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  • Disable digit substitution

    - by Uwe
    How can I disable the digit substitution (for example for Hindi numerals instead of Arabic ones) for my application (native c++) completely? I want all the numbers displayed with 0123 instead of ???? There is an option in localization options in windows, but I don't want to change that for the user. Only for my app. Thank you!

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  • Retrieving the first digit of a number

    - by Michoel
    Hi, I am just learning Java and am trying to get my program to retrieve the first digit of a number - for example 543 should return 5, etc. I thought to convert to a string, but I am not sure how I can convert it back? Thanks for any help. int number = 534; String numberString = Integer.toString(number); char firstLetterChar = numberString.charAt(0); int firstDigit = ????

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  • Define a regex, which matches one digit twice and all others once

    - by Amin
    As part of a larger regex I would like to match the following restrictions: The string has 11 digits All digits are numbers Within the first 10 digits one number [0-9] (and one only!) must be listed twice This means the following should match: 12345678914 12235879600 Whereas these should not: 12345678903 -> none of the numbers at digits 1 to 10 appears twice 14427823482 -> one number appears more than twice 72349121762 -> two numbers appear twice I have tried to use a lookahead, but all I'm managing is that the regex counts a certain digit, i.e.: (?!.*0\1{2}) That does not do what I need. Is my query even possible with regex?

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  • How to get user input for 2 digit data

    - by oneMinute
    In a HTML form user is expect to fill / select some data and trigger an action probably a http-post. If your only requested data field is a "2 digit" you can use html text input element get some data. Then you want to make it useful; enable user easily select data from a 'html select' But not all of your data is well-ordered so eye-searching within these data is somehow cumbersome. Because your data is meaningful with its relations. If there is no primary key for foreign key "12" it should not be shown. Vice versa if this foreign key occurs a lot, then it has some weight and could be displayed with more importance. So, what will be your way? a) Use text input to get data and validate it with regex, javascript, ... b) Use some dropdown select. c) Any other way ? Any answer will appreciated :)

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  • Unknown http requests of type http://<domain>/cache/<32-digit-alphanumeric-key>

    - by Siva Bathula
    I am getting a lot of incoming requests with this structure: //domain_name/cache/22092e9b25c40809dfb94b6179166b26. I am running a .NET 4.0 website served from IIS 7.5. A lot of these URLs have no referrer URLs and come in randomly with a different 32 digit alphanumeric key. And I do not have any resource like '.../cache/...' on my website. I just want to eliminate such requests and want to understand where these are coming from at all. Any help would be appreciated.

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  • Sorting 1 million 8-digit numbers in 1MB of RAM

    - by Favourite Chigozie Onwuemene
    I have a computer with 1M of RAM and no other local storage. I must use it to accept 1 million 8-digit decimal numbers over a TCP connection, sort them, and then send the sorted list out over another TCP connection. The list of numbers may contain duplicates, which I must not discard. The code will be placed in ROM, so I need not subtract the size of my code from the 1M. I already have code to drive the ethernet port and handle TCP/IP connections, and it requires 2k for its state data, including a 1k buffer via which the code will read and write data. Is there a solution to this problem?

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  • mySQL removes first digit

    - by kielie
    Hi guys, I am inputting data into a mySQL database via a PHP script, but for some reason when I check the database, all of the phone numbers have their first digit removed, like so, 0123456789 shows up as 123456789 in the database, but if I change the data type from INT to TEXT, it shows correctly, I am very hesitant to keep it as TEXT though, as I am sure this will cause complications further down the road as the database app starts to become more complicated, here is the PHP code. <?php $gender = $_POST['gender']; $first_name = $_POST['first_name']; $second_name = $_POST['second_name']; $id_number = $_POST['id_number']; $home_number = $_POST['home_number']; $cell_work = $_POST['cell_work']; $email_address = $_POST['email_address']; $curDate = date("Y-m-d"); mysql_connect ("server", "user", "pass") or die ('Error: ' . mysql_error()); mysql_select_db ("database"); $query = "INSERT INTO table (id,gender,first_name,second_name,id_number,home_number,cell_work,email_address,date) VALUES('NULL','".$gender."','".$first_name."','".$second_name."','".$id_number."','".$home_number."','".$cell_work."','".$email_address."','".$curDate."' )"; mysql_query($query) or die (mysql_error()); ?> Thanx in advance!

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  • Decimal To Octal Converter, last digit issue

    - by Srishan Supertramp
    I tried making a C program to convert a user entered decimal number to octal. I wrote the C code with my own logic without any research of how other users try to do it. It works fine for the number 601 and some other numbers but for most numbers it returns the octal equivalent with the last digit being 1 less than it should be. For 75 it returns 112 instead of 113. I realize using printf with %o gets the job done but it's kind of defeating the purpose of learning to program. Here's my code: #include <stdio.h> #include <math.h> /* converting decimal to octal */ int main() { int n,x,y,p,s; printf("Enter a decimal number "); scanf("%d",&x); s=0;p=0; while (x!=0) { y=x%8; s=s+y*pow(10,p); x=(x-y)/8; p=p+1; } printf("the octal equivalent is: %d\n",s); getch(); return 0; }

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  • How to turn this simple 10 digit hex number back into 8 digits?

    - by Babil
    The algorithm to convert input 8 digit hex number into 10 digit are following: Given that the 8 digit number is: '12 34 56 78' x1 = 1 * 16^8 * 2^3 x2 = 2 * 16^7 * 2^2 x3 = 3 * 16^6 * 2^1 x4 = 4 * 16^4 * 2^4 x5 = 5 * 16^3 * 2^3 x6 = 6 * 16^2 * 2^2 x7 = 7 * 16^1 * 2^1 x8 = 8 * 16^0 * 2^0 Final 10 digit hex is: = x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 = '08 86 42 98 E8' The problem is - how to go back to 8 digit hex from a given 10 digit hex (for example: 08 86 42 98 E8 to 12 34 56 78) Some sample input and output are following: input output 11 11 11 11 08 42 10 84 21 22 22 33 33 10 84 21 8C 63 AB CD 12 34 52 D8 D0 88 64 45 78 96 32 21 4E 84 98 62 FF FF FF FF 7B DE F7 BD EF

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  • checking every digit in a number for oddness

    - by tryt
    Hello. I was writing a function which checks if every digit in a number is odd. I came accross this weird behaviour. Why does the second function return different (incorrect) results, eventhough its basically the same? (implemented in an opposite way) #include <stdio.h> int all_odd_1(int n) { if (n == 0) return 0; if (n < 0) n = -n; while (n > 0) { if (n&1 == 1) n /= 10; else return 0; } return 1; } int all_odd_2(int n) { if (n == 0) return 0; if (n < 0) n = -n; while (n > 0) { if (n&1 == 0) return 0; else n /= 10; } return 1; } int main() { printf("all_odd_1\n"); printf("%d\n", all_odd_1(-131)); printf("%d\n", all_odd_1(121)); printf("%d\n", all_odd_1(2242)); printf("-----------------\n"); printf("all_odd_2\n"); printf("%d\n", all_odd_2(131)); printf("%d\n", all_odd_2(121)); printf("%d\n", all_odd_2(2242)); return 0; }

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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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  • 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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  • 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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  • 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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  • 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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