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  • Read number from text file using C++ stream

    - by Yongwei Xing
    Hi all I have a text file like below 2 1 2 5 10 13 11 12 14 2 0 1 2 99 2 200 2 1 5 5 1 2 3 4 5 1 0 0 0 I want to read file line by line, and read the umbers from each line. I know how to use the stream to read a fixed field line, but what about the non-fixed line? Best Regards,

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  • use printf("text %d", number) type format to assign value to variable

    - by rksprst
    I would like to use the syntax that printf uses, using the %d, %s and adding values after to assign a value to a char[]. Is this possible? e.g. Given an output of: printf("now: %d-%d-%d %d:%d:%d\n", tm.tm_year + 1900, tm.tm_mon + 1, tm.tm_mday, tm.tm_hour, tm.tm_min, tm.tm_sec); I'd like to assign that to char[] output; How can this be done? I tried: sprintf(output, "now: %d-%d-%d %d:%d:%d\n", tm.tm_year + 1900, tm.tm_mon + 1, tm.tm_mday, tm.tm_hour, tm.tm_min, tm.tm_sec); but that didn't seem to work. Is sprintf used differently... or is that not what I should be using? Thanks!

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  • Generate 3 random number that sum to 1 in R

    - by user1034797
    I am hoping to create 3 (non-negative) quasi-random numbers that sum to one, and repeat over and over. Basically I am trying to partition something into three random parts over many trials. While I am aware of a= runif(3,0,1) I was thinking that I could use 1-a as the max in the next run if, but it seems messy. But these of course don't sum to one. Any thoughts, oh wise stackoverflow-ers?

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  • how to number the datas from mysql

    - by udaya
    Hi This is a doubt on mysql select query let me axplain my doubt with a simple example consider this is my query SELECT dbCountry from tableCountry tableCountry has fields dbCuntryId dbCountry and dbState I have the result as dbCountry india america england kenya pakisthan I need the result as 1 india 2 america 3 england 4 kenya 5 pakisthan the numbers 12345 must be generated with the increase in data and itis not a autoincrement id How can i get it is it something like loop

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  • Quickest way to compute the number of shared elements between two vectors

    - by shn
    Suppose I have two vectors of the same size vector< pair<float, NodeDataID> > v1, v2; I want to compute how many elements from both v1 and v2 have the same NodeDataID. For example if v1 = {<3.7, 22>, <2.22, 64>, <1.9, 29>, <0.8, 7>}, and v2 = {<1.66, 7>, <0.03, 9>, <5.65, 64>, <4.9, 11>}, then I want to return 2 because there are two elements from v1 and v2 that share the same NodeDataIDs: 7 and 64. What is the quickest way to do that in C++ ? Just for information, note that the type NodeDataIDs is defined as I use boost as: typedef adjacency_list<setS, setS, undirectedS, NodeData, EdgeData> myGraph; typedef myGraph::vertex_descriptor NodeDataID; But it is not important since we can compare two NodeDataID using the operator == (that is, possible to do v1[i].second == v2[j].second)

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  • what math do i need to convert this number

    - by Uberfuzzy
    given an X, what math is needed to find its Y, using this table? x->y 0->1 1->0 2->6 3->5 4->4 5->3 6->2 language agnostic problem and no, i dont/cant just store the array, and do the lookup. yes, the input will always be the finite set of 0 to 6. it wont be scaling later.

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  • Jquery incrementing number +1 0n page refresh.

    - by sameast
    Hi guys i am trying to modify this code here. $(document).ready(function() { var randomImages = ['img-1','img-2','img-3','img-4']; var rndNum = Math.floor(Math.random() * randomImages.length); $("div#bg-image").css({ background: "url(http://example.com/images/" + randomImages[rndNum] + ".jpg) no-repeat" }); }); This works fine and gives me a random image each time which is great but what i need is for the image to increment +1 each time on page refresh. This is because sometimes i can refresh 3 times and will still get the same image show when using Math.random(). I need to cancel the random and set +1 each time. Any help

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  • construct a unique number for a string in java

    - by praveen
    We have a requirement of reading/writing more than 10 million strings into a file. Also we do not want duplicates in the file. Since the strings would be flushed to a file as soon as they are read we are not maintaining it in memory. We cannot use hashcode because of collisions in the hash code due to which we might miss a string as duplicate. Two other approaches i found in my googling: 1.Use a message digest algorithm like MD5 - but it might be too costly to calculate and store. 2.Use a checksum algorithm. [i am not sure if this produces a unique key for a string- can someone please confirm] Is there any other approach avaiable. Thanks.

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  • PHP number range menu from array

    - by Julious
    Hello people! I'm a bit confused here. I have a php array like this Array(2010,2009,2008 ...1992) and i want to create a loop to print a menu with a four year range counting down like this 2010-2006 2005-2001 2000-1996 etc.. How can i do this Everything i tried end up in an endless loop. THnx in advance. J.

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  • How much does precomputation (matching a series of strings and their permutations with a set number

    - by nipun
    Consider a typical slots machine with n reels(say reel1: a,b,c,d,w1,d,b, ..etc). On play we generate a concatenated string of n objects (like for above, chars) We have a paytable which lists winning strings with payout amounts. The problem is a wild character (list of wilds: w1,w2) which can replace {w1:a,b,c},{w2:a} ..etc. Is it really worthwhile to have all possible winning strings permutations with the wilds precomputed and used or simply at the time of occurance, generate all combinations with the pattern in hand accordingly. I did'nt really see much difference initially, but now if I need to scale the machine to handle 11+ reels with a much higher concentration of wilds than previously, I need to figure out the exact approach for this particular bit. Any ideas will be really appreciated :)

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  • Determining smallest number of samples for 99% accuracy

    - by test
    I'm trying to compare 100,000 records on a local database (L) with 100,000 records on a remote database (R). Basically I want to know if an elment in L exusts in R. To determine that, I have to make a request against the R for each L, which takes a long time (I know, there should be a better way, there isn't, that's the API I've got). So I would like to test a small sample of L against R, and then infer with some level of confidence how many are present in the whole R. How many do I have to test to have a 99% confidence level?

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  • 1136: Incorrect number of arguments. Expected 0.? AS3 Flash Cs4

    - by charmaine
    Basically i am working through a book called..Foundation Actionscript 3.0 Animation, making things move. i am now on Chapter 9 - collision detection. On two lines of my code i get the 1135 error, letting me know that i have an incorrect number of arguments. Can anybody help me out on why this may be? package { import flash.display.Sprite; import flash.events.Event; public class Bubbles extends Sprite { private var balls:Array; private var numBalls:Number = 10; private var centerBall:Ball; private var bounce:Number = -1; private var spring:Number = 0.2; public function Bubbles() { init(); } private function init():void { balls = new Array(); centerBall = new Ball(100, 0xcccccc); addChild(centerBall); centerBall.x = stage.stageWidth / 2; centerBall.y = stage.stageHeight / 2; for(var i:uint = 0; i < numBalls; i++) { var ball:Ball = new Ball(Math.random() * 40 + 5, Math.random() * 0xffffff); ball.x = Math.random() * stage.stageWidth; ball.y = Math.random() * stage.stageHeight; ball.vx = Math.random() * 6 - 3; ball.vy = Math.random() * 6 - 3; addChild(ball); balls.push(ball); } addEventListener(Event.ENTER_FRAME, onEnterFrame); } private function onEnterFrame(event:Event):void { for(var i:uint = 0; i < numBalls; i++) { var ball:Ball = balls[i]; move(ball); var dx:Number = ball.x - centerBall.x; var dy:Number = ball.y - centerBall.y; var dist:Number = Math.sqrt(dx * dx + dy * dy); var minDist:Number = ball.radius + centerBall.radius; if(dist < minDist) { var angle:Number = Math.atan2(dy, dx); var tx:Number = centerBall.x + Math.cos(angle) * minDist; var ty:Number = centerBall.y + Math.sin(angle) * minDist; ball.vx += (tx - ball.x) * spring; ball.vy += (ty - ball.y) * spring; } } } ***private function move(ball:Ball):void*** { ball.x += ball.vx; ball.y += ball.vy; if(ball.x + ball.radius > stage.stageWidth) { ball.x = stage.stageWidth - ball.radius; ball.vx *= bounce; } else if(ball.x - ball.radius < 0) { ball.x = ball.radius; ball.vx *= bounce; } ***if(ball.y + ball.radius > stage.stageHeight)*** { ball.y = stage.stageHeight - ball.radius; ball.vy *= bounce; } else if(ball.y - ball.radius < 0) { ball.y = ball.radius; ball.vy *= bounce; } } } } The bold parts are the lines im having trouble with! please help..thanks in advance!!

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  • [0-9a-zA-Z]* string expressed with primes or prime-factorization-style way to break it into parts?

    - by HH
    Suppose a string consists of numbers and alphabets. You want to break it into parts, an analogy is primes' factorization, but how can you do similar thing with strings [0-9a-zA-Z]* or even with arbitrary strings? I could express it in alphabets and such things with octal values and then prime-factorize it but then I need to keep track of places where I had the non-numbers things. Is there some simple way to do it? I am looking for simple succinct solutions and don't want too much side-effects. [Update] mvds has the correct idea, to change the base, how would you implement it?

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  • Project Euler 10: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 10.  As always, any feedback is welcome. # Euler 10 # http://projecteuler.net/index.php?section=problems&id=10 # The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. # Find the sum of all the primes below two million. import time start = time.time() def primes_to_max(max): primes, number = [2], 3 while number < max: isPrime = True for prime in primes: if number % prime == 0: isPrime = False break if (prime * prime > number): break if isPrime: primes.append(number) number += 2 return primes primes = primes_to_max(2000000) print sum(primes) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • 1136: Incorrect number of arguments. Expected 0 AS3 Flash CS5.5 [on hold]

    - by Erick
    how do I solve this error? I've been trying to get the answer online but have not been successful. I'm trying to learn As3 for flash so I decided to try making a preloader for a game. Preloader.as package com.game.moran { import flash.display.LoaderInfo; import flash.display.MovieClip; import flash.events.*; public class ThePreloader extends MovieClip { private var fullWidth:Number; public var ldrInfo:LoaderInfo; public function ThePreloader (fullWidth:Number = 0, ldrInfo:LoaderInfo = null) { this.fullWidth = fullWidth; this.ldrInfo = ldrInfo; addEventListener(Event.ENTER_FRAME, checkLoad); } private function checkLoad (e:Event) : void { if (ldrInfo.bytesLoaded == ldrInfo.bytesTotal && ldrInfo.bytesTotal !=0) { dispatchEvent (new Event ("loadComplete")); phaseOut(); } updateLoader (ldfInfo.bytesLoaded / ldrInfo.bytesTotal); } private function updateLoader(num:Number) : void { mcPreloaderBar.Width = num * fullWidth; } private function phaseOut() : void { removeEventListener (Event.ENTER_FRAME, checkLoad); phaseComplete(); } private function phaseComplete() : void { dispatchEvent (new Event ("preloaderFinished")); } } } Engine.as package com.game.moran { import flash.display.MovieClip; import flash.display.Sprite; import flash.events.Event; public class Engine extends MovieClip { private var preloader:ThePreloader; public function Engine() { preloader = new ThePreloader(732, this.loaderInfo); stage.addChild(preloader); preloader.addEventListener("loadComplete", loadAssets); preloader.addEventListener("preloaderFinished", showSponsors); } private function loadAssets (e:Event) : void { this.play(); } private function showSponsors(e:Event) : void { stage.removeChild(preloader); trace("show sponsors") } } } The line being flagged as an error is line 13 in Engine.as.

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  • How to count each digit in a range of integers?

    - by Carlos Gutiérrez
    Imagine you sell those metallic digits used to number houses, locker doors, hotel rooms, etc. You need to find how many of each digit to ship when your customer needs to number doors/houses: 1 to 100 51 to 300 1 to 2,000 with zeros to the left The obvious solution is to do a loop from the first to the last number, convert the counter to a string with or without zeros to the left, extract each digit and use it as an index to increment an array of 10 integers. I wonder if there is a better way to solve this, without having to loop through the entire integers range. Solutions in any language or pseudocode are welcome. Edit: Answers review John at CashCommons and Wayne Conrad comment that my current approach is good and fast enough. Let me use a silly analogy: If you were given the task of counting the squares in a chess board in less than 1 minute, you could finish the task by counting the squares one by one, but a better solution is to count the sides and do a multiplication, because you later may be asked to count the tiles in a building. Alex Reisner points to a very interesting mathematical law that, unfortunately, doesn’t seem to be relevant to this problem. Andres suggests the same algorithm I’m using, but extracting digits with %10 operations instead of substrings. John at CashCommons and phord propose pre-calculating the digits required and storing them in a lookup table or, for raw speed, an array. This could be a good solution if we had an absolute, unmovable, set in stone, maximum integer value. I’ve never seen one of those. High-Performance Mark and strainer computed the needed digits for various ranges. The result for one millon seems to indicate there is a proportion, but the results for other number show different proportions. strainer found some formulas that may be used to count digit for number which are a power of ten. Robert Harvey had a very interesting experience posting the question at MathOverflow. One of the math guys wrote a solution using mathematical notation. Aaronaught developed and tested a solution using mathematics. After posting it he reviewed the formulas originated from Math Overflow and found a flaw in it (point to Stackoverflow :). noahlavine developed an algorithm and presented it in pseudocode. A new solution After reading all the answers, and doing some experiments, I found that for a range of integer from 1 to 10n-1: For digits 1 to 9, n*10(n-1) pieces are needed For digit 0, if not using leading zeros, n*10n-1 - ((10n-1) / 9) are needed For digit 0, if using leading zeros, n*10n-1 - n are needed The first formula was found by strainer (and probably by others), and I found the other two by trial and error (but they may be included in other answers). For example, if n = 6, range is 1 to 999,999: For digits 1 to 9 we need 6*105 = 600,000 of each one For digit 0, without leading zeros, we need 6*105 – (106-1)/9 = 600,000 - 111,111 = 488,889 For digit 0, with leading zeros, we need 6*105 – 6 = 599,994 These numbers can be checked using High-Performance Mark results. Using these formulas, I improved the original algorithm. It still loops from the first to the last number in the range of integers, but, if it finds a number which is a power of ten, it uses the formulas to add to the digits count the quantity for a full range of 1 to 9 or 1 to 99 or 1 to 999 etc. Here's the algorithm in pseudocode: integer First,Last //First and last number in the range integer Number //Current number in the loop integer Power //Power is the n in 10^n in the formulas integer Nines //Nines is the resut of 10^n - 1, 10^5 - 1 = 99999 integer Prefix //First digits in a number. For 14,200, prefix is 142 array 0..9 Digits //Will hold the count for all the digits FOR Number = First TO Last CALL TallyDigitsForOneNumber WITH Number,1 //Tally the count of each digit //in the number, increment by 1 //Start of optimization. Comments are for Number = 1,000 and Last = 8,000. Power = Zeros at the end of number //For 1,000, Power = 3 IF Power 0 //The number ends in 0 00 000 etc Nines = 10^Power-1 //Nines = 10^3 - 1 = 1000 - 1 = 999 IF Number+Nines <= Last //If 1,000+999 < 8,000, add a full set Digits[0-9] += Power*10^(Power-1) //Add 3*10^(3-1) = 300 to digits 0 to 9 Digits[0] -= -Power //Adjust digit 0 (leading zeros formula) Prefix = First digits of Number //For 1000, prefix is 1 CALL TallyDigitsForOneNumber WITH Prefix,Nines //Tally the count of each //digit in prefix, //increment by 999 Number += Nines //Increment the loop counter 999 cycles ENDIF ENDIF //End of optimization ENDFOR SUBROUTINE TallyDigitsForOneNumber PARAMS Number,Count REPEAT Digits [ Number % 10 ] += Count Number = Number / 10 UNTIL Number = 0 For example, for range 786 to 3,021, the counter will be incremented: By 1 from 786 to 790 (5 cycles) By 9 from 790 to 799 (1 cycle) By 1 from 799 to 800 By 99 from 800 to 899 By 1 from 899 to 900 By 99 from 900 to 999 By 1 from 999 to 1000 By 999 from 1000 to 1999 By 1 from 1999 to 2000 By 999 from 2000 to 2999 By 1 from 2999 to 3000 By 1 from 3000 to 3010 (10 cycles) By 9 from 3010 to 3019 (1 cycle) By 1 from 3019 to 3021 (2 cycles) Total: 28 cycles Without optimization: 2,235 cycles Note that this algorithm solves the problem without leading zeros. To use it with leading zeros, I used a hack: If range 700 to 1,000 with leading zeros is needed, use the algorithm for 10,700 to 11,000 and then substract 1,000 - 700 = 300 from the count of digit 1. Benchmark and Source code I tested the original approach, the same approach using %10 and the new solution for some large ranges, with these results: Original 104.78 seconds With %10 83.66 With Powers of Ten 0.07 A screenshot of the benchmark application: If you would like to see the full source code or run the benchmark, use these links: Complete Source code (in Clarion): http://sca.mx/ftp/countdigits.txt Compilable project and win32 exe: http://sca.mx/ftp/countdigits.zip Accepted answer noahlavine solution may be correct, but l just couldn’t follow the pseudo code, I think there are some details missing or not completely explained. Aaronaught solution seems to be correct, but the code is just too complex for my taste. I accepted strainer’s answer, because his line of thought guided me to develop this new solution.

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  • How do I send DTMF tones and pauses using Android ACTION_CALL Intent with commas in the number?

    - by Rob Kent
    I have an application that calls a number stored by the user. Everything works okay unless the number contains commas or hash signs, in which case the Uri gets truncated after the digits. I have read that you need to encode the hash sign but even doing that, or without a hash sign, the commas never get passed through. However, they do get passed through if you just pick the number from your contacts. I must be doing something wrong. For example: String number = "1234,,,,4#1"; Uri uri = Uri.parse(String.format("tel:%s", number)); try { startActivity(new Intent(callType, uri)); } catch (ActivityNotFoundException e) { ... Only the number '1234' would end up in the dialer.

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  • What are the Tags Around Default iPhone Address Book People Phone Number Labels?

    - by rnistuk
    My question concerns markup that surrounds some of the default phone number labels in the Person entries of the Contact list on the iPhone. I have created an iPhone contact list address book entry for a person, "John Smith" with the following phone number entries: Mobile (604) 123-4567 iPhone (778) 123-4567 Home (604) 789-4561 Work (604) 456-7891 Main (604) 789-1234 megaphone (234) 567-8990 Note that the first five labels are default labels provided by the Contacts application and the last label, "megaphone", is a custom label. I wrote the following method to retrieve and display the labels and phone numbers for each person in the address book: -(void)displayPhoneNumbersForAddressBook { ABAddressBookRef book = ABAddressBookCreate(); CFArrayRef people = ABAddressBookCopyArrayOfAllPeople(book); ABRecordRef record = CFArrayGetValueAtIndex(people, 0); ABMultiValueRef multi = ABRecordCopyValue(record, kABPersonPhoneProperty); NSLog(@"---------" ); NSLog(@"displayPhoneNumbersForAddressBook" ); CFStringRef label, phone; for (CFIndex i = 0; i < ABMultiValueGetCount(multi); ++i) { label = ABMultiValueCopyLabelAtIndex(multi, i); phone = ABMultiValueCopyValueAtIndex(multi, i); NSLog(@"label: \"%@\" number: \"%@\"", (NSString*)label, (NSString*)phone); CFRelease(label); CFRelease(phone); } NSLog(@"---------" ); CFRelease(multi); CFRelease(people); CFRelease(book); } and here is the output for the address book entry that I entered: 2010-03-08 13:24:28.789 test2m[2479:207] --------- 2010-03-08 13:24:28.789 test2m[2479:207] displayPhoneNumbersForAddressBook 2010-03-08 13:24:28.790 test2m[2479:207] label: "_$!<Mobile>!$_" number: "(604) 123-4567" 2010-03-08 13:24:28.790 test2m[2479:207] label: "iPhone" number: "(778) 123-4567" 2010-03-08 13:24:28.791 test2m[2479:207] label: "_$!<Home>!$_" number: "(604) 789-4561" 2010-03-08 13:24:28.791 test2m[2479:207] label: "_$!<Work>!$_" number: "(604) 456-7891" 2010-03-08 13:24:28.792 test2m[2479:207] label: "_$!<Main>!$_" number: "(604) 789-1234" 2010-03-08 13:24:28.792 test2m[2479:207] label: "megaphone" number: "(234) 567-8990" 2010-03-08 13:24:28.793 test2m[2479:207] --------- What are the markup characters _$!< and >!$_ surrounding most, save for iPhone, of the default labels for? Can you point me to where in the "Address Book Programming Guide for iPhone OS" I can find the information? Thank you for your help.

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  • Most Elegant Way to write isPrime in java

    - by Anantha Kumaran
    public class Prime { public static boolean isPrime1(int n) { if (n <= 1) { return false; } if (n == 2) { return true; } for (int i = 2; i <= Math.sqrt(n) + 1; i++) { if (n % i == 0) { return false; } } return true; } public static boolean isPrime2(int n) { if (n <= 1) { return false; } if (n == 2) { return true; } if (n % 2 == 0) { return false; } for (int i = 3; i <= Math.sqrt(n) + 1; i = i + 2) { if (n % i == 0) { return false; } } return true; } } public class PrimeTest { public PrimeTest() { } @Test public void testIsPrime() throws IllegalArgumentException, IllegalAccessException, InvocationTargetException { Prime prime = new Prime(); TreeMap<Long, String> methodMap = new TreeMap<Long, String>(); for (Method method : Prime.class.getDeclaredMethods()) { long startTime = System.currentTimeMillis(); int primeCount = 0; for (int i = 0; i < 1000000; i++) { if ((Boolean) method.invoke(prime, i)) { primeCount++; } } long endTime = System.currentTimeMillis(); Assert.assertEquals(method.getName() + " failed ", 78498, primeCount); methodMap.put(endTime - startTime, method.getName()); } for (Entry<Long, String> entry : methodMap.entrySet()) { System.out.println(entry.getValue() + " " + entry.getKey() + " Milli seconds "); } } } I am trying to find the fastest way to check whether the given number is prime or not. This is what is finally came up with. Is there any better way than the second implementation(isPrime2).

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  • dmidecode showing more ram slots than available?

    - by Jestep
    I have some failing RAM in a server and I ran dmidecode to figure out what tyoe of RAM I needed to replace it with. The server has 6 RAM slots, 4 of which are in use. When I run dmidecode this is what I get. dmidecode 2.10 SMBIOS 2.4 present. Handle 0x001F, DMI type 17, 27 bytes Memory Device Array Handle: 0x001E Error Information Handle: No Error Total Width: 72 bits Data Width: 64 bits Size: 2048 MB Form Factor: DIMM Set: 1 Locator: JXXX Bank Locator: DIMM 00 Type: DDR2 Type Detail: Synchronous Speed: 667 MHz Manufacturer: Not Specified Serial Number: Not Specified Asset Tag: Not Specified Part Number: Not Specified Handle 0x0020, DMI type 17, 27 bytes Memory Device Array Handle: 0x001E Error Information Handle: No Error Total Width: 72 bits Data Width: 64 bits Size: 2048 MB Form Factor: DIMM Set: 1 Locator: JXXX Bank Locator: DIMM 01 Type: DDR2 Type Detail: Synchronous Speed: 667 MHz Manufacturer: Not Specified Serial Number: Not Specified Asset Tag: Not Specified Part Number: Not Specified Handle 0x0021, DMI type 17, 27 bytes Memory Device Array Handle: 0x001E Error Information Handle: No Error Total Width: Unknown Data Width: Unknown Size: No Module Installed Form Factor: DIMM Set: 1 Locator: JXXX Bank Locator: DIMM 02 Type: DDR2 Type Detail: Synchronous Speed: 667 MHz Manufacturer: Not Specified Serial Number: Not Specified Asset Tag: Not Specified Part Number: Not Specified Handle 0x0022, DMI type 17, 27 bytes Memory Device Array Handle: 0x001E Error Information Handle: No Error Total Width: Unknown Data Width: Unknown Size: No Module Installed Form Factor: DIMM Set: 1 Locator: JXXX Bank Locator: DIMM 03 Type: DDR2 Type Detail: Synchronous Speed: 667 MHz Manufacturer: Not Specified Serial Number: Not Specified Asset Tag: Not Specified Part Number: Not Specified Handle 0x0023, DMI type 17, 27 bytes Memory Device Array Handle: 0x001E Error Information Handle: No Error Total Width: 72 bits Data Width: 64 bits Size: 2048 MB Form Factor: DIMM Set: 1 Locator: JXXX Bank Locator: DIMM 10 Type: DDR2 Type Detail: Synchronous Speed: 667 MHz Manufacturer: Not Specified Serial Number: Not Specified Asset Tag: Not Specified Part Number: Not Specified Handle 0x0024, DMI type 17, 27 bytes Memory Device Array Handle: 0x001E Error Information Handle: No Error Total Width: 72 bits Data Width: 64 bits Size: 2048 MB Form Factor: DIMM Set: 1 Locator: JXXX Bank Locator: DIMM 11 Type: DDR2 Type Detail: Synchronous Speed: 667 MHz Manufacturer: Not Specified Serial Number: Not Specified Asset Tag: Not Specified Part Number: Not Specified Handle 0x0025, DMI type 17, 27 bytes Memory Device Array Handle: 0x001E Error Information Handle: No Error Total Width: Unknown Data Width: Unknown Size: No Module Installed Form Factor: DIMM Set: 1 Locator: JXXX Bank Locator: DIMM 12 Type: DDR2 Type Detail: Synchronous Speed: 667 MHz Manufacturer: Not Specified Serial Number: Not Specified Asset Tag: Not Specified Part Number: Not Specified Handle 0x0026, DMI type 17, 27 bytes Memory Device Array Handle: 0x001E Error Information Handle: No Error Total Width: Unknown Data Width: Unknown Size: No Module Installed Form Factor: DIMM Set: 1 Locator: JXXX Bank Locator: DIMM 13 Type: DDR2 Type Detail: Synchronous Speed: 667 MHz Manufacturer: Not Specified Serial Number: Not Specified Asset Tag: Not Specified Part Number: Not Specified Does anyone know why it would show 8 slots, with 4 empty instead of 6 slots with 2 empty? Also, but my records and by other tools, the server has 16Gb and not 8Gb in it currently. grep MemTotal /proc/meminfo MemTotal: 16435808 kB The board is a Tyan S5372-LC, running CentOS 5.4 x64. Also, my error log is showing errors in bank 6. Is there any way to determine which slot bank 6 is in via: dmidecode?

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  • How can I generate the Rowland prime sequence idiomatically in J?

    - by Gregory Higley
    If you're not familiar with the Rowland prime sequence, you can find out about it here. I've created an ugly, procedural monadic verb in J to generate the first n terms in this sequence, as follows: rowland =: monad define result =. 0 $ 0 t =. 1 $ 7 while. (# result) < y do. a =. {: t n =. 1 + # t t =. t , a + n +. a d =. | -/ _2 {. t if. d > 1 do. result =. ~. result , d end. end. result ) This works perfectly, and it indeed generates the first n terms in the sequence. (By n terms, I mean the first n distinct primes.) Here is the output of rowland 20: 5 3 11 23 47 101 7 13 233 467 941 1889 3779 7559 15131 53 30323 60647 121403 242807 My question is, how can I write this in more idiomatic J? I don't have a clue, although I did write the following function to find the differences between each successive number in a list of numbers, which is one of the required steps. Here it is, although it too could probably be refactored by a more experienced J programmer than I: diffs =: monad : '}: (|@({.-2&{.) , $:@}.) ^: (1 < #) y'

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  • Beginner Question ; About Prime Generation in "C" - What is wrong with my code ? -

    - by alorsoncode
    I'm a third year irregular CS student and ,i just realized that i have to start coding. I passed my coding classes with lower bound grades so that i haven't a good background in coding&programming. I'm trying to write a code that generates prime numbers between given upper and lower bounds. Not knowing C well, enforce me to write a rough code then go over it to solve. I can easily set up the logic for intended function but i probably create a wrong algorithm through several different ways. Here I share my last code, i intend to calculate that when a number gives remainder Zero , it should be it self and 1 , so that count==2; What is wrong with my implementation and with my solution generating style? I hope you will warm me up to programming world, i couldn't find enough motivation and courage to get deep into programming. Thanks in Advance :) Stdio and Math.h is Included int primegen(int down,int up) { int divisor,candidate,count=0,k; for(candidate=down;candidate<=up;candidate++) { for(divisor=1;divisor<=candidate;divisor++) { k=(candidate%divisor); } if (k==0) count++; if(count==2) { printf("%d\n", candidate); count=0; } else { continue; } } } int main() { primegen(3,15); return 0; }

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  • Runtime error of TASM language help!

    - by dominoos
    .model small .stack 400h .data message db "hello. ", 0ah, 0dh, "$" firstdigit db ? seconddigit db ? thirddigit db ? number dw ? newnumber db ? anumber dw 0d bnumber dw 0d Firstn db 0ah, 0dh, "Enter first 3 digit number: ","$" secondn db 0ah, 0dh, "Enter second 3 digit number: ","$" messageB db 0ah, 0dh, "HCF of two number is: ","$" linebreaker db 0ah, 0dh, ' ', 0ah, 0dh, '$' .code Start: mov ax, @data ; establish access to the data segment mov ds, ax ; mov number, 0d mov dx, offset message ; print the string "yob choi 0648293" mov ah, 9h int 21h num: mov dx, offset Firstn ; print the string "put 1st 3 digit" mov ah, 9h int 21h ;run JMP FirstFirst ; jump to FirstFirst FirstFirst: ;first digit mov ah, 1d ;bios code for read a keystroke int 21h ;call bios, it is understood that the ascii code will be returned in al mov firstdigit, al ;may as well save a copy sub al, 30h ;Convert code to an actual integer cbw ;CONVERT BYTE TO WORD. This takes whatever number is in al and ;extends it to ax, doubling its size from 8 bits to 16 bits ;The first digit now occupies all of ax as an integer mov cx, 100d ;This is so we can calculate 100*1st digit +10*2nd digit + 3rd digit mul cx ;start to accumulate the 3 digit number in the variable imul cx ;it is understood that the other operand is ax ; the result will use both dx::ax ;dx will contain only leading zeros add anumber, ax ;save ;Second Digit mov ah, 1d ;bios code for read a keystroke int 21h ;call bios, it is understood that the ascii code will be returned in al mov seconddigit, al ;may as well save a copy sub al, 30h ;Convert code to an actual integer cbw ;CONVERT BYTE TO WORD. This takes whatever number is in al and ;extends it to ax, boubling its size from 8 bits to 16 bits ;The first digit now occupies all of ax as an integer mov cx, 10d ;continue to accumulate the 3 digit number in the variable mul cx ;it is understood that the other operand is ax, containing first digit ;the result will use both dx::ax ;dx will contain only leading zeros. add anumber, ax ;save ;third Digit mov ah, 1d ;samething as above int 21h ; mov thirddigit, al ; sub al, 30h ; cbw ; add anumber, ax ; jmp num2 ;go to checks Num2: mov dx, offset secondn ; print the string "put 2nd 3 digits" mov ah, 9h int 21h ;run JMP SecondSecond SecondSecond: ;first digit mov ah, 1d ;bios code for read a keystroke int 21h ;call bios, it is understood that the ascii code will be returned in al mov firstdigit, al ;may as well save a copy sub al, 30h ;Convert code to an actual integer cbw ;CONVERT BYTE TO WORD. This takes whatever number is in al and ;extends it to ax, doubling its size from 8 bits to 16 bits ;The first digit now occupies all of ax as an integer mov cx, 100d ;This is so we can calculate 100*1st digit +10*2nd digit + 3rd digit mul cx ;start to accumulate the 3 digit number in the variable imul cx ;it is understood that the other operand is ax ; the result will use both dx::ax ;dx will contain only leading zeros add bnumber, ax ;save ;Second Digit mov ah, 1d ;bios code for read a keystroke int 21h ;call bios, it is understood that the ascii code will be returned in al mov seconddigit, al ;may as well save a copy sub al, 30h ;Convert code to an actual integer cbw ;CONVERT BYTE TO WORD. This takes whatever number is in al and ;extends it to ax, boubling its size from 8 bits to 16 bits ;The first digit now occupies all of ax as an integer mov cx, 10d ;continue to accumulate the 3 digit number in the variable mul cx ;it is understood that the other operand is ax, containing first digit ;the result will use both dx::ax ;dx will contain only leading zeros. add bnumber, ax ;save ;third Digit mov ah, 1d ;samething as above int 21h ; mov thirddigit, al ; sub al, 30h ; cbw ; add bnumber, ax ; jmp compare ;go to compare compare: CMP ax, anumber ;comparing numbB and Number JA comp1 ;go to comp1 if anumber is bigger CMP ax, anumber ; JB comp2 ;go to comp2 if anumber is smaller CMP ax, anumber ; JE equal ;go to equal if two numbers are the same JMP compare ;go to compare (avioding error) comp1: SUB ax, anumber; subtract smaller number from bigger number JMP compare ; comp2: SUB anumber, ax; subtract smaller number from bigger number JMP compare ; equal: mov ah, 9d ;make linkbreak after the 2nd 3 digit number mov dx, offset linebreaker int 21h mov ah, 9d ;print "HCF of two number is:" mov dx, offset messageB int 21h mov ax,anumber ;copying 2nd number into ax add al,30h ; converting to ascii mov newnumber,al ; copying from low part of register into newnumb mov ah, 2d ;bios code for print a character mov dl, newnumber ;we had saved the ascii code here int 21h ;call to bios JMP exit; exit: mov ah, 4ch int 21h ;exit the program End hi, this is a program that finds highest common factor of 2 different 3digit number. if i put 200, 235,312 (low numbers) it works fine. but if i put 500, 550, 654(bigger number) the program crashes after the 2nd 3digit number is entered. can you help me to find out what problem is?

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