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通讯编程文档 根据有无固定基础设施
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 ...
软件设计/软件工程 本文档提供JSR120无线消息API规范的说明
本文档提供JSR120无线消息API规范的说明,也描述Sum的RI的API.
matlab例程 Waterfilling algorithm (from [Palomar and Fonollosa, Trans-SP2004]) to compute: pi = (mu*ai - bi
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).
数值算法/人工智能 北京大学ACM比赛题目 In 1742, Christian Goldbach, a German amateur mathematician, sent a letter to Leonhard
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 n ...
汇编语言 设计一个程序完成求1-100的累加和
设计一个程序完成求1-100的累加和,结果送到SUM单元中
数值算法/人工智能 KMEANS Trains a k means cluster model.CENTRES = KMEANS(CENTRES, DATA, OPTIONS) uses the batch K-mean
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 ...
数值算法/人工智能 杭电 ACM 1002 I have a very simple problem for you. Given two integers A and B, your job is to calcul
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.
数学计算 中南赛区ACM竞赛题 Description Given a two-dimensional array of positive and negative integers, a sub-rec
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-rectangl ...
汇编语言 利用凌阳单片机中计算从从1到100进行累加
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
中间件编程 动态规划的方程大家都知道
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 ...