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.
标签： represented integers group items
上传时间： 2016-01-17
上传用户：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.
标签： solution software mankind deliver
上传时间： 2014-01-11
上传用户：sclyutian
分治法解决最近对问题 画一条垂直线x=c，把这些给定点分为两个包含n/2个点的子集S1和S2，使得n/2个点位于直线的左侧或直线上，另外n/2个点位于直线的右侧或直线上；遵循分治法的思想，递归地求出左子集S1和右子集S2中的最近对，分别为d1与d2；之后d=min{d1,d2}。合并过程：在以垂线x=c为对称轴，2d为宽度的区域内求最近两个点的距离，记为d3；求D=min{d,d3}；
上传时间： 2013-12-26
上传用户：源码3
MATLAB对11中调度算法进行比较，包括min-min、max-min、遗传算法等等。
上传时间： 2014-01-06
上传用户：lnnn30
假设在一个ad hoc网络中，移动节点的发射功率PTx总是恒定的。要发送数据的移动节点总是先监听信道，测量接收到的信号功率X，其中X= I + N, I为接收到的干扰，N是噪声。移动节点只有在X＜INThre时，才可以发射。式中，INThre为背景噪声门限。 在仿真中，我们规定每个移动节点的发射功率是常数，PTx = 1W；接收节点接收机的灵敏度Smin = -80 dBm；信号质量 min = 2 dB；系统的背景噪声门限INThre = 1.2e-10。
上传时间： 2016-03-16
上传用户：sevenbestfei
根据有无固定基础设施，无线局域网又可分为BSS (Basic Service Set)和IBSS (Independent Basic Service Set)。我们要研究的ad hoc网络属于后者。假设在一个ad hoc网络中，移动节点的发射功率PTx总是恒定的。要发送数据的移动节点总是先监听信道，测量接收到的信号功率X，其中X= I + N, I为接收到的干扰，N是噪声。移动节点只有在X＜INThre时，才可以发射。式中，INThre为背景噪声门限。 在仿真中，我们规定每个移动节点的发射功率是常数，PTx = 1W；接收节点接收机的灵敏度Smin = -80 dBm；信号质量 min = 2 dB；系统的背景噪声门限INThre = 1.2e-10。
标签：
上传时间： 2013-12-19
上传用户：顶得柱
本文档提供JSR120无线消息API规范的说明，也描述Sum的RI的API.
上传时间： 2014-12-07
上传用户：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).
标签： Waterfilling Fonollosa algorithm Trans-SP
上传时间： 2014-01-10
上传用户：liansi
北京大学ACM比赛题目 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.
标签： mathematician Christian Goldbach Leonhard
上传时间： 2016-04-22
上传用户：wangchong
设计一个程序完成求1－100的累加和，结果送到SUM单元中
上传时间： 2013-12-25
上传用户：1051290259
