《Neural Network Design》 Martin T. Hagan, Howard B. Demuth, Mark H. Beale 一書(shū)中相關(guān)matlab演示程序軟件包。對(duì)于學(xué)習(xí)很有參考。
上傳時(shí)間： 2016-03-19
上傳用戶(hù)：zhanditian
~{JGR 8vQ IzWwR5SC5D2V?bD#DbO5M3~} ~{3v?b~} ~{Hk?b~} ~{2iQ/5H9&D\~} ~{?IRTWw@)3d~} ~{TZ~}JDK1.4.2~{OBM(9}~}
上傳時(shí)間： 2015-02-22
上傳用戶(hù)：ommshaggar
b to b 模式 電子商務(wù)系統(tǒng) ，c# 開(kāi)發(fā) , B/S結(jié)構(gòu)
標(biāo)簽： to 模式 電子商務(wù)系統(tǒng)
上傳時(shí)間： 2014-01-20
上傳用戶(hù)：hanli8870
樣板 B 樹(shù) ( B - tree ) 規(guī)則 : (1) 每個(gè)節(jié)點(diǎn)內(nèi)元素個(gè)數(shù)在 [MIN,2*MIN] 之間, 但根節(jié)點(diǎn)元素個(gè)數(shù)為 [1,2*MIN] (2) 節(jié)點(diǎn)內(nèi)元素由小排到大, 元素不重複 (3) 每個(gè)節(jié)點(diǎn)內(nèi)的指標(biāo)個(gè)數(shù)為元素個(gè)數(shù)加一 (4) 第 i 個(gè)指標(biāo)所指向的子節(jié)點(diǎn)內(nèi)的所有元素值皆小於父節(jié)點(diǎn)的第 i 個(gè)元素 (5) B 樹(shù)內(nèi)的所有末端節(jié)點(diǎn)深度一樣
上傳時(shí)間： 2017-05-14
上傳用戶(hù)：日光微瀾
歐幾里德算法：輾轉(zhuǎn)求余 原理: gcd(a,b)=gcd(b,a mod b) 當(dāng)b為0時(shí),兩數(shù)的最大公約數(shù)即為a getchar()會(huì)接受前一個(gè)scanf的回車(chē)符
標(biāo)簽： gcd getchar scanf mod
上傳時(shí)間： 2014-01-10
上傳用戶(hù)：2467478207
數(shù)據(jù)結(jié)構(gòu)課程設(shè)計(jì) 數(shù)據(jù)結(jié)構(gòu)B+樹(shù) B+ tree Library
標(biāo)簽： Library tree 數(shù)據(jù)結(jié)構(gòu) 樹(shù)
上傳時(shí)間： 2013-12-31
上傳用戶(hù)：semi1981
給定兩個(gè)集合A、B，集合內(nèi)的任一元素x滿(mǎn)足1 ≤ x ≤ 109，并且每個(gè)集合的元素個(gè)數(shù)不大于105。我們希望求出A、B之間的關(guān)系。 任 務(wù) ：給定兩個(gè)集合的描述，判斷它們滿(mǎn)足下列關(guān)系的哪一種： A是B的一個(gè)真子集，輸出“A is a proper subset of B” B是A的一個(gè)真子集，輸出“B is a proper subset of A” A和B是同一個(gè)集合，輸出“A equals B” A和B的交集為空，輸出“A and B are disjoint” 上述情況都不是，輸出“I m confused!”
標(biāo)簽：
上傳時(shí)間： 2017-03-15
上傳用戶(hù)：yulg
* 高斯列主元素消去法求解矩陣方程AX=B,其中A是N*N的矩陣,B是N*M矩陣 * 輸入: n----方陣A的行數(shù) * a----矩陣A * m----矩陣B的列數(shù) * b----矩陣B * 輸出: det----矩陣A的行列式值 * a----A消元后的上三角矩陣 * b----矩陣方程的解X
上傳時(shí)間： 2015-07-26
上傳用戶(hù)：xauthu
Implemented BFS, DFS and A* To compile this project, use the following command: g++ -o search main.cpp Then you can run it: ./search The input is loaded from a input file in.txt Here is the format of the input file: The first line of the input file shoud contain two chars indicate the source and destination city for breadth first and depth first algorithm. The second line of input file shoud be an integer m indicate the number of connections for the map. Following m lines describe the map, each line represents to one connection in this form: dist city1 city2, which means there is a connection between city1 and city2 with the distance dist. The following input are for A* The following line contains two chars indicate the source and destination city for A* algorithm. Then there is an integer h indicate the number of heuristic. The following h lines is in the form: city dist which means the straight-line distance from the city to B is dist.
標(biāo)簽： Implemented following compile command
上傳時(shí)間： 2014-01-01
上傳用戶(hù)：lhc9102
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(biāo)簽： represented integers group items
上傳時(shí)間： 2016-01-17
上傳用戶(hù)：jeffery
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