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Original URL: http://www.cnblogs.com/fhefh/archive/2011/04/15/2017613.htmlThe meaning of shell variable $#,[email protected],$0,$1,$2 in Linux is explained:Variable Description:$$The PID of the shell itself (ProcessID)$!PID of the Shell's last running background process$?End code of the last Run command (return value)$-Flag at a glance using the SET command$*All parameter lists. such as "$*" in the Case of

HTML Definition and usageA simple HTML table consists of a TABLE element and one or more tr, TH, or TD elements.The TR element defines the table row, the th element defines the header, and the TD element defines the table cell.More complex HTML tables may also include caption, Col, Colgroup, THEAD, TFOOT, and tbody elements.1 DOCTYPE HTML>2 HTMLLang= "en">3 Head>4 MetaCharSet= "UTF-8">5 title>Table

1MariaDB [zabbix]>Selectauthentication_type from config;2+---------------------+3| Authentication_type |4+---------------------+5|2|6+---------------------+7 1RowinchSet (0.00sec)8 9MariaDB [zabbix]> Update config Set authentication_type =0 ;TenQuery OK,1Row affected (0.00sec) OneRows matched:1Changed:1Warnings:0 A -MariaDB [zabbix]>Selectauthentication_type fro

CSS3 Border-radius PropertiesDefinition and usageThe Border-radius property is a shorthand property for setting four Border-*-radius properties.Tip: This property allows you to add rounded borders to elements!1 DOCTYPE HTML>2 HTMLLang= "en">3 Head>4 MetaCharSet= "UTF-8">5 title>Documenttitle>6 style>7 Div{8 width:200px;9 Height:200px;Ten background:Re

The new year is over. In retrospect, the only fun thing to do is to play with friends at home.This article has been accordingly.I'm sorry, but I have no idea at this time.Let me take a stroke.。。。。。。A smooth stroke ...。。。All right. Let's do it today:(1) The size of all the cards will be sorted out first.(2) Arrange all the cards in the suit.(3) to produce a random size and random color of the thing, to generate three cards per person.//All of the cards

# Include # Include # Include # Include # Define t0 1000# Define tf0.01# Define t 0.95# Define n 1000# Define M 50 Int weight [m] = {80, 82, 85, 70, 72, 70, 66, 50, 55, 25,50, 55, 40, 48, 50, 32, 22, 60, 30, 32,40, 38, 35, 32, 25, 28, 30, 22, 50, 30,45, 30, 60, 50, 20, 65, 20, 25, 30, 10,20, 25, 15, 10, 10, 10, 4, 4, 2, 1 },Profit [m] = {220,208,198,192,180,180,165,162,160,158,155,130,125,122,120,118,115,110,105,101,100,100, 98, 96, 95, 90, 88, 82, 8

);System. Out. println ("****** backtracking *****");System. Out. println ("***** number of items: n = 50 ");System. Out. println ("***** backpack capacity: c = 1000 ");System. Out. println ("***** optimal value: =" + bestp );T1.settext ("" + bestp );System. Out. println ("***** the selected item is :");} Public static void main (string [] ARGs ){Int P [] = {1, 220,208,198,192,180,180,165,162,160,158,155,130,125,122,120,118,115,110,105,101,100,100, 9

At present, I have 12 interfaces on hand in the outdoor 1/4 sweep Unit board, the initial understanding of the interface: OE is display enable, AB is row selectable, default high = 4 lines, all low = 1 rows, a low = 2 rows, b low = 3 rows. R is serial data, S is serial clock, L is latch and shift. To lsbfirst low-level output light, that is, Common Yang. It is worth mentioning that you can send many bytes o

In fact, I saw this in LINUX as early as I saw it. Later I also saw something similar in the MFC memory pool. I also wrote a similar small memory pool according to MFC, (MFC uses return this + 1) later, I saw a similar definition in Li xianjing's System Programmer growth plan, So I thought about it as a summary, the results showed that some cool people on the Internet had already summarized it very well, so it turned around. Thank you for your sharing

# Include # Include # Include # Define popsize 200# Define PC 0.618# Define PM 0.03.# Define lchrom 50# Define maxgen1000 Struct population{Int chrom [lchrom];Double weight;Double fitness;} Oldpop [popsize], newpop [popsize]; Int weight [lchrom] = {80, 82, 85, 70, 72, 70, 66, 50, 55, 25,50, 55, 40, 48, 50, 32, 22, 60, 30, 32,40, 38, 35, 32, 25, 28, 30, 22, 50, 30,45, 30, 60, 50, 20, 65, 20, 25, 30, 10,20, 25, 15, 10, 10, 10, 4, 4, 2, 1 },Profit [lchr

1 //0-1 knapsack problem solved by dynamic programming2 //Example:3 //Item Type n=5, backpack capacity c=10,4 //Item weight Vector w={2,2,6,5,4}, item value vector v={6,3,5,4,6}5 //O (Min{n*c,2^n})6#include"stdafx.h"7#include 8#include 9 Ten using namespacestd; OnetemplateclassType> AType Knapsack (intN,type C,type V[]

sequence pairs, backtracking constructs the optimal decision sequenceSequential couple:　　Dominance rule: if one of the Si-1 and Si1 has (PJ,WJ), the other has (PK,WK), and there is wj>=wk at the same time as PJThe generation of the sequence pairs of SiA sequence couple (PP,WW) in each generated Si1 to examine which order pairs are added to the SI① to add a sequence of w② by the dominant rule (PP,WW) whether to join SI, to examine the last sequence of

0-1 Backpack:0-1 Backpack is the basis of knapsack problem, it derived from the knapsack problem roughly with the same idea.And since it belongs to the dynamic programming problem, its method lies in two steps: 1) define the state: Ans[i][t] represents the maximum value of t

1 /*2 depending on the relationship, you can build a tree3 Dp[i][j] Indicates the maximum value of the first node selected J4 Dp[i][j]=max (Sigma (Dp[c[i][ki]))5 Sigma (Dp[c[i][ki]]) indicates the maximum number of J points selected from the son node of I6 */7 /*#include 8 #include 9 #include Ten #include One using namespace std; A - const int maxn=205; - int DP[MAXN][MAXN]; the int VAL[MAXN],N,M,NUM[MAXN

the kernel state, it uses the TSS.SS0, TSS.ESP0, or kernel stack, of the privilege level 0 given in its TSS segment. The original user stack pointer is saved in the kernel stack. When the user state is returned from the kernel state, the stack using the user's state is restored. They are described separately below. 5.8.1 initialization phase (1) When power-on initialization (bootsect. S,SETUP.S) When the B

0-1 questions about BackpacksKnapsack problem (knapsack Problem) is a np-complete problem of combinatorial Optimization. The problem can be described as: given a set of items, each item has its own weight and price, within a limited total weight, we choose how to make the total price of the item the Highest. The name of the problem comes from how to choose the most suitable item to be placed in a given back

0 or 1 Time Limit: 4000/2000 MS (Java/others) memory limit: 32768/32768 K (Java/Others)Total submission (s): 692 accepted submission (s): 185 Problem descriptiongiven a n * n matrix CIJ (1 Besides, Xij meets the following conditions: 1. X12 + x13 +... x1n = 12. x1n + x2n +... Xn-1n = 13. For each I (

Code Cache 4X32KB 8-Way Collection Association L1 Data Cache 4X32KB 4-Way Collection Association L2 Cache 4X256KB 8-Way Collection Association L3 Cache 8MB 16-Way Collection Association Server Board Motherboard model HP ProLiant DL120 G7 Motherboard chipset H

Why is the last $ a value 0 in the recursion below? Isn't it 1? PHPcode lt ;? PhpechoTest (); functionTest () {static $ a0; here is the 4th line echo $. lt; br gt; $ a ++; if ($ a lt; 2) {Test () ;}about the recursion below, why is the last $ a value 0? Isn't it 1? PHP code '; $ A ++; if ($ a Why is th

PL/0 language compilation interpreter Implementation 1 (C language), pl URL: http://www.cnblogs.com/archimedes/p/pl0-compiler1.html.PL/0 Overview The following content is taken from Wikipedia: PL/0, Is similar to but much simpler than the general-purpose programming language Pascal, intended as an educational programmi