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OpenMP tutorial learning notes (2) OpenMP Programming Model

OpenMP Tutorial: https://computing.llnl.gov/tutorials/openMP/#ProgrammingModel Understanding the OpenMP programming model is the first step in learning OpenMP. (1) shared memory, thread based Parallelism: OpenMP is a parallel model based on shared memory and threads. (2) Explain It Parallelism: OpenMP is clear about parallelism rather than automatic parallelism. Provide parallelFull Control.

Java multithreaded Programming tutorial thread operation and control

code, the thread can call the object's Wait () method, release the object lock flag, enter the wait state, and call the Notify () or the Notifyall () method to notify other threads that are waiting. Notify () notifies the first thread in the wait queue. Notifyall () Notifies all threads waiting on the queue.The Wait () and notify () mechanisms provided in Java are as follows:Called by thread 1 synchronized method1 (...){//Access data variable available = true; Notify ();} Called by thread 2 syn

Python programming for loop statement learning tutorial, pythonfor

Python programming for loop statement learning tutorial, pythonfor A Python for loop can traverse any series of projects, such as a list or a string.Syntax:The syntax format of the for Loop is as follows: for iterating_var in sequence: statements(s) Flowchart:Instance: #! /Usr/bin/python #-*-coding: UTF-8-*-for letter in 'python': # First Instance print 'current letter :', letterfruits = ['bana', 'apple',

Unix/linux Environment C Programming Novice Tutorial (opensuseccpp) and Linux kernel-driven development environment building

. Start the terminal40. Enter the pathwatermark/2/text/ahr0cdovl2jsb2cuy3nkbi5uzxqvaxrjyxn0y3bw/font/5a6l5l2t/fontsize/400/fill/i0jbqkfcma==/ Dissolve/70/gravity/southeast "/>41. Compile and run42. Write a C + + code 1,CC43. Print statements, C + +44. Print out the results45. Drive HelloWorld#include #include Module_license ("Dual BSD/GPL");//Follow the Linux protocolstatic int hello_init (void){PRINTK (Kern_alert "Hello, world\n");//Print driver informationreturn 0;}static void Hello_exit (void

Unix/linux surrounding Environment C programming Tutorial (1) Solaris 64bit environment structure

, select 1 24.F2 Next 25. Select the disk, continue F2 and use the entire disk to continue F2 26. Set the name, use the TAB key to choose to actively configure the network by F2 27. Choose Asia to continue F2 28. Choose China to continue F2 29. Select PRC to continue F2 30. Use the default time to continue F2 31. Enter the root password a

Shell Programming Basics Tutorial 7--script parameter delivery

7. Passing of script parameters7.1.shift commandBrief introduction:Shift n Shifts the parameter position to the left n bits at a timeExample#!/bin/bashusage () { echo "usage: ' basename ' filenames"}totalline=0# The following statement $# the number of arguments if [$#-lt 2]then usagefiwhile [$#-ne 0]do line= ' cat $ | wc-l ' echo ' $1:${line} ' totalline=$ [$totalline + $line]shiftdoneecho '----- ----"echo" Total:${totalline} "7.2.getoptsBrief introduction:Get multiple command-line arg

C Language Programming Case Tutorial (2nd edition) code note (ii)

parameter. Call to function The basic format of the function call statement is: The The parameters in the Example: 1 intdataarray[ -];2 intN;3 DoubleAve;4 5 //Custom Functions6 DoubleAverageintValue[],intnum) {7 intI, sum =0;8 for(i =0; i ){9Sum + =Value[i];Ten returnsum*1.0/num; One } A } - - //function Call theAve = Average (DataArray, n);Application examples of custom functions Output multiplication Ta

C Language Programming Case Tutorial (2nd edition) code note (iii)

].birthday.month, S[i].birthday.day, - s[i].department, s[i].major); WuPutchar ('\ n'); - } About } $ - voidSearchInfo (Studentinfo s[], date date) - { - inti; A for(i =0; i ){ + if(s[i].birthday.month>date.month) { theprintf"\n%4d%16s%2d/%2d", S[i].num,s[i].name, S[i].birthday.month, - s[i].birthday.day); $ Continue; the } the if(S[i].birthday.month = = Date.month s[i].birthday.day>date.day) { theprintf"\n%4d%16s%2d/%2d", S[i].num, S[i].name, S

1016:c Language Programming Tutorial (third Edition) after class exercise 6.5

Title DescriptionFor the sum of the following three digits, the sum of 2 decimal 1~a and the reciprocal of the square and 1~c of the 1~b are reserved.InputA b COutput1+2+...+a + 1^2+2^2+...+b^2 + 1/1+1/2+...+1/cSample input100 50 10Sample output47977.93TipsNote the output format, do not output extra space!Source#includemain(){int i,a,b,c;float s=0;scanf("%d%d%d",a,b,c);for(i=1;is+=i;for(i=1;is+=i*i;for(i=1;is+=1.0/i;printf("%.2f\n",s);}1016:c Language Progra

Wuhan University of Science and Technology acm:1007: Chinese Edition C language Programming tutorial (second edition) Exercise 5.7

Problem DescriptionThis day the teacher gave small Hao a very simple topic relaxed: Enter a score, let you find their simplest score. InputThe first line includes a T, which represents the number of groups of test data.The next line of T lines includes a fraction. (The numerator denominator is within the int range)OutputFor each test sample, the output line has its simplest fraction.Sample Input32/63/111/36Sample Output1/33/111/36My Code:1#include 2ProcessintAintb/*greatest common divisor, so th

Wuhan University of Science and Technology Acm:1004:c Language Programming Tutorial (third Edition) after class exercise 6.3

Problem DescriptionBeg sn=2+22+222+...+22 ... A value of 222 (with N 2). For example: 2+22+222+2222+22222 (n=5), n is entered by the keyboard.InputNOutputAndSample Input5Sample Output24690My Code:1#include 2 voidMain ()3 {4 intN,a=2, t,sum=0, I;5scanf"%d",n);6t=A;7 for(i=1; i)8 {9sum=sum+T;Tent=t*Ten+A; One } Aprintf"%d", sum); -}Other code:1#include 2 3 using namespacestd;4 5 intMain ()6 {7 Longsum=0;8 intt=2;9 intN;TenCin>>N; One for(intI=0; ii) A { -Sum + =T;

Wuhan University of Science and Technology acm:1005: Chinese Edition C language Programming tutorial (second Edition) example 5.8

Problem DescriptionThe teacher gave a title to Xiao Hao: give you two integers x and n ( -10now that little Howe is in a quandary, please help him solve the problem. InputEnter two integers x and n per line.Outputfor each set of test data, the output line evaluates. Sample Input1 12 2Sample Output16HINT1#include 2 3 4 Long LongSumintXintN)5 {6 Long Longs=0;7 inti,j;8 Long LongS1;9 for(i=1; i)Ten { Ones1=1; A for(j=1; j) - { -s1*=x; the } -s+=S1; - } -

Wuhan University of Science and Technology acm:1001: Chinese Edition C language Programming tutorial (second edition) Exercise 6.7

Problem DescriptionThe output Yang Hui the first n rows of the triangle.InputEnter a number n (n OutputThe output Yang Hui the first n rows of the triangle. (Note that there is no extra space at the end of the line, the number is output in%3d format)Sample Input34Sample Output 1 1 1 1 2 1 1 1 1 1 2 1 1 3 3 1HINTNote that there are multiple sets of inputs. Output a blank line after each set of test data.while (scanf ("%d", n)! = EOF){......}1#include 2 i

Wuhan University of Science and Technology acm:1009: Chinese Edition C language Programming tutorial (second edition) Exercise 6.11

Problem Descriptionn individuals in a circle, numbered from 1 to n sequentially. From the person numbered 1 starts 1 to K, where the number of people who count K out of the circle, the output of the last left a person's original numbers.InputFirst enter a T, which indicates that there is a T group of data (1Then there are t-lines, each with 2 positive integers n and K. (1OutputFor each set of test data, output a number that represents the number of the last person left.Sample Input310 37 15 4Sam

Wuhan University of Science and Technology acm:1007: Chinese Edition C language Programming tutorial (second edition) Exercise 7.10

Problem DescriptionEnter n (nInputExample of multi-group test sample. The first row of each group has an integer n indicating that there are n strings. Next there are n rows, one string per line.OutputOutput a sequential string, with each line outputting a string.Sample Input3abaaabcabSample OutputAababacab1#include 2#include string.h>3 voidSortChar* str[],intsize)4 {5 inti,j;6 Char*tmp;7 for(i=0; i1; i++)8 {9 for(j=i+1; j)Ten { One if(strcmp (str[i]

Wuhan University of Science and Technology acm:1001: Chinese Edition C language Programming tutorial (second Edition) after class exercise 3.12

Problem DescriptionEnter n, and the output corresponds to a hollow positive hexagon with a side length of N.For easy viewing, the sample midpoint '. ' Represents a space and prints a space instead of a small dot when plotting a drawing.InputEdge length N. (NOutputSide length is a positive hexagon of nSample Input5Sample Output.....*****....*.....*...*.......*.. *.........*.*...........*.. *.........*...*.......*....*.....*.....*****1#include 2#include 3 voidPrtCharCintcount)4 { 5 while(cou

Wuhan University of Science and Technology acm:1006: Chinese Edition C language Programming tutorial (second edition) Exercise 7.15

Problem DescriptionEnter n strings (nInputMultiple sets of test data. The first row of each set of test data contains an integer n, representing a total of n strings. Next, each line contains a string that consists of printable characters.OutputEach set of test sample output one row. The output finds a string that satisfies the test instructions.Sample Input3djdlkfjsadfjwedlkfjdlkfjl;jf;sfjdsl;al/dljfd2dlkasfjmmlld;femflsad;fiwejdifSample OutputDlkfjdlkfjl;jf;sfjdsl;al/ld;femflsad;fiwejdif1#incl

Wuhan University of Science and Technology acm:1009: Chinese Edition C language Programming tutorial (second Edition) example 4.18

Problem DescriptionVerify Goldbach conjecture that any even number of sufficiently large (>=4) can be represented by a sum of two primes.InputEnter an even n. (2OutputFind A, B make n=a+bWhere A and B are two primes, and aSample Input4100Sample Output2 23 971#include 2 3#include 4 5 using namespacestd;6 7 intPrimeintm)8 9 {Ten One intI,n; A - if(m==1)return 0; - theN= (int) sqrt (Double) m); - - for(i=2; i) - + if(m%i==0)return 0; - +

Wuhan University of Science and Technology acm:1007: Chinese Edition C language Programming tutorial (second Edition) example 4.13

+ if((m%i==0) (n%i==0)) - $ { $ -printf"%d%d\n", i,m*n/i); - the Break; - Wuyi } the - } Wu - } About $ } - -}Other code:1#include 2#include 3 intgcdintAintb)4 {5 if(a%b==0)6 returnb;7 Else8 returnGCD (b,a%b);9 }Ten intLcmintAintb) One { A returna

1046:c Language Programming Tutorial (third Edition) after class exercise 10.4

The title description has n integers, so that the previous number in the order to move backward m position, the last m number into the front number of m, see figure. Write a function: To achieve the above function, in the main function input n number and output adjusted n number. Enter the number of input data n n integer moved position m output moving N number sample input101 2 3 4 5 6 7 8 9 102Sample output1#include 2 #defineN 1003 4 //Move Once5MoveintA[],intN)6 {7 intI, temp = a[n-1];8

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