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。
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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
KMEANS Trains a k means cluster model.CENTRES = KMEANS(CENTRES, DATA, OPTIONS) uses the batch K-means algorithm to set the centres of a cluster model. The matrix DATA represents the data which is being clustered, with each row corresponding to a vector. The sum of squares error function is used. The point at which a local minimum is achieved is returned as CENTRES.
标签： CENTRES KMEANS OPTIONS cluster
上传时间： 2014-01-07
上传用户：zhouli
杭电 ACM 1002 I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.
标签： integers problem calcul simple
上传时间： 2014-01-12
上传用户：dave520l
中南赛区ACM竞赛题 Description Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1*1 or greater located within the whole array. The sum of a rectangle is the sum of all the elements in that rectangle. In this problem the sub-rectangle with the largest sum is referred to as the maximal sub-rectangle.
标签： two-dimensional Description negative integers
上传时间： 2013-12-22
上传用户：lijianyu172
利用凌阳单片机中计算从从1到100进行累加，并把计算结果保存在[Sum]单元里
上传时间： 2016-07-01
上传用户：alan-ee
动态规划的方程大家都知道，就是 f[i,j]=min{f[i-1,j-1],f[i-1,j],f[i,j-1],f[i,j+1]}+a[i,j] 但是很多人会怀疑这道题的后效性而放弃动规做法。 本来我还想做Dijkstra，后来变了没二十行pascal就告诉我数组越界了……（dist:array[1..1000*1001 div 2]...） 无奈之余看了xj_kidb1的题解，刚开始还觉得有问题，后来豁然开朗…… 反复动规。上山容易下山难，我们可以从上往下走，最后输出f[n][1]。 xj_kidb1的一个技巧很重要，每次令f[i][0]=f[i][i],f[i][i+1]=f[i][1]（xj_kidb1的题解还写错了）
上传时间： 2014-07-16
上传用户：libinxny
