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Search found 102 results on 5 pages for 'acm icpc'.

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  • Access Control Management Tool ACM.exe

    - by kaleidoscope
    The Access Control Management Tool (Acm.exe) is a command-line tool you can use to perform management operations (CREATE, UPDATE, GET, GET ALL, and DELETE) on the AppFabric Access Control entities (scopes, issuers, token policies, and rules). Basic Syntax The command line for Acm.exe follows the basic pattern of verb-noun. For example: acm.exe <command> <resource> [-option:<option value>] This tool will automatically generate random keys, which helps ensure that they can't easily be guessed by an attacker. Note that ACM.EXE is a thin wrapper around a REST Web Service (the AC management service). That helps to remember the commands it accepts, which are the typical resource management commands for a REST service: · Get(All) · Create · Update · Delete ACM.EXE.config file can be used to configure Host, Service and the Management key for a Service Namespace. Geeta, G

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  • cdc-acm driver: This device cannot do calls on its own. It is not a modem

    - by Sorcrer
    I am using Beagleboard-xm with 3.12 Kernel and ubuntu rootfs from Robert Nelson's site. I use a Telit HE910 GPS+GSM modem along with my project .So as per the HW user guide i have to apply a logic high for 5s on the input of this modem for enabling it So when I does this by toggling the gpio pin for 5s using a script I'm getting some messages on the terminal I am sure this message comes from the driver in usb/class/cdc-acm.c but couldn't find the reason behind this? How can I solve this issue?? root@arm:~# ./modem_on.sh Turning on Telit modem ...... going to sleep and toggle [ 70.791381] cdc_acm 1-2:1.0: This device cannot do calls on its own. It is not a modem. [ 74.390258] cdc_acm 1-2:1.0: This device cannot do calls on its own. It is not a modem. [ 74.406890] cdc_acm 1-2:1.2: This device cannot do calls on its own. It is not a modem. [ 74.462188] cdc_acm 1-2:1.4: This device cannot do calls on its own. It is not a modem. [ 74.478363] cdc_acm 1-2:1.6: This device cannot do calls on its own. It is not a modem. [ 74.495269] cdc_acm 1-2:1.8: This device cannot do calls on its own. It is not a modem. [ 74.510040] cdc_acm 1-2:1.10: This device cannot do calls on its own. It is not a modem. [ 74.530090] cdc_acm 1-2:1.12: This device cannot do calls on its own. It is not a modem. [ 74.619720] cdc_acm 1-2:1.0: This device cannot do calls on its own. It is not a modem. [ 74.634429] cdc_acm 1-2:1.2: This device cannot do calls on its own. It is not a modem. [ 74.649475] cdc_acm 1-2:1.4: This device cannot do calls on its own. It is not a modem. [ 74.664459] cdc_acm 1-2:1.6: This device cannot do calls on its own. It is not a modem. [ 74.678741] cdc_acm 1-2:1.8: This device cannot do calls on its own. It is not a modem. [ 74.693389] cdc_acm 1-2:1.10: This device cannot do calls on its own. It is not a modem. [ 74.708099] cdc_acm 1-2:1.12: This device cannot do calls on its own. It is not a modem. Script complete .......... The realted necessary portion of dmesg is below [ 30.623107] init: plymouth-upstart-bridge main process ended, respawning [ 70.629943] usb 1-2: new high-speed USB device number 2 using ehci-omap [ 70.782501] usb 1-2: config 1 interface 0 altsetting 0 endpoint 0x81 has an invalid bInterval 255, changing to 11 [ 70.782592] usb 1-2: New USB device found, idVendor=058b, idProduct=0041 [ 70.782623] usb 1-2: New USB device strings: Mfr=0, Product=0, SerialNumber=0 [ 70.791381] cdc_acm 1-2:1.0: This device cannot do calls on its own. It is not a modem. [ 70.801483] cdc_acm 1-2:1.0: ttyACM0: USB ACM device [ 73.041625] usb 1-2: USB disconnect, device number 2 [ 74.209930] usb 1-2: new high-speed USB device number 3 using ehci-omap [ 74.369049] usb 1-2: New USB device found, idVendor=1bc7, idProduct=0021 [ 74.369110] usb 1-2: New USB device strings: Mfr=1, Product=2, SerialNumber=3 [ 74.369140] usb 1-2: Product: Telit Wireless Module [ 74.369171] usb 1-2: Manufacturer: Telit wireless solutions [ 74.369201] usb 1-2: SerialNumber: 357164042197668 [ 74.390258] cdc_acm 1-2:1.0: This device cannot do calls on its own. It is not a modem. [ 74.400207] cdc_acm 1-2:1.0: ttyACM0: USB ACM device [ 74.406890] cdc_acm 1-2:1.2: This device cannot do calls on its own. It is not a modem. [ 74.416900] cdc_acm 1-2:1.2: ttyACM1: USB ACM device [ 74.462188] cdc_acm 1-2:1.4: This device cannot do calls on its own. It is not a modem. [ 74.472259] cdc_acm 1-2:1.4: ttyACM2: USB ACM device [ 74.478363] cdc_acm 1-2:1.6: This device cannot do calls on its own. It is not a modem. [ 74.488372] cdc_acm 1-2:1.6: ttyACM3: USB ACM device [ 74.495269] cdc_acm 1-2:1.8: This device cannot do calls on its own. It is not a modem. [ 74.505279] cdc_acm 1-2:1.8: ttyACM4: USB ACM device [ 74.510040] cdc_acm 1-2:1.10: This device cannot do calls on its own. It is not a modem. [ 74.520141] cdc_acm 1-2:1.10: ttyACM5: USB ACM device [ 74.530090] cdc_acm 1-2:1.12: This device cannot do calls on its own. It is not a modem. [ 74.540283] cdc_acm 1-2:1.12: ttyACM6: USB ACM device [ 74.619720] cdc_acm 1-2:1.0: This device cannot do calls on its own. It is not a modem. [ 74.629455] cdc_acm 1-2:1.0: ttyACM0: USB ACM device [ 74.634429] cdc_acm 1-2:1.2: This device cannot do calls on its own. It is not a modem. [ 74.644042] cdc_acm 1-2:1.2: ttyACM1: USB ACM device [ 74.649475] cdc_acm 1-2:1.4: This device cannot do calls on its own. It is not a modem. [ 74.659027] cdc_acm 1-2:1.4: ttyACM2: USB ACM device [ 74.664459] cdc_acm 1-2:1.6: This device cannot do calls on its own. It is not a modem. [ 74.674133] cdc_acm 1-2:1.6: ttyACM3: USB ACM device [ 74.678741] cdc_acm 1-2:1.8: This device cannot do calls on its own. It is not a modem. [ 74.688415] cdc_acm 1-2:1.8: ttyACM4: USB ACM device [ 74.693389] cdc_acm 1-2:1.10: This device cannot do calls on its own. It is not a modem. [ 74.703186] cdc_acm 1-2:1.10: ttyACM5: USB ACM device [ 74.708099] cdc_acm 1-2:1.12: This device cannot do calls on its own. It is not a modem. [ 74.717895] cdc_acm 1-2:1.12: ttyACM6: USB ACM device `

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  • What are the IEEE and ACM good for?

    - by Joshua Fox
    Membership in the IEEE and ACM is sometimes portrayed as a sign of professionalism. But all that is involved, as far as I can tell, is sending them your money. In return, besides the potential resume line, these organizations sponsor conferences and journals. I can always attend a conference or subscribe to or submit a paper to a journal, whether I am a member or not. If being a member makes some of that cheaper, or is a prerequisite for admission then OK, but I still don't see the purpose of these organizations. The answer, as far as I can gather, is that their most important value is to provide some reading material. I'd suggest that this is not worth the money given the wide availability of other valuable reading materials.

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  • For Programmers familiar with ACM API? Drawing Initials

    - by user71992
    I came across an exercise (in the book "The Art and Science of Java" by Eric Roberts) that requires using only GArc and GLine classes to create a lettering library which draws your initials on the canvas. This should be made independent of the GLabel class. I'd like to know the correct approach to use in solving this problem. I'm not sure what I have so far is good enough (I'm thinking it's too long). The questions requires that I use a good Top-Down approach.

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  • java bubblesort with acm dialog

    - by qzar
    Hi, the program gives following exception: Exception in thread "main" java.lang.NullPointerException at myclasses.BubbleSort.run(BubbleSort.java:42) at acm.program.Program.runHook(Program.java:1519) at acm.program.Program.startRun(Program.java:1508) at acm.program.Program.start(Program.java:729) at myclasses.BubbleSort.main(BubbleSort.java:49) what is wrong? thank you very much! package myclasses; import acm.program.DialogProgram; public class BubbleSort extends DialogProgram { int[] array; public int[] getArray() { return array; } public void setArray(int[] array) { this.array = array; } void swap(int firstPos, int secondPos) { int temp = array[firstPos]; array[firstPos] = array[secondPos]; array[secondPos] = temp; } public void bubblesort() { int i, j, k; for (i = 1; i < array.length; i++) { j = i; k = array[i]; while (j > 0 && array[j - 1] > k) { array[j] = array[j - 1]; --j; } array[j] = k; } } public void run() { BubbleSort a = new BubbleSort(); a.setArray(new int[] {1, 3, 5, 7, 6, 2}); a.bubblesort(); StringBuffer sb = new StringBuffer(a.array.length * 2); for (int i = 0; i < getArray().length; i++) sb.append(getArray()[i]).append(" "); println(sb); } public static void main(String[] args) { new BubbleSort().start(args); } }

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  • Join ACM, join IEEE, or read programming books?

    - by Laura
    I read blogs and listen to podcasts, and I own many of the "classic" programming books. For the money, what kind of printed material would you say is the most valuable to keep current in software engineering -- books, or magazines from professional organizations such as ACM and IEEE? Which organization has the best periodicals?

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  • For Programmers familiar with ACM API? Drawing Initials [closed]

    - by user71992
    Possible Duplicate: For Programmers familiar with ACM API? Drawing Initials I came across an exercise (in the book "The Art and Science of Java" by Eric Roberts) that requires using only GArc and GLine classes to create a lettering library which draws your initials on the canvas. This should be made independent of the GLabel class. I'd like to know the correct approach to use in solving this problem. I'm not sure what I have so far is good enough (I'm thinking it's too long). The questions requires that I use a good Top-Down approach. Here's my code so far: //Passes letters to GLetter objects and draws them on the canvas package artScienceJavaExercises.chapter8; import acm.program.*; //import acm.graphics.*; public class DrawInitials extends GraphicsProgram{ public void init(){ resize(400,400); } public void run(){ //String let = readLine("Letter?: "); letter = new GLetter("l"); add(letter, (getWidth()-letter.getWidth()*2)/2, (getHeight()-letter.getHeight())/2); add(new GLetter("o"), (letter.getX()+letter.getWidth()), letter.getY()); } private GLetter letter; } //GLetter Class package artScienceJavaExercises.chapter8; import acm.graphics.*; import java.awt.*; public class GLetter extends GCompound{ private static final int ONE_THIRD = 30; private static final int ROW_2_HEIGHT = 40; private GArc[] arc = new GArc[4]; private GLine[] line = new GLine[24]; public GLetter(String s){ line[0] = new GLine(0,0, ONE_THIRD, 0); line[1] = new GLine(ONE_THIRD,0, ONE_THIRD*2, 0); line[2] = new GLine(ONE_THIRD*2,0, ONE_THIRD*3, 0); line[3] = new GLine(0,0, 0,ONE_THIRD); line[4] = new GLine(ONE_THIRD,0, ONE_THIRD, ONE_THIRD); line[5] = new GLine(ONE_THIRD*2,0, ONE_THIRD*2, ONE_THIRD); line[6] = new GLine(ONE_THIRD*3,0, ONE_THIRD*3, ONE_THIRD); line[7] = new GLine(0,ONE_THIRD, ONE_THIRD*2, ONE_THIRD); line[8] = new GLine(ONE_THIRD,ONE_THIRD, ONE_THIRD*2, ONE_THIRD); line[9] = new GLine(ONE_THIRD*2,ONE_THIRD, ONE_THIRD*3, ONE_THIRD); line[10] = new GLine(0,ONE_THIRD, 0, ONE_THIRD+ROW_2_HEIGHT); line[11] = new GLine(ONE_THIRD, ONE_THIRD, ONE_THIRD, ONE_THIRD+ROW_2_HEIGHT); line[12] = new GLine(ONE_THIRD*2,ONE_THIRD, ONE_THIRD*2, ONE_THIRD+ROW_2_HEIGHT); line[13] = new GLine(ONE_THIRD*3,ONE_THIRD, ONE_THIRD*3, ONE_THIRD+ROW_2_HEIGHT); line[14] = new GLine(0, ONE_THIRD+ROW_2_HEIGHT, ONE_THIRD, ONE_THIRD+ROW_2_HEIGHT); line[15] = new GLine(ONE_THIRD, ONE_THIRD+ROW_2_HEIGHT, ONE_THIRD*2, ONE_THIRD+ROW_2_HEIGHT); line[16] = new GLine(ONE_THIRD*2, ONE_THIRD+ROW_2_HEIGHT, ONE_THIRD*3, ONE_THIRD+ROW_2_HEIGHT); line[17] = new GLine(0, ONE_THIRD+ROW_2_HEIGHT, 0, ONE_THIRD*2+ROW_2_HEIGHT); line[18] = new GLine(ONE_THIRD, ONE_THIRD+ROW_2_HEIGHT, ONE_THIRD, ONE_THIRD*2+ROW_2_HEIGHT); line[19] = new GLine(ONE_THIRD*2, ONE_THIRD+ROW_2_HEIGHT, ONE_THIRD*2, ONE_THIRD*2+ROW_2_HEIGHT); line[20] = new GLine(ONE_THIRD*3, ONE_THIRD+ROW_2_HEIGHT, ONE_THIRD*3, ONE_THIRD*2+ROW_2_HEIGHT); line[21] = new GLine(0,ONE_THIRD*2+ROW_2_HEIGHT, ONE_THIRD, ONE_THIRD*2+ROW_2_HEIGHT); line[22] = new GLine(ONE_THIRD, ONE_THIRD*2+ROW_2_HEIGHT, ONE_THIRD*2, ONE_THIRD*2+ROW_2_HEIGHT); line[23] = new GLine(ONE_THIRD*2,ONE_THIRD*2+ROW_2_HEIGHT, ONE_THIRD*3, ONE_THIRD*2+ROW_2_HEIGHT); for(int i = 0; i<line.length; i++){ add(line[i]); line[i].setColor(Color.BLACK); line[i].setVisible(false); } arc[0] = new GArc(getWidth(), getHeight(), 106.699, 49.341); arc[1] = new GArc(getWidth(), getHeight(), 23.96, 49.341); arc[2] = new GArc(getWidth(), getHeight(), -23.96, -49.341); arc[3] = new GArc(0,0,getWidth(), getHeight(), -106.699, -49.341); for(int i = 0; i<arc.length; i++){ add(arc[i],0,0); arc[i].setColor(Color.BLACK); arc[i].setVisible(false); } paintLetter(s); } private void paintLetter(String s){ if (s.equalsIgnoreCase("l")){ turnOn(line[3]); turnOn(line[10]); turnOn(line[17]); turnOn(line[21]); turnOn(line[22]); turnOn(line[23]); } else if(s.equalsIgnoreCase("o")){ for(int i = 0; i<4; ++i){ turnOn(arc[i]); } turnOn(line[1]); turnOn(line[10]); turnOn(line[13]); turnOn(line[22]); } } private void turnOn(GObject g){ g.setVisible(true); } } I created a class (GLetter.java) with arrays for GArc and GLine objects. They are positioned in certain ways thereby turning certain Glines and/or GArcs on or off (changing visiblity) would create a pattern for a letter. This Gletter uses the if/else statements to determine which pattern to create - this makes me feel my code is too long. There is another class (DrawInitials.java) that simulates a GraphicsProgram and allows the user to pass certain letters as arguments to the GLetter object. I've used 'L' and 'O' as examples. However, I posted this because I'm not sure I'm using the right approach. That's why I need your help. I feel MY CODE IS TOO LONG! The code above is not the complete project...it only draws letters 'L' and 'O' for now.

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  • Java Code for calculating Leap Year, is this code correct ?

    - by Ibn Saeed
    Hello I am following "The Art and Science of Java" book and it shows how to calculate a leap year. The book uses ACM Java Task Force's library. Here is the code the books uses: import acm.program.*; public class LeapYear extends ConsoleProgram { public void run() { println("This program calculates leap year."); int year = readInt("Enter the year: "); boolean isLeapYear = ((year % 4 == 0) && (year % 100 != 0) || (year % 400 == 0)); if (isLeapYear) { println(year + " is a leap year."); } else println(year + " is not a leap year."); } } Now, this is how I calculated the leap year. import acm.program.*; public class LeapYear extends ConsoleProgram { public void run() { println("This program calculates leap year."); int year = readInt("Enter the year: "); if ((year % 4 == 0) && year % 100 != 0) { println(year + " is a leap year."); } else if ((year % 4 == 0) && (year % 100 == 0) && (year % 400 == 0)) { println(year + " is a leap year."); } else { println(year + " is not a leap year."); } } } Is there anything wrong with my code or should i use the one provided by the book ? EDIT :: Both of the above code works fine, What i want to ask is which code is the best way to calculate the leap year.

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  • CodeGolf: Brothers

    - by John McClane
    Hi guys, I just finished participating in the 2009 ACM ICPC Programming Conest in the Latinamerican Finals. These questions were for Brazil, Bolivia, Chile, etc. My team and I could only finish two questions out of the eleven (not bad I think for the first try). Here's one we could finish. I'm curious to seeing any variations to the code. The question in full: ps: These questions can also be found on the official ICPC website available to everyone. In the land of ACM ruled a greeat king who became obsessed with order. The kingdom had a rectangular form, and the king divided the territory into a grid of small rectangular counties. Before dying the king distributed the counties among his sons. The king was unaware of the rivalries between his sons: The first heir hated the second but not the rest, the second hated the third but not the rest, and so on...Finally, the last heir hated the first heir, but not the other heirs. As soon as the king died, the strange rivaly among the King's sons sparked off a generalized war in the kingdom. Attacks only took place between pairs of adjacent counties (adjacent counties are those that share one vertical or horizontal border). A county X attacked an adjacent county Y whenever X hated Y. The attacked county was always conquered. All attacks where carried out simultanously and a set of simultanous attacks was called a battle. After a certain number of battles, the surviving sons made a truce and never battled again. For example if the king had three sons, named 0, 1 and 2, the figure below shows what happens in the first battle for a given initial land distribution: INPUT The input contains several test cases. The first line of a test case contains four integers, N, R, C and K. N - The number of heirs (2 <= N <= 100) R and C - The dimensions of the land. (2 <= R,C <= 100) K - Number of battles that are going to take place. (1 <= K <= 100) Heirs are identified by sequential integers starting from zero. Each of the next R lines contains C integers HeirIdentificationNumber (saying what heir owns this land) separated by single spaces. This is to layout the initial land. The last test case is a line separated by four zeroes separated by single spaces. (To exit the program so to speak) Output For each test case your program must print R lines with C integers each, separated by single spaces in the same format as the input, representing the land distribution after all battles. Sample Input: Sample Output: 3 4 4 3 2 2 2 0 0 1 2 0 2 1 0 1 1 0 2 0 2 2 2 0 0 1 2 0 0 2 0 0 0 1 2 2 Another example: Sample Input: Sample Output: 4 2 3 4 1 0 3 1 0 3 2 1 2 2 1 2

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  • what is the task of a coach in acm programming contests?

    - by Layla
    In the university that I am working they have decided to participate in the ACM regionals for the first time, they would like to appoint me like a coach. I have never been into that situation before and have not found so much information about it, so what is the real work of a coach in those contests? Sometimes I have found experienced programmers like coaches, but others are just people with no so good programming skills; so what is all about?

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  • Where can you find fun/educational programming challenges?

    - by tj9991
    I've searched around for different challenge sites, and most of them seem to be geared towards difficulty in problem solving logically, rather than trying to use your language of choice to do something you haven't used it for. Their center is around mathematics rather than function design. Some kind of point system for correctly solving challenges, or solving them the most efficient/smallest would be neat as well. Listed sites Project Euler TopCoder UVa Online Judge Challenges with Python Google Code Jam Programming Challenges Less Than Dot ACM's Programing Contest archive USACO problems ITA Software's puzzle page Refactor My Code Ruby Quiz

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  • being able to solve google code jam problem sets

    - by JPro
    This is not a homework question, but rather my intention to know if this is what it takes to learn programming. I keep loggin into TopCoder not to actually participate but to get the basic understand of how the problems are solved. But to my knowledge I don't understand what the problem is and how to translate the problem into an algorithm that can solve it. Just now I happen to look at ACM ICPC 2010 World Finals which is being held in china. The teams were given problem sets and one of them is this: Given at most 100 points on a plan with distinct x-coordinates, find the shortest cycle that passes through each point exactly once, goes from the leftmost point always to the right until it reaches the rightmost point, then goes always to the left until it gets back to the leftmost point. Additionally, two points are given such that the the path from left to right contains the first point, and the path from right to left contains the second point. This seems to be a very simple DP: after processing the last k points, and with the first path ending in point a and the second path ending in point b, what is the smallest total length to achieve that? This is O(n^2) states, transitions in O(n). We deal with the two special points by forcing the first path to contain the first one, and the second path contain the second one. Now I have no idea what I am supposed to solve after reading the problem set. and there's an other one from google code jam: Problem In a big, square room there are two point light sources: one is red and the other is green. There are also n circular pillars. Light travels in straight lines and is absorbed by walls and pillars. The pillars therefore cast shadows: they do not let light through. There are places in the room where no light reaches (black), where only one of the two light sources reaches (red or green), and places where both lights reach (yellow). Compute the total area of each of the four colors in the room. Do not include the area of the pillars. Input * One line containing the number of test cases, T. Each test case contains, in order: * One line containing the coordinates x, y of the red light source. * One line containing the coordinates x, y of the green light source. * One line containing the number of pillars n. * n lines describing the pillars. Each contains 3 numbers x, y, r. The pillar is a disk with the center (x, y) and radius r. The room is the square described by 0 = x, y = 100. Pillars, room walls and light sources are all disjoint, they do not overlap or touch. Output For each test case, output: Case #X: black area red area green area yellow area Is it required that people who program should be should be able to solve these type of problems? I would apprecite if anyone can help me interpret the google code jam problem set as I wish to participate in this years Code Jam to see if I can do anthing or not. Thanks.

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  • A fatal exception 0E occured at 0028:xxxxxx in VxD IOS(01)

    - by winlin
    I get a blue screen of death in my windows 98 machine every time I boot it. I can't reach to my desktop. The error is like this: A fatal exception 0E occured at 0028:C003CC2F in VxD IOS(01) + 0000156B This was called from 0028:C0082E60 in VxD VKD(01) + 000001D0 I have to then give it a three finger salute to restart the system. There is no other way to shut down the system at this point except pressing the CPU power button. What could be the problem? My windows system.ini is: [boot] oemfonts.fon=vgaoem.fon shell=Explorer.exe system.drv=system.drv drivers=mmsystem.dll power.drv user.exe=user.exe gdi.exe=gdi.exe sound.drv=mmsound.drv dibeng.drv=dibeng.dll comm.drv=comm.drv mouse.drv=mouse.drv keyboard.drv=keyboard.drv *DisplayFallback=0 fonts.fon=vgasys.fon fixedfon.fon=vgafix.fon 386Grabber=vgafull.3gr display.drv=pnpdrvr.drv [keyboard] keyboard.dll= oemansi.bin= subtype= type=4 [boot.description] system.drv=Standard PC mouse.drv=Standard mouse keyboard.typ=Standard 101/102-Key or Microsoft Natural Keyboard aspect=100,96,96 display.drv=Standard PCI Graphics Adapter (VGA) [386Enh] ;device=tddebug.386 ;device=D:\TC\TASM\BIN\WINDPMI.386 ebios=*ebios woafont=dosapp.fon mouse=*vmouse, msmouse.vxd device=*dynapage device=*vcd device=*vpd device=*int13 keyboard=*vkd display=*vdd,*vflatd ConservativeSwapfileUsage=0 Paging=on [NonWindowsApp] TTInitialSizes=4 5 6 7 8 9 10 11 12 13 14 15 16 18 20 22 [power.drv] [drivers] wavemapper=*.drv MSACM.imaadpcm=*.acm ;msvideo.STV680=STV680sg.drv midi=mmsystem.dll wave=mmsystem.dll MSACM.msadpcm=*.acm [iccvid.drv] [mciseq.drv] [mci] cdaudio=mcicda.drv sequencer=mciseq.drv waveaudio=mciwave.drv avivideo=mciavi.drv videodisc=mcipionr.drv vcr=mcivisca.drv MPEGVideo=mciqtz.drv MPEGVideo2=mciqtz.drv [vcache] [MSNP32] [DISPLAY] BusThrottle=1 [network] SSID=1438661605 [vicax] msacm711=74603 msacm811=148933 msacm911=42405 [Sessew] VideoManufacturer=Standard VGA VideoBoard=Standard Display Adapter (VGA) MouseType=0 VidType=0 Mono=0 Ddraw=1 [drivers32] msacm.lhacm=lhacm.acm VIDC.IV50=ir50_32.dll msacm.iac2=C:\WINDOWS\SYSTEM\IAC25_32.AX VIDC.YUY2=msyuv.dll VIDC.UYVY=msyuv.dll VIDC.YVYU=msyuv.dll msacm.msaudio1=msaud32.acm msacm.vorbis=vorbis.acm msacm.l3acm=C:\WINDOWS\SYSTEM\L3CODECA.ACM msacm.sl_anet=sl_anet.acm VIDC.TSCC=tsccvid.dll VIDC.IV41=IR41_32.AX vidc.mpg4=mpg4c32.dll vidc.mp43=mpg4c32.dll msacm.voxacm160=vct3216.acm MSACM.msadpcm=msadp32.acm [TTFontDimenCache] 0 4=2 4 0 5=3 5 0 6=4 6 0 7=4 7 0 8=5 8 0 9=5 9 0 10=6 10 0 11=7 11 0 12=7 12 0 13=8 13 0 14=8 14 0 15=9 15 0 16=10 16 0 18=11 18 0 20=12 20 0 22=13 22

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  • What should be the potential reason to get runtime error for this program?

    - by MiNdFrEaK
    #include<iostream> #include<stack> #include<vector> #include<string> #include<fstream> #include<cstdlib> /*farnaws,C++,673,08/12/2012*/ using namespace std; string verifier(string input_line) { stack <char> braces; for(int i=0; i<input_line.size(); i++) { if(input_line[i]=='(' || input_line[i]=='[') { braces.push(input_line[i]); } else if(input_line[i]==')' || input_line[i]==']') { braces.pop(); } } if(braces.size()==0) { return "YES"; } else { return "NO"; } } int main() { ifstream file_input("input.in"); string read_file; vector<string> file_contents; if(file_input.is_open()) { while(file_input>>read_file) { file_contents.push_back(read_file); } } else { cout<<"File cant be open!"<<endl; } int limit=atoi(file_contents[0].c_str()); //cout<< limit; ofstream file_output("output.out"); if(file_output.is_open()) { for(int i=1; i<=limit; i++ ) { file_output<<verifier(file_contents[i])<<endl; } } else { cout<<"File cant be open!"<<endl; } return 0; }

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  • Adaptive Case Management Modeling with CMMN by Jessica Ray

    - by JuergenKress
    A new version of Oracle BPM Suite 11.1.1.7 with Adaptive Case Management (ACM) is now available, so what will that mean for requirements gathering? BPM project requirements can be documented using Business Process Model and Notation (BPMN 2.0). For ACM, there is a new notation in the works. It is called Case Management Model and Notation (CMMN). For now, this notation isn’t included as a modeling tool in the new version of Oracle BPM Suite 11.1.1.7 with ACM, but it is possible that a modeling tool could be included in a future release. What is CMMN? CMMN is a standard intended to capture the common elements that Case Management Products use, the same way that BPMN is a standard for BPM products (such as Oracle BPM). CMMN is created by the Object Management Group (OMG) and is still in the beta version. In April 2014, OMG released the second beta version the CMMN 1.0, and the most recent version is available here. CMMN captures some of the elements that are commonly used when talking about ACM such as Cases, Milestones, and Tasks. It also introduces some elements that you may not automatically hear when talking about ACM such as Stages, Events, and Decorators. Here is a quick summary at a few (but not all) of the elements of CMMN taken from the CMMN spec. A Few CMMN Elements Read the complete article here SOA & BPM Partner Community For regular information on Oracle SOA Suite become a member in the SOA & BPM Partner Community for registration please visit www.oracle.com/goto/emea/soa (OPN account required) If you need support with your account please contact the Oracle Partner Business Center. Blog Twitter LinkedIn Facebook Wiki Technorati Tags: Jessica Ray,Avio,Adaptive Case Management,ACM,CMMN,SOA Community,Oracle SOA,Oracle BPM,Community,OPN,Jürgen Kress

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  • How to extract a Vorbis stream from a WAVE file?

    - by H.B.
    I would like to move the Vorbis stream into an ogg container but ffmpeg does not seem to recognize the stream. Even though MPlayer gives this output upon playback: Opening audio decoder: [acm] Win32/ACM decoders Loading codec DLL: 'vorbis.acm' Loaded DLL driver vorbis.acm at 10000000 Warning! ACM codec reports srcsize=0 AUDIO: 44100 Hz, 2 ch, s16le, 128.0 kbit/9.07% (ratio: 16000-176400) Selected audio codec: [vorbisacm] afm: acm (OggVorbis ACM) ffmpeg: ffmpeg -i Source.wav -acodec copy Target.ogg Input #0, wav, from 'Source.wav': Duration: 00:02:15.17, bitrate: 128 kb/s Stream #0.0: Audio: qg[0][0] / 0x6771, 44100 Hz, 2 channels, 128 kb/s [ogg @ 00000000003096C0] Unsupported codec id in stream 0 Output #0, ogg, to 'Target.ogg': Metadata: encoder : Lavf53.6.0 Stream #0.0: Audio: qg[0][0] / 0x6771, 44100 Hz, 2 channels, 128 kb/s Stream mapping: Stream #0.0 -> #0.0 Could not write header for output file #0 (incorrect codec parameters ?) Of course this does not necessarily need to be done via ffmpeg, any method that is workable would be fine... I have cut down one of the files to 512KB: sample.wav (Changed two chunk size fields in the wave header to account for this, the embedded stream is cut "without notice")

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  • Adaptive Case Management – Exposing the API – part 1 by Roger Goossens

    - by JuergenKress
    One of the most important building blocks of Adaptive Case Management is the ACM API. At one point or another you’re gonna need a way to get information (think about a list of stakeholders, available activities, milestones reached, etc.) out of the case. Since there’s no webservice available yet that exposes the internals of the case, your only option right now is the ACM API. ACM evangelist Niall Commiskey has put some samples online to give you a good feeling of the power of the ACM API. The examples show how you can access the API by means of RMI. You first need to obtain a BPMServiceClientFactory that gives access to the important services you’ll mostly be needing, i.e. the IBPMUserAuthenticationService (needed for obtaining a valid user context) and the ICaseService (the service that exposes all important case information). Now, obtaining an instance of the BPMServiceClientFactory involves some boilerplate coding in which you’ll need the RMI url and user credentials: Read the complete article here. SOA & BPM Partner Community For regular information on Oracle SOA Suite become a member in the SOA & BPM Partner Community for registration please visit www.oracle.com/goto/emea/soa (OPN account required) If you need support with your account please contact the Oracle Partner Business Center. Blog Twitter LinkedIn Facebook Wiki Technorati Tags: ACM,API,Adaptive Case Management,Community,Oracle SOA,Oracle BPM,OPN,Jürgen Kress

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  • Is "send us a page with code" a typical interview requirement?

    - by acm
    Recently I was asked to show "a page with code" for a job interview. Being mainly a back-end programmer, and that's the position I applied for, I first said to the person I was talking to exactly that: PHP is executed at the server and therefore not visible by just giving a "page". However, following their desire, I sent links to the pages I've worked on before. Obviously they couldn't see anything except for the HTML, CSS, JS... They said it was not enough, they could not see the PHP. Understanding that they probably just wanted to know my skills and/or interest I sent them my Stack Overflow profile. Among all my questions and answers, most of them with code, certainly the PHP is there. But it seems this is not what they wanted. Well, I don't have any code put together that I can simply publish for someone to see. And I would never do it for the code I have deployed, obviously. So my question is/are: What does "send us a page with code" mean? What should I send? Is this a typical interview requirement?

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  • Combining two data sets and plotting in matlab

    - by bautrey
    I am doing experiments with different operational amplifier circuits and I need to plot my measured results onto a graph. I have two data sets: freq1 = [.1 .2 .5 .7 1 3 4 6 10 20 35 45 60 75 90 100]; %kHz Vo1 = [1.2 1.6 1.2 2 2 2.4 14.8 20.4 26.4 30.4 53.6 68.8 90 114 140 152]; %mV V1 = 19.6; Acm = Vo1/(1000*V1); And: freq2 = [.1 .5 1 30 60 70 85 100]; %kHz Vo1 = [3.96 3.96 3.96 3.84 3.86 3.88 3.88 3.88]; %V V1 = .96; Ad = Vo1/(2*V1); (I would show my plots but apparently I need more reps for that) I need to plot the equation, CMRR vs freq: CMRR = 20*log10(abs(Ad/Acm)); The size of Ad and Acm are different and the frequency points do not match up, but the boundaries of both of these is the same, 100Hz to 100kHz (x-axis). On the line of CMRR, Matlab says that Ad and Acm matrix dimensions do not agree. How I think I would solve this is using freq1 as the x-axis for CMRR and then taking approximated points from Ad according to the value on freq1. Or I could do function approximations of Ad and Acm and then do the divide operator on those. I do not know how I would code up this two ideas. Any other ideas would helpful, especially simpler ones. Thanks

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  • SOA Community Newsletter June 2013

    - by JuergenKress
    Dear SOA partner community member Thanks for showing us your interest to rerun the Fusion Middleware Summer Camps! After knowing your suggestions we are happy to announce the 3rd edition of our advanced Fusion Middleware training. The camps will take place from August 26th - 30th 2013 in Lisbon Portugal. Topics will include Adaptive Case Management (ACM) as part of BPM Suite, b2b, Advanced SOA and SOA Governance. Please make sure you plan and book your seat in advance - (Booking is on the basis of first come first seat!). Thanks for all your efforts to become certified and Specialized. For all the experts who achieved the SOA Suite 11g Essentials or BPM Suite 11g Certified Implementation Specialist, you can download a logo for your blog or business card at the Competence Center. For all the companies who achieved a SOA or BPM specialization you can request a nice Plaques for your office. As part of our Industrial SOA article services we published “Canonizing a Language for Architecture” in the Service Technology Magazine and on Oracle Technology Network. If you write books or a blog - make sure you share it with us! Cloud Computing is the hottest topic in IT, specially as an architect you should be aware of the concepts and technology, therefore I highly recommend you Thomas Erl’s latest book named “Cloud Computing”. In the BPM space, Adaptive Case Management (ACM) is the hottest topic, with BPM PS6 the backend ACM functionality and an ACM sample application are available. You can even combine this hype with Customer Experience. The BPM section in this newsletter reflects the high importance of the topic and includes BPM PS6 video showing process lifecycle,BPM Resource Kit, Functional Testing, Introduction to Web Forms, Customized Workspace Application and Instance Patching Demo. B2B also become more and more popular in the Oracle SOA Suite. If you could not attend the training organized in the month May, we offer you an additional B2B training as a part of the Summer Camps or you can download the B2B training material from our SOA Community Workspace (SOA Community membership required). Thanks to all for sharing the valuable SOA content with our community! Special thanks to ec4u for the new reference of SOA Suite and AIA Foundation Pack at a Swiss insurance company. It is time to submit a SOA and BPM  reference request today! In this edition of the newsletter you will see Guido and Ronald's second part of OSB article series and Kathiravan Udayakumar's published an exclusive article on SOA Suite best practice. If you want to submit your content for the next edition of the Newsletter then please feel free to submit it to myself. The A-Team is an excellent contributor to the best practice - make sure you visit the new A-Team page and read their articles such as Getting to know Maven. Also on the SOA side, we have published many new articles from the community Oracle SOA Suite for the Busy IT Professional by Frank Munz, SOA Suite Knowledge - Polyglot Service Implementation with Groovy by Alexander Suchier, QA82 Analyzer - Automated Quality Assurance for Oracle SOA Suite Projects, Verifying the Target by Anthony Reynolds and a new book called Oracle SOA Governance 11g Implementation book by Luis Augusto Weir. Two new SOA on-demand training courses NEW - Oracle Business Rules Self-Study Course & Introduction Human Workflow online course are available now! Make use of the Summer Time and get trained - hope to see you in Lisbon for the Summer Camps! Jürgen Kress Oracle SOA & BPM Partner Adoption EMEA To read the newsletter please visit http://tinyurl.com/soanewsJune2013 (OPN Account required) To become a member of the SOA Partner Community please register at http://www.oracle.com/goto/emea/soa (OPN account required) If you need support with your account please contact the Oracle Partner Business Center. Blog Twitter LinkedIn Mix Forum Technorati Tags: SOA Community newsletter,SOA Community,Oracle,OPN,Jürgen Kress,SOA,BPM

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  • SOA Community Newsletter May 2014

    - by JuergenKress
    Registration for the Fusion Middleware Summer Camps 2014 is open – Register asap for one of our bootcamps August 4th – 8th 2014 in Lisbon. Please read details and pre-requisitions careful before you register. We expect that like in the past, the conference will be booked out soon! If you can’t make it to Lisbon attend our SOA Suite 11c free on-demand Bootcamp or  Managing the Complexity of IoT online trainings. With more than 5000 customers, SOA Suite Achieves Significant Customer Adoption and Industry Recognition.Thanks to all our SOA Specialized partners for making our joins SOA customers successful! As a summary of the Industrial SOA series we published the Podcast Show Notes: SOA and Cloud - Where's This Relationship Going? Make sure you use the Oracle Demo Systems for your customer presentations. The demo systems are hosted by Oracle and include complete scenarios based on the latest Middleware version like the new B2B SOA Suite Demo System! For local presentations without fast internet use the SOA/BPM 11.1.1.7.1 Virtual Machine and Case Management Sample. At our SOA Community Workspace (SOA Community membership required) you can get new IoT presentations for Location Based Offers for Banking & Whitepaper and online Webcast & Utility presentation. In this newsletter you will find many articles about OSB: OSB 11g – A Hands-on Tutorial & Using Split-Joins in OSB Services for parallel processing of messages & OSB, Service Callouts and OQL & Working with Oracle Security Token Service. Thanks for sharing all the additional SOA articles within the community: How to configure Oracle SOA/BPM task auto release & Controlling BPEL process flow at runtime & Upgrading to Oracle SOA Suite 11g PS6 (11.1.1.7)? Do this. & BPEL and BPM's performance monitoring using DMS & SOA 11g - Create RESTful Service In Oracle SOA & Wrong timezone causes TopLink warning in SOA suite. Highlight of the BPM and ACM section is the IDC BPM vendor report. The new bundle Patch including the ACM UI is now available. If you want to learn more about ACM, get the ACM training material at our SOA Community Workspace (SOA Community membership required). A great demo for your next BPM presentation is the BPM iPad app. It’s simpleMobile BPM is Not An Option. It’s a Necessity. Thanks for sharing all the additional BPM articles within the community: BPM update adds Case Management Web Interface and REST APIs & Implementing deadline functionality with Oracle Adaptive Case Management & BPM 11g Timeout Heuristics & Humantask Assignment: Names and Expressions Assignment via Rules. In our last section Architecture, it is all about design. Usability is a key factor for customer satisfaction, worth to spend some time and read the Simplified User Experience Design Patterns eBook. Great blueprint for your project! See you in Lisbon! To read the newsletter please visit www.tinyurl.com/soaNewsMay2014 (OPN Account required) To become a member of the SOA Partner Community please register at http://www.oracle.com/goto/emea/soa (OPN account required) If you need support with your account please contact the Oracle Partner Business Center. Blog Twitter LinkedIn Facebook Wiki Mix Forum Technorati Tags: newsletter,SOA Community newsletter,SOA Community,Oracle,OPN,Jürgen Kress

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  • How to find out memory layout of your data structure implementation on Linux 64bit machine

    - by ajay
    In this article, http://cacm.acm.org/magazines/2010/7/95061-youre-doing-it-wrong/fulltext the author talks about the memory layouts of 2 data structures - The Binary Heap and the B-Heap and compares how one has better memory layout than the other. http://deliveryimages.acm.org/10.1145/1790000/1785434/figs/f5.jpg http://deliveryimages.acm.org/10.1145/1790000/1785434/figs/f6.jpg I want to get hands on experience on this. I have an implementation of a N-Ary Tree and I want to find out the memory layout of my data structure. What is the best way to come up with a memory layout like the one in the article? Secondly, I think it is easier to identify the memory layout if it is an array based implementation. If the implementation of a Tree uses pointers then what Tools do we have or what kind of approach is required to map it's memory layout? Thanks!

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