其他

 This problem requires us to find two non-negative numbers m1 and m2 that satisfying  n1*m1 + n2*m2 = nand minimize the cost  c = c1*m1 + c2*m2. First, we can use extended Euclid's algorithm to find m1' and m2' that give  n1*m1' + n2*m2' = gwhere g

Let g(n) denote the n'th number that satifies f(k) > f(k-1). We can know that     g(0) = 1, g(1) = 2, g(2) = 4, ... From the definition of g(n), we can know that all the values of f(k) where k belongs to the range of [g(n-1), g(n)) is n. Since

This is the hardest problem ever since in this book. After some thinking, I realized that this problem can be solved with backtracking. After some more thinking, I realized that we should backtrack the cipher, i.e., the decrypt table along with the

花了兩三周，每天10點半到家後，花上兩小時看上一部分，這周3終於看完了。這本書比《對象設計》結構整齊一些，更容易跟上作者的思路。前面部分寫了領域驅動設計的一些基礎方面，包括如何交流，如何調整結構，哪些是領域構造塊等等。後面用部分開始跳出細節，從一個很高的視角看待前面的所有內容在整個體系中的位置及作用，並把前面的內容放在一起來示範了大比例結構上的應用。看完後，突然覺得，其實一些方法論，在不同的層次上都是通用的。你可以在非常大的體系粒度上來應用一些設計，對系統進行劃分層次、劃分模組、限定系統邊界；同

 Imagine that a chessboard with black and white squares. If a bishop is on a white square, can it attack bishops on black squares? No.So you can divide the chessboard into two independent ones. Then rotate each board for PI/4 and now the diagonal

類比題，根據題目描述的內容直接寫程式就可以。需要注意的是非法輸入的情況：對於a==b或是a和b在同一堆的輸入命令直接忽略。 返回 Volume I 索引返回總索引 // 101_The_Blocks_Problem.cpp : Defines the entry point for the console application.//#include <set>#include <map>#include <list>#include

計劃翻譯The Old New

  最近自己製作了幾個iphone4使用的離線地圖，總結一下經驗： 1. 原始地圖下載使用Global Map DownloadTool下載，網上教程很多。 2. 轉為iPhone格式使用GMDL的‘->iPhone' 功能，選擇firmware2.2 & 3，產生MapTiles.sqlitedb。 3.

我們的程式用到Flash控制項進行互操作以提供更友好的介面。為了屏蔽右鍵菜單，直接子類化了Flash視窗，用自己的視窗函數，在右鍵時不發送給Flash控制項，自己處理。測試中發現，一些情況下（主要是點擊後開IE），關閉視窗銷毀控制項會導致崩潰。 1. 發現在ShellExecute這個API中會有訊息迴圈，導致WM_CLOSE會在ShellExecute返回之前被處理2. 改成非同步ShellExecute後，發現還是會有問題，而且總是在WM_TIMER(0x0113)這個訊息上。3.

The close form of fibs does not help in the problem. The input range is too large that the precision of long double won't be sufficient to handle it correctly. Just generate all the fibs up to 100 digits and use string compare to find the

I originally used the straight forward method and got a RT of 1.250s. I believe that if I were using Java, I would get a TL. Thus, I switch to the method of precomputing all the cycle-lengths in range (0, 1000000). Let al, al-1, ..., a1 donate the

This problem is much easier than its brother, "843 - Crypt Kicker". Just find the key line and get the translate table from it will do all the job. I generate a "signature" for the key line to make the search easier. By the way, pay attention to

ight:450px;width:100%;">  有很多函數返回被稱為“顯示名稱（display names）”的字串。對於這些顯示名稱你可以做的唯一一件事就是拿來顯示。不要假設這些字串有任何實際的意義，因為它們的確沒有意義。理論上，一個類似於SHGetFileInfo這樣的函數可以被實現成： ... if (uFlags & SHGFI_DISPLAYNAME) {    StringCchCopy(psfi->szDisplayName, MAX_PATH, TEXT(

使用C++的RTTI，可以在運行期顯示多態指標的類型名 typeid(p).name()是多態指標的當前類型typeid(*p).name()是多態指標的most

I solve this problem in a manner that seems not straight forward. I let the fragments to "vote" on each bit if it is '0' or '1'. Specifically, for each fragment, it is either the head or tail of the file. We let it to vote as both. Thus, for each

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 在GDI+中，使用Bitmap建立一個離屏緩衝，必須要使用以下建構函式版本：Bitmap::Bitmap(INT width, INT height, Graphics* target)而不要使用以下建構函式版本：Bitmap::Bitmap(INT width, INT height, PixelFormat format)否則，在調用Graphics::DrawImage把位元影像繪製到螢幕表面時，會有嚴重的效能下降。我以下列幾種PixelFormat嘗試過後者：PixelFormat24

前面我們瞭解了“協助文檔不是設定邏輯迷題的地方”。對話方塊也是一樣。

General100 - The 3n+1 problem | Solution10189 - Minesweeper | Solution10137 - The Trip | Solution706 - LC-Display | Solution10267 - Graphical Editor | Solution10033 - Interpreter | Solution10196 - Check The Check | Solution10142 - Australian Voting |

  UVa Online Judge - Volume I 題目和解答索引。前面為原題連結，後面為我的解答連結。 返回總目錄  100 - The 3n + 1 problemAd hocSolution101 - The Blocks ProblemSimulationSolution102 - Ecological Bin PackingAd hocSolution103 - Stacking BoxesGraph: DAGSolution104 - ArbitrageGraph: