". See Theorem 3.1 on http://chern.ie.nthu.edu.tw/alg2003/Suppl_3_rec-sol.PDF or http://en.wikipedia.org/wiki/Master_theorem. 圖片參考：http://tw.yimg.com/i/tw/ugc/rte/smiley_35.gif

其他解答:

• 【雙11來了！限時下殺】SONY Xperia XZ攜碼至台灣大哥大 4G 上網月繳 $1399 手機1元 【贈32G記憶卡-1+Q Style10400行動-移動電源-1+空壓氣墊殼-1】
• 【雙11來了！限時下殺】LG系列 - GOR 9H G2 玻璃 鋼化 保護貼 全透明 2片裝 下標區 全館299免運費
• Lenovo 聯想 IdeaPad V310 專用第二顆硬碟轉接盒 筆電光碟機位硬碟托架 免拆面板尾翼【HDC-LV1】
• QNAP 威聯通 TS-831X-4G 8-Bay NAS 網路儲存伺服器
• HP OJ-4650雲端無線傳真事務機【三井3C】
• Moomin 嚕嚕米正版授權【 荒島上的Moomin(黑) 】《 iPad Mini-Air-Pro 》水晶殼＋Smart Cover（磁桿）