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  • LCS(最長公共子序列)問題可以簡單地描述如下: 一個(gè)給定序列的子序列是在該序列中刪去若干元素后得到的序列。給定兩個(gè)序列X和Y

    LCS(最長公共子序列)問題可以簡單地描述如下: 一個(gè)給定序列的子序列是在該序列中刪去若干元素后得到的序列。給定兩個(gè)序列X和Y,當(dāng)另一序列Z既是X的子序列又是Y的子序列時(shí),稱Z是序列X和Y的公共子序列。例如,若X={A,B,C,B,D,B,A},Y={B,D,C,A,B,A},則序列{B,C,A}是X和Y的一個(gè)公共子序列,但它不是X和Y的一個(gè)最長公共子序列。序列{B,C,B,A}也是X和Y的一個(gè)公共子序列,它的長度為4,而且它是X和Y的一個(gè)最長公共子序列,因?yàn)閄和Y沒有長度大于4的公共子序列。 最長公共子序列問題就是給定兩個(gè)序列X={x1,x2,...xm}和Y={y1,y2,...yn},找出X和Y的一個(gè)最長公共子序列。對于這個(gè)問題比較容易想到的算法是窮舉,對X的所有子序列,檢查它是否也是Y的子序列,從而確定它是否為X和Y的公共子序列,并且在檢查過程中記錄最長的公共子序列。X的所有子序列都檢查過后即可求出X和Y的最長公共子序列。X的每個(gè)子序列相應(yīng)于下標(biāo)集{1,2,...,m}的一個(gè)子集。因此,共有2^m個(gè)不同子序列,從而窮舉搜索法需要指數(shù)時(shí)間。

    標(biāo)簽: 序列 LCS 元素

    上傳時(shí)間: 2015-06-09

    上傳用戶:氣溫達(dá)上千萬的

  • c語言版的多項(xiàng)式曲線擬合。 用最小二乘法進(jìn)行曲線擬合. 用p-1 次多項(xiàng)式進(jìn)行擬合

    c語言版的多項(xiàng)式曲線擬合。 用最小二乘法進(jìn)行曲線擬合. 用p-1 次多項(xiàng)式進(jìn)行擬合,p<= 10 x,y 的第0個(gè)域x[0],y[0],沒有用,有效數(shù)據(jù)從x[1],y[1] 開始 nNodeNum,有效數(shù)據(jù)節(jié)點(diǎn)的個(gè)數(shù)。 b,為輸出的多項(xiàng)式系數(shù),b[i] 為b[i-1]次項(xiàng)。b[0],沒有用。 b,有10個(gè)元素ok。

    標(biāo)簽: 多項(xiàng)式 曲線擬合 c語言 最小二乘法

    上傳時(shí)間: 2014-01-12

    上傳用戶:變形金剛

  • 高精度乘法基本思想和加法一樣。其基本流程如下: ①讀入被乘數(shù)s1

    高精度乘法基本思想和加法一樣。其基本流程如下: ①讀入被乘數(shù)s1,乘數(shù)s2 ②把s1、s2分成4位一段,轉(zhuǎn)成數(shù)值存在數(shù)組a,b中;記下a,b的長度k1,k2; ③i賦為b中的最低位; ④從b中取出第i位與a相乘,累加到另一數(shù)組c中;(注意:累加時(shí)錯(cuò)開的位數(shù)應(yīng)是多少位 ?) ⑤i:=i-1;檢測i值:小于k2則轉(zhuǎn)⑥,否則轉(zhuǎn)④ ⑥打印結(jié)果

    標(biāo)簽: 高精度 乘法 加法 基本流程

    上傳時(shí)間: 2015-08-16

    上傳用戶:源弋弋

  • This ActiveX control and Demo will allow you to dock your toolbars/forms to a mdiform much in the wa

    This ActiveX control and Demo will allow you to dock your toolbars/forms to a mdiform much in the way in which Visual C++ allows you to. 象Vc++一樣使用多文本窗體

    標(biāo)簽: toolbars ActiveX control mdiform

    上傳時(shí)間: 2015-09-14

    上傳用戶:asdfasdfd

  • ejb3.0 in action的全部配套源代碼

    ejb3.0 in action的全部配套源代碼,相信讀過spring in action,ajax in action的讀者應(yīng)該已經(jīng)領(lǐng)閱到action從書的魅力,這本書是最新的action從書,對于正在學(xué)習(xí)ejb3.0的同志無疑是最好的幫助,源代碼與書配套,相得益彰。

    標(biāo)簽: action ejb 3.0 in

    上傳時(shí)間: 2015-10-03

    上傳用戶:woshiayin

  • In this first-ever paperback edition of his long-time best-seller, motivational speaker Steve Chandl

    In this first-ever paperback edition of his long-time best-seller, motivational speaker Steve Chandler helps you create an action plan for living your vision in business and in life. It features 100 proven methods to positively change the way you think and act-methods based on feedback from the hundreds of thousands of corporate and public seminar attendees Chandler speaks to each year. 100 Ways to Motivate Yourself will help you break through the negative barriers and banish the pessimistic thoughts that are preventing you from fulfilling your lifelong goals and dreams. Whether you re self-employed, a manager, or a high-level executive, it s still easy to get stuck in the daily routines of life, fantasizing about what could have been. Steve Chandler helps you turn that way of thinking around and make what could have been into what can and will be.

    標(biāo)簽: motivational best-seller first-ever paperback

    上傳時(shí)間: 2015-10-26

    上傳用戶:牛津鞋

  • [輸入] 圖的頂點(diǎn)個(gè)數(shù)N

    [輸入] 圖的頂點(diǎn)個(gè)數(shù)N,圖中頂點(diǎn)之間的關(guān)系及起點(diǎn)A和終點(diǎn)B [輸出] 若A到B無路徑,則輸出“There is no path” 否則輸出A到B路徑上個(gè)頂點(diǎn) [存儲(chǔ)結(jié)構(gòu)] 圖采用鄰接矩陣的方式存儲(chǔ)。 [算法的基本思想] 采用廣度優(yōu)先搜索的方法,從頂點(diǎn)A開始,依次訪問與A鄰接的頂點(diǎn)VA1,VA2,...,VAK, 訪問遍之后,若沒有訪問B,則繼續(xù)訪問與VA1鄰接的頂點(diǎn)VA11,VA12,...,VA1M,再訪問與VA2鄰接頂點(diǎn)...,如此下去,直至找到B,最先到達(dá)B點(diǎn)的路徑,一定是邊數(shù)最少的路徑。實(shí)現(xiàn)時(shí)采用隊(duì)列記錄被訪問過的頂點(diǎn)。每次訪問與隊(duì)頭頂點(diǎn)相鄰接的頂點(diǎn),然后將隊(duì)頭頂點(diǎn)從隊(duì)列中刪去。若隊(duì)空,則說明到不存在通路。在訪問頂點(diǎn)過程中,每次把當(dāng)前頂點(diǎn)的序號作為與其鄰接的未訪問的頂點(diǎn)的前驅(qū)頂點(diǎn)記錄下來,以便輸出時(shí)回溯。 #include<stdio.h> int number //隊(duì)列類型 typedef struct{ int q[20]

    標(biāo)簽: 輸入

    上傳時(shí)間: 2015-11-16

    上傳用戶:ma1301115706

  • The initial planning and thinking about this book began during a discussion of SQL Server futures in

    The initial planning and thinking about this book began during a discussion of SQL Server futures in July 2001. The discussion was with Rob Howard during a trip to Microsoft to discuss the first book I was working on at that time. After that, I stayed involved in what was happening in ADO.NET by going to the SQL Server Yukon Technical Preview in Bellevue, Washington, in February 2002 and by working with the ASP.NET and SQL Server teams at Microsoft since July 2003.

    標(biāo)簽: discussion planning thinking initial

    上傳時(shí)間: 2014-01-08

    上傳用戶:cjf0304

  • the calculator s usage! after you have inputed 2 operators,choose + - * / function! But the only

    the calculator s usage! after you have inputed 2 operators,choose + - * / function! But the only situation I did t deal with is that when you choos + fuction ,and the operaters signs is like this -A+B,just turn it to B-A!

    標(biāo)簽: calculator the operators function

    上傳時(shí)間: 2016-02-12

    上傳用戶:lili123

  • CRC16算法的Java實(shí)現(xiàn)

    CRC16算法的Java實(shí)現(xiàn),使用方法如下: CRC16 crc16 = new CRC16() byte[] b = new byte[] { // (byte) 0xF0,(byte)0xF0,(byte)0xF0,(byte)0x72 (byte) 0x2C, (byte) 0x00, (byte) 0xFF, (byte) 0xFE, (byte) 0xFE, (byte) 0x04, (byte) 0x00, (byte) 0x00, (byte) 0x00, (byte) 0x00 } for (int k = 0 k < b.length k++) { crc16.update(b[k]) } System.out.println(Integer.toHexString(crc16.getValue())) System.out.println(Integer.toHexString(b.length))

    標(biāo)簽: Java CRC 16 算法

    上傳時(shí)間: 2014-12-20

    上傳用戶:ve3344

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