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  • Award-Winning Architects at Oracle OpenWorld

    - by Bob Rhubart
    "The Winner," a sculpture by John J. Seward Jr. The role of the IT architect may be the most hotly debated and unjustly maligned role in IT. But at this year's Oracle OpenWorld in San Francisco several architects will enjoy some much-deserved recognition through the Oracle Magazine Technologist of the Year Awards. Part of the Oracle Excellence Awards, the Technologist of the Year Awards "honor Oracle technologists for their cutting-edge solutions using Oracle products and services." Seven of the ten Technologist of the Year categories honor architects: Technologist of the Year: Big Data Architect Technologist of the Year: Cloud Architect Technologist of the Year: Enterprise Architect Technologist of the Year: Mobile Architect Technologist of the Year: Security Architect Technologist of the Year: Social Architect Technologist of the Year: Virtualization Architect If you or one of your colleagues is an architect deserving of this recognition, click the appropriate link above to find the nomination form. Deadline for nominations is Tuesday, July 17, 2012. For more information see: Technologist of the Year Awards. See last year's winners here.

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  • [C#] Convert string to double with 2 digit after decimal separator

    - by st.stoqnov
    All began with these simple lines of code: string s = "16.9"; double d = Convert.ToDouble(s); d*=100; The result should be 1690.0, but it's not. d is equal to 1689.9999999999998. All I want to do is to round a double to value with 2 digit after decimal separator. Here is my function. private double RoundFloat(double Value) { float sign = (Value < 0) ? -0.01f : 0.01f; if (Math.Abs(Value) < 0.00001) Value = 0; string SVal = Value.ToString(); string DecimalSeparator = System.Globalization.CultureInfo.CurrentCulture.NumberFormat.CurrencyDecimalSeparator; int i = SVal.IndexOf(DecimalSeparator); if (i > 0) { int SRnd; try { // ????? ??????? ????? ???? ?????????? ?????????? SRnd = Convert.ToInt32(SVal.Substring(i + 3, 1)); } catch { SRnd = 0; } if (SVal.Length > i + 3) SVal = SVal.Substring(0, i + 3); //SVal += "00001"; try { double result = (SRnd >= 5) ? Convert.ToDouble(SVal) + sign : Convert.ToDouble(SVal); //result = Math.Round(result, 2); return result; } catch { return 0; } } else { return Value; } But again the same problem, converting from string to double is not working as I want. A workaround to this problem is to concatenate "00001" to the string and then use the Math.Round function (commented in the example above). This double value multiplied to 100 (as integer) is send to a device (cash register) and this values must be correct. I am using VS2005 + .NET CF 2.0 Is there another more "elegant" solution, I am not happy with this one.

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  • PHP strtotime without leap-years

    - by Christian Sciberras
    With regards to this thread, I've developed a partial solution: function strtosecs($time,$now=null){ static $LEAPDIFF=86400; $time=strtotime($time,$now); return $time-((date('Y',$time)-1968)/4*$LEAPDIFF); } The function is supposed to get the number of seconds given a string without checking leap-years. It does this calculating the number of leap-years 1970 [(year-1986)/4], multiplying it by the difference in seconds between a leap-year and a normal year (which in the end, it's just the number of seconds in a day). Finally, I simply remove all those excess leap-year seconds from the calculated time. Here's some examples of the inputs/outputs: // test code echo strtosecs('+20 years',0).'=>'.(strtosecs('+20 years',0)/31536000); echo strtosecs('+1 years',0).'=>'.(strtosecs('+1 years',0)/31536000); // test output 630676800 => 19.998630136986 31471200 => 0.99794520547945 You will probably ask why am I doing a division on the output? It's to test it out; 31536000 is the number of seconds in a year, so that 19.99... should be 20 and 0.99... should be a 1. Sure, I could round it all and get "correct" answer, but I'm worried about the inaccuracies.

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  • Export 1 year of CVS to another repo?

    - by John Dibling
    We have a CVS repo with many years of history. It has become huge and unwieldly, so we would like to split this singe repo in to two repos: The main repo would have 1 year's worth of history, up to and including present day. This is where all dev work would take place. An archive repo would have the complete history, up to the point where the main repo would take over. This would be read-only, and only used to look at historical changes. Given that we are starting with one huge, monolithic CVS repo, is it possible to split it up in this way? How can this be accomplished?

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  • Advice for last year college graduates

    - by Tomh
    Hey guys, I know there are many "advice" questions around this site. But I wanted to to narrow mine down to last year college students, in my case my last year as Master student in computer science. So far is a list of things I've done during my time in college (which I can recommend others to do aswell): Code a lot I've written several hobby projects, had part time jobs, entered the Imagine cup from Microsoft, took programming extensive courses and did freelance gigs. Read a lot I've bought most top books from the recommended book topics here, to be honest I have not read them all. learn different languages I've tried several languages including Haskell, Java, Python, Ruby, Lisp, Prolog, C#, PHP, JS, AS3 and possibly some more I forgot. Tried to start a blog Joel recommends to learn how to write, I tried starting a couple of blogs to improve upon this, I gave up on all instances after writing about three posts. It was just not my thing... Have a portfolio of launched projects/programs I'm busy with this, have a couple of finished, working projects I worked on to show to people. So this is my last year. Is there anything else you can recommend a last year college student to do before hitting the job market? Personally I'm tempted to spend my time on the following: Practice algorithm design Learn and memorize the usage of the low level API's of your favorite language Polish your portfolio Why? Because those first two will make sure you pass the majority of the interviews, here in Holland (I could be wrong). I rather not spend my time on those first two points, but I have to be realistic and thats just my experience on what kind of questions you'll get when you apply. The third point is my hope that I won't have to answer questions about the amount of standard types in c# for example if they can see I get projects done and launched. But I'm still graduating, so I don't know anything :), and many of you might be hiring grads on a recent base and could tell me and other interested people what you wish that the recent grads you interviewed would have done before they applied.

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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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  • help, php calendar links

    - by Ray
    Below is a partial table row that will create a line row of 3 columns. Two columns in each side is links, left is previous year and right is next year links. Center is links for january thru december. When you click a month links, the calendar will show that month you've clicked in the current year the calendar is on. For example, the calendar will open current month and year by default...March 2010. If you clicked previous year (2009), it will display current month (March) of last year (2009)...and then if you click Jun, the calendar will display June of whatever the year the calendar is currently on, which is June of 2009. my question is, what can i do to following code to do such thing. $Calendar.= "</tr><tr><td>"."<a "."href=\"".$_SERVER["PHP_SELF"]."?year=" . $LastYear["year"] ."\"> $LastY </a></td>\n"; $Calendar.= "<td colspan=\"5\">"."<a "."href=\"".$_SERVER["PHP_SELF"]."?year=".$ThisYear["year"]."&month=" . $MonthArray[0] ."\">Jan.</a> | "."<a "."href=\"".$_SERVER["PHP_SELF"]."?year=".$ThisYear["year"]."&month=" . $MonthArray[1] ."\">Feb.</a> | "."<a "."href=\"".$_SERVER["PHP_SELF"]."?year=".$ThisYear["year"]."&month=" . $MonthArray[2] ."\">Mar.</a> | "."<a "."href=\"".$_SERVER["PHP_SELF"]."?year=".$ThisYear["year"]."&month=" . $MonthArray[3] ."\">Apr.</a> | "."<a "."href=\"".$_SERVER["PHP_SELF"]."?year=".$ThisYear["year"]."&month=" . $MonthArray[4] ."\">May</a> | "."<a "."href=\"".$_SERVER["PHP_SELF"]."?year=".$ThisYear["year"]."&month=" . $MonthArray[5] ."\">Jun.</a> | "."<a "."href=\"".$_SERVER["PHP_SELF"]."?year=".$ThisYear["year"]."&month=" . $MonthArray[6] ."\">Jul.</a> | "."<a "."href=\"".$_SERVER["PHP_SELF"]."?year=".$ThisYear["year"]."&month=" . $MonthArray[7] ."\">Aug.</a> | "."<a "."href=\"".$_SERVER["PHP_SELF"]."?year=".$ThisYear["year"]."&month=" . $MonthArray[8] ."\">Sep.</a> | "."<a "."href=\"".$_SERVER["PHP_SELF"]."?year=".$ThisYear["year"]."&month=" . $MonthArray[9] ."\">Oct.</a> | "."<a "."href=\"".$_SERVER["PHP_SELF"]."?year=".$ThisYear["year"]."&month=" . $MonthArray[10] ."\">Nov.</a> | "."<a "."href=\"".$_SERVER["PHP_SELF"]."?year=".$ThisYear["year"]."&month=" . $MonthArray[11] ."\">Dec.</a> </td>"; $Calendar.= "<td>"."<a "."href=\"".$_SERVER["PHP_SELF"]."?year=" . $NextYear["year"] ."\"> $NextY </a></td>\n"; Thanks in advance.

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  • A query for date within a year

    - by Fabiano PS
    My table is like this on Postgres, note that all days start by 01, there is only 1 entry a month+year SELECT * FROM "fis_historico_receita" +----+------------+---------------+ | id | data | receita_bruta | +----+------------+---------------+ | 1 | 2010-02-01 | 100000.0 | | 2 | 2010-01-01 | 100000.0 | | 3 | 2009-12-01 | 100000.0 | | 4 | 2009-11-01 | 100000.0 | | 5 | 2009-10-01 | 100000.0 | | 6 | 2009-09-01 | 100000.0 | | 7 | 2009-08-01 | 100000.0 | | 8 | 2009-07-01 | 100000.0 | | 9 | 2009-06-01 | 100000.0 | | 10 | 2009-05-01 | 100000.0 | | 11 | 2009-04-01 | 100000.0 | | 12 | 2009-03-01 | 100000.0 | | 13 | 2009-02-01 | 100000.0 | | 14 | 2009-01-01 | 100000.0 | | 15 | 2008-12-01 | 100000.0 | +----+------------+---------------+ What I want is to find 12 months starting right from before the current. I tried this: select * from fis_historico_receita where data in interval '1 year' I really would like an answer using Interval, +1 goes for everyone that runs on Postgres

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  • Convert local time (10 digit number) to a readable datetime format

    - by djerry
    Hey all, I'm working with pbx for voip calls. One aspect of pbx is that you can choose to receive CDR packages. Those packages have 2 timestamps : "utc" and "local", but both seem to always be the same. Here's an example of a timestamp : "1268927156". At first sight, there seems to be no logic in it. So i tried converting it several ways, but with no good result. That value should provide a time around 11am (+1GMT) today. Things i tried: Datetime dt = new Datetime(number); Timespan ts = new Timespan(number); DateTime utc = new DateTime(number + 504911232000000000, DateTimeKind.Utc) and some others i can't remember right now. Am i missing something stupid here? Thanks in advance

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  • Round date to fiscal year

    - by Dave Jarvis
    The following database view rounds the date back to the closest fiscal year (April 1st): CREATE OR REPLACE VIEW FISCAL_YEAR_VW AS SELECT CASE WHEN to_number(to_char( SYSDATE, 'MM' )) < 4 THEN to_date('1-APR-'||to_char(add_months(SYSDATE, -12), 'YYYY'), 'dd-MON-yyyy') ELSE to_date('1-APR-'||to_char(SYSDATE, 'YYYY'), 'dd-MON-yyyy') END AS fiscal_year FROM dual; This allows us to calculate the current fiscal year based on today's date. How can this calculation be simplified or optimized?

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  • Fibonacci sequence subroutine returning one digit too high...PERL

    - by beProactive
    #!/usr/bin/perl -w use strict; sub fib { my($num) = @_; #give $num to input array return(1) if ($num<=1); #termination condition return($num = &fib($num-1) + &fib($num-2)); #should return sum of first "n" terms in the fibonacci sequence } print &fib(7)."\n"; #should output 20 This subroutine should be outputting a summation of the first "x" amount of terms, as specified by the argument to the sub. However, it's one too high. Does this have something to do with the recursion? Thanks.

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  • Split number and put each digit to separate html element

    - by Seven
    The problem is that I do not know how to break the loop for the first number and start again for the next one. Currently, one span.nNumber has a total digits of the two numbers (123456) and the next span.nNumber contains digits only from another number (456). Goal is to create sequence 123 and 456: <span class='nNumber'> <span>1</span> <span>2</span> <span>3</span> </span> and <span class='nNumber'> <span>4</span> <span>5</span> <span>6</span> </span> Example script: http://jsfiddle.net/PZ8Pt/2/

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  • First Year Computer Science Programming Languages

    - by Jon
    I was reading this article earlier regarding C/C#/PHP being dropped as first languages in Advanced Level (pre-university) Computer Science courses: http://www.theregister.co.uk/2010/05/12/aqa_c_php/ It also goes on to say: Teachers planning to use Java are warned that many universities are considering dropping it from their first year computer science programmes, "as has happened n the US". Does anybody know, what the language predominantly used in US first year Comp Science programs is currently?

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  • Is it possible to create a DateFormatter which converts a two-digit year into a four-digit year?

    - by DR
    In my Java application I use a DateFormat instance to parse date inputs. DateFormat fmt; fmt = DateFormat.getDateInstance(DateFormat.DEFAULT) // dd.MM.yyyy for de_DE The problem is that the user insists to enter dates in the form 31.12.11. Unfortunately this is parsed to 31.12.11. (0011-12-31 in ISO format) Instead I want the parsed date to become 31.12.2011 (2011-12-31 in ISO format). Can I modify the date format to somehow parse inputs that way?

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  • Selecting records with specific month and year in SQL Server 2005

    - by John
    I want to list records with a particular month and year. The table name is 'Arrival' and 'date' is the field that stores the date that the record was added. This is to be done from a C# application. For example, if the user selects month as 'April' and year as '2009' in the application, it will list all the records that were added on April,2009. (I only need the query, hope I can figure out the rest :) )

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  • Enumerating large (20-digit) [probable] prime numbers

    - by Paul Baker
    Given A, on the order of 10^20, I'd like to quickly obtain a list of the first few prime numbers greater than A. OK, my needs aren't quite that exact - it's alright if occasionally a composite number ends up on the list. What's the fastest way to enumerate the (probable) primes greater than A? Is there a quicker way than stepping through all of the integers greater than A (other than obvious multiples of say, 2 and 3) and performing a primality test for each of them? If not, and the only method is to test each integer, what primality test should I be using?

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  • How to compare the year with the current year in iphone?

    - by Warrior
    I am new to iphone development.I am parsing a XML URL and display the title ,date and summary in a table view.I noticed some of the date were very old like "Wed, 31 Dec 1969 19:00:00 -0500" ,So i don't want to display the dates which are 1 year older than the current year.How to do that? I used the sample code from this site for parsing and display the details.Please help me out.Thanks.

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  • find nth digit in C

    - by kokkch
    i spend more than a day to solve this and i can't. I have a function name int get_nth_digit (int x, int pos); which takes as entering the number given by the user(x) and a number that represents the position in which the user wishes to return the item. How can do this with C program? can you help me pls???? thx

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  • Thanks All the readers and community and Happy new year to all of you.

    - by Jalpesh P. Vadgama
    This is my first blog post for new year 2011 and I would like to take this opportunity to thank all the readers for making my blog very successful and accepting me a community member. As year 2010 has lots of up down in IT filed it was recession period and now we almost recovered from it. Personally year 2010 has been very successful to me as I have been awarded as Microsoft Most Valuable Professional for visual C#. And It was one of the greatest achievement of my life. I would like to take this opportunity to thanks Microsoft for this and thanks all friends specially Jacob Sebastian who has given me guidance any time I required it. I have been also awarded dzone most valuable blogger this year and it was a nice surprise from dzone. I would like thanks dzone for this. Once again I am wishing you happy new year and may this year will bring success to all of you. One more thing I have found that I have met lots of people who is quite intelligent and exceptional developers and IT professionals but they are not blogging their stuff. I would say please my blog post a why a developer should write blog and Start blogging immediately because unless and until you don’t blog community will not know what you are doing.  Till then happy blogging and programming ... Stay tuned for more..

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