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We have a group of N items (represented by integers from 1 to N), and we know that there is some total order defined for these items. You may assume that no two elements will be equal (for all a, b: a<b or b<a). However, it is expensive to compare two items. Your task is to make a number of comparisons, and then output the sorted order. The cost of determining if a < b is given by the bth integer of element a of costs (space delimited), which is the same as the ath integer of element b. Naturally, you will be judged on the total cost of the comparisons you make before outputting the sorted order. If your order is incorrect, you will receive a 0. Otherwise, your score will be opt/cost, where opt is the best cost anyone has achieved and cost is the total cost of the comparisons you make (so your score for a test case will be between 0 and 1). Your score for the problem will simply be the sum of your scores for the individual test cases.
標簽： represented integers group items
上傳時間： 2016-01-17
上傳用戶：jeffery
To date tests are still the best solution mankind has found to deliver working software. This book is the sum of four years of research and practice in the testing field. The practice comes from my IT consulting background, first at Octo Technology and then at Pivolis the research comes from my involvement with open source development at night and on weekends.
標簽： solution software mankind deliver
上傳時間： 2014-01-11
上傳用戶：sclyutian
分治法解決最近對問題 畫一條垂直線x=c，把這些給定點分為兩個包含n/2個點的子集S1和S2，使得n/2個點位于直線的左側或直線上，另外n/2個點位于直線的右側或直線上；遵循分治法的思想，遞歸地求出左子集S1和右子集S2中的最近對，分別為d1與d2；之后d=min{d1,d2}。合并過程：在以垂線x=c為對稱軸，2d為寬度的區域內求最近兩個點的距離，記為d3；求D=min{d,d3}；
上傳時間： 2013-12-26
上傳用戶：源碼3
MATLAB對11中調度算法進行比較，包括min-min、max-min、遺傳算法等等。
上傳時間： 2014-01-06
上傳用戶：lnnn30
假設在一個ad hoc網絡中，移動節點的發射功率PTx總是恒定的。要發送數據的移動節點總是先監聽信道，測量接收到的信號功率X，其中X= I + N, I為接收到的干擾，N是噪聲。移動節點只有在X＜INThre時，才可以發射。式中，INThre為背景噪聲門限。 在仿真中，我們規定每個移動節點的發射功率是常數，PTx = 1W；接收節點接收機的靈敏度Smin = -80 dBm；信號質量 min = 2 dB；系統的背景噪聲門限INThre = 1.2e-10。
上傳時間： 2016-03-16
上傳用戶：sevenbestfei
根據有無固定基礎設施，無線局域網又可分為BSS (Basic Service Set)和IBSS (Independent Basic Service Set)。我們要研究的ad hoc網絡屬于后者。假設在一個ad hoc網絡中，移動節點的發射功率PTx總是恒定的。要發送數據的移動節點總是先監聽信道，測量接收到的信號功率X，其中X= I + N, I為接收到的干擾，N是噪聲。移動節點只有在X＜INThre時，才可以發射。式中，INThre為背景噪聲門限。 在仿真中，我們規定每個移動節點的發射功率是常數，PTx = 1W；接收節點接收機的靈敏度Smin = -80 dBm；信號質量 min = 2 dB；系統的背景噪聲門限INThre = 1.2e-10。
標簽：
上傳時間： 2013-12-19
上傳用戶：頂得柱
本文檔提供JSR120無線消息API規范的說明，也描述Sum的RI的API.
上傳時間： 2014-12-07
上傳用戶：baitouyu
Waterfilling algorithm (from [Palomar and Fonollosa, Trans-SP2004]) to compute: pi = (mu*ai - bi)^+ sum(pi) = Pt By Daniel Perez Palomar (last revision: May 10, 2004).
標簽： Waterfilling Fonollosa algorithm Trans-SP
上傳時間： 2014-01-10
上傳用戶：liansi
北京大學ACM比賽題目 In 1742, Christian Goldbach, a German amateur mathematician, sent a letter to Leonhard Euler in which he made the following conjecture: Every even number greater than 4 can be written as the sum of two odd prime numbers. For example: 8 = 3 + 5. Both 3 and 5 are odd prime numbers. 20 = 3 + 17 = 7 + 13. 42 = 5 + 37 = 11 + 31 = 13 + 29 = 19 + 23. Today it is still unproven whether the conjecture is right. (Oh wait, I have the proof of course, but it is too long to write it on the margin of this page.) Anyway, your task is now to verify Goldbach s conjecture for all even numbers less than a million.
標簽： mathematician Christian Goldbach Leonhard
上傳時間： 2016-04-22
上傳用戶：wangchong
設計一個程序完成求1－100的累加和，結果送到SUM單元中
上傳時間： 2013-12-25
上傳用戶：1051290259
