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。
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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
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.
標簽： CENTRES KMEANS OPTIONS cluster
上傳時間： 2014-01-07
上傳用戶：zhouli
杭電 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.
標簽： integers problem calcul simple
上傳時間： 2014-01-12
上傳用戶：dave520l
中南賽區ACM競賽題 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.
標簽： two-dimensional Description negative integers
上傳時間： 2013-12-22
上傳用戶：lijianyu172
利用凌陽單片機中計算從從1到100進行累加，并把計算結果保存在[Sum]單元里
上傳時間： 2016-07-01
上傳用戶：alan-ee
動態規劃的方程大家都知道，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] 但是很多人會懷疑這道題的后效性而放棄動規做法。 本來我還想做Dijkstra，后來變了沒二十行pascal就告訴我數組越界了……（dist:array[1..1000*1001 div 2]...） 無奈之余看了xj_kidb1的題解，剛開始還覺得有問題，后來豁然開朗…… 反復動規。上山容易下山難，我們可以從上往下走，最后輸出f[n][1]。 xj_kidb1的一個技巧很重要，每次令f[i][0]=f[i][i],f[i][i+1]=f[i][1]（xj_kidb1的題解還寫錯了）
上傳時間： 2014-07-16
上傳用戶：libinxny
