UVa OJ 147 – Dollars (錢幣)

Time limit: 3.000 seconds限時:3.000秒 Problem問題New Zealand currency consists of $100, $50, $20, $10, and $5 notes and $2, $1, 50c, 20c, 10c and 5c coins. Write a program that will determine, for any given amount, in how many ways that amount may be

UVa OJ 143 – Orchard Trees (果樹林)

Time limit: 3.000 seconds限時:3.000秒 Problem問題An Orchardist has planted an orchard in a rectangle with trees uniformly spaced in both directions. Thus the trees form a rectangular grid and we can consider the trees to have integer coordinates. The

UVa OJ 139 – Telephone Tangles (話費之糾結)

Problem問題A large company wishes to monitor the cost of phone calls made by its personnel. To achieve this the PABX logs, for each call, the number called (a string of up to 15 digits) and the duration in minutes. Write a program to process this data

UVa OJ 138 – Street Numbers (門牌號)

Time limit: 3.000 seconds限時3.000秒 Problem問題A computer programmer lives in a street with houses numbered consecutively (from 1) down one side of the street. Every evening she walks her dog by leaving her house and randomly turning left or right and

UVa OJ 137 – Polygons (多邊形)

Time limit: 3.000 seconds限時:3.000秒Problem問題Given two convex polygons, they may or may not overlap. If they do overlap, they will do so to differing degrees and in different ways. Write a program that will read in the coordinates of the corners of

全排列產生演算法:next_permutation

概念全排列的產生演算法有很多種,有遞迴遍例,也有迴圈移位法等等。但C++/STL中定義的next_permutation和prev_permutation函數則是非常靈活且高效的一種方法,它被廣泛的應用於為指定序列產生不同的排列。本文將詳細的介紹prev_permutation函數的內部演算法。按照STL文檔的描述,next_permutation函數將按字母表順序產生給定序列的下一個較大的排列,直到整個序列為降序為止。prev_permutation函數與之相反,是產生給定序列的上一個較小的排

UVa OJ 135 – No Rectangles (沒有矩形)

Problem問題Consider a grid such as the one shown. We wish to mark k intersections in each of n rows and n columns in such a way that no 4 of the selected intersections form a rectangle with sides parallel to the grid. Thus for k = 2 and n = 3, a

UVa OJ 134 – Loglan-A Logical Language (Loglan邏輯語言)

Problem問題Loglan is a synthetic speakable language designed to test some of the fundamental problems of linguistics, such as the Sapir Whorf hypothesis. It is syntactically unambiguous, culturally neutral and metaphysically parsimonious. What follows

UVa OJ 145 – Gondwanaland Telecom (岡瓦納電信)

Time limit: 3.000 seconds限時:3.000秒 Problem問題Gondwanaland Telecom makes charges for calls according to distance and time of day. The basis of the charging is contained in the following schedule, where the charging step is related to the

解決QQ無法在沙盤中啟動並執行問題

可能不少在沙盤(Sandboxie)裡運行QQ的同學都會遇到這樣的問題:先把QQ裝在沙盤裡,然後把QQ的安裝資料夾複製到另一個地方,清空沙盤,再從沙盤中啟動QQ:經過排除法尋找,確定這個問題是由於兩個安全性群組件(COM)沒有註冊引起的。這兩個組件的動態連結程式庫檔案分別是安裝在以下目錄的SSOCommon.dll和SSOPlatform.dllc:\program

Win32環境下代碼注入與API鉤子的實現

API鉤子不是新技術了,但很慚愧的是我最近才真正的掌握它,並想出了一些笨辦法來達到自己的目的。文中可能有很多錯誤或漏洞,希望您能夠及時指正。本文詳細的介紹了在Visual

UVa OJ 732 – Anagrams by Stack (堆棧字迷)

Time limit: 3.000 seconds限時:3.000秒 Background背景How can anagrams result from sequences of stack operations? There are two sequences of stack operators which can convert "TROT" to

UVa OJ 641 – Do the Untwist (解密工作)

Time limit: 3.000 seconds BackgroundCryptography deals with methods of secret communication that transform a message (the plaintext) into a disguised form (the ciphertext) so that no one seeing the ciphertext will be able to figure out the plaintext

平面內兩條線段的位置關係(相交)判定與交點求解

概念平面內兩條線段位置關係的判定在很多領域都有著廣泛的應用,比如遊戲、CAD、圖形處理等,而兩線段交點的求解又是該演算法中重要的一環。本文將儘可能用通俗的語言詳細的描述一種主流且效能較高的判定演算法。外積,又稱叉積,是向量代數(解析幾何)中的一個概念。兩個二維向量v1(x1, y1)和v2(x2,

UVa OJ 571 – Jugs (罐子)

Time limit: 3.000 seconds限時:3.000秒 Background背景In the movie "Die Hard 3", Bruce Willis and Samuel L. Jackson were confronted with the following puzzle. They were given a 3-gallon jug and a 5-gallon jug and were asked to fill the 5-gallon jug with

UVa OJ 199 – Partial differential equations (偏微分方程)

Time limit: 3.000 seconds限時:3.000秒 Problem問題In engineering sciences, partial differential equations play an important and central role. For example, the temperature of a metal plate can be expressed as a partial differential equation if the

VC6.0開發OCX按鈕控制項

VC6.0開發OCX按鈕控制項 0 前言1、OCX是典型的ActiveX控制項,常見的OCX控制項有很多,IE瀏覽器、Word、Excel等都提供各種各樣的OCX控制項。2、OCX控制項的優點是可以在其他應用程式中嵌入使用。OCX控制項做為外掛程式、外掛式的應用非常靈活方便。3、一般用的比較多的是做為視頻、圖片、圖表等顯示地區控制項。最近碰到幾個按鈕的功能要封裝成OCX控制項,供其它程式使用。網上大概搜尋了下,資料不多,於是自己寫了下來,以供尋找。 1

UVa OJ 150 – Double Time (雙時間)

Time limit: 3.000 seconds限時3.000秒 Background and Problem背景與問題In 45 BC a standard calendar was adopted by Julius Caesar--each year would have 365 days, and every fourth year have an extra day--the 29th of February. However this calendar was not quite

太有才了!史上最有創意的404頁面設計集錦

  404頁面在我們平常瀏覽網頁的時候並不會經常碰到,也因此很容易被忽略,但是有一些無意中出現的404頁面的確很有意思。404頁面也有一些特別的用途,例如顯示公益廣告,推廣網站自身的內容等等,自訂404頁面也是改善網站使用者體驗的一個方面。  這篇文章向大家分享25個史上最有創意的404頁面設計集錦,同時推薦:《一組漂亮的404錯誤頁面模板下載》。您可能感興趣的相關文章讓人驚歎的的創意404錯誤頁面設計33個非常有創意的404錯誤頁面設計推薦35個非常有創意的404錯誤頁面40個非常有創意的40

UVa OJ 149 – Forests (森林)

Time limit: 3.000 seconds限時:3.000秒Background背景The saying "You can't see the wood for the trees" is not only a cliche, but is also incorrect. The real problem is that you can't see the trees for the wood. If you stand in the middle of a "wood" (in NZ

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