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  • Javascript Regular expression to enforce 2 digits after decimal points

    - by Manas Saha
    I need to validate a text box where users are supposed to enter numeric values with 2 decimal places. I am validating the text in client side javascript. The validation will pass only if the number has precisely 2 decimal places, and there is at least 1 digit before the decimal point. (could be zero) the number before the decimal point can be 0, but can not be multiple zeroes. like 00 or 000 The number before the decimal point can not begin with more than 1 zero. Example of Passed validation: 0.01 0.12 111.23 1234.56 012345.67 123.00 0.00 Example of failed validation .12 1.1 0.0 00.00 1234. 1234.567 1234 00123.45 abcd.12 12a4.56 1234.5A I have tried with [0-9][.][0-9][0-9]$ But it is allowing alphabets before decimal point. like 12a4.56 Can anyone please fix the expression? thanks a lot in advance.

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  • iteration on numbers with no 2 same digits

    - by rahmivolkan
    I dont know if it is asked (I couldn't find any). I want to iterate on this kind of numbers implemented on array; int a[10]; int i = 0; for( ; i < 10; i++ ) a[i] = i+1; now the array has "1 2 3 4 5 6 7 8 9 10" and I want to get "1 2 3 4 5 6 7 8 10 9" and then "1 2 3 4 5 6 7 9 8 10" "1 2 3 4 5 6 7 9 10 8" . . . . I tried to get an algorithm but I couldn't figure it out. Is there an easy way to implement "next" iterator for this kind of problems? Thanks in advance

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  • C++: recursively computer all permutaions of digits 0 - 9

    - by Nate
    I have a homework assignment where part of the requirement is to recursively compute all the permutations of integers 0 - 9. The professor actually gave us the algorithm for this part of the question. I've finished the rest of the assignment, but I can't get the permute function working...I'm implementing it exactly like it was shown on the assignment information. However, when I run it each permutation is repeated multiple times (and I'm not sure if I'm even getting all the correct permutations.) I think he must've made a mistake on the assignment instructions. I've been working on this for a couple of hours and can't seem to figure out where I'm going wrong. Can anybody help point me in the right direction? Here's the current code: void permute(int v[], int curr) { for (int i = curr; i < MAX; i++) { swap(v[i], v[curr]); permute(v, curr + 1); swap(v[curr], v[i]); } } EDIT: Actually, right after posting this I realized it has to do with the swap, right? Because right now i and curr are the same, so I'm swapping identical numbers. Hm, should it be swap(v[i], v[curr+1])?

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  • Python sort strings started with digits

    - by vlad
    I have the next list: a = ['1th Word', 'Another Word', '10th Word'] print a.sort() >>> ['10th Word', '1th Word', 'Another Word'] But I need: ['1th Word', '10th Word','Another Word'] Is there an easy way to do this? I tried: r = re.compile(r'(\d+)') def sort_by_number(s): m = r.match(s) return m.group(0) x.sort(key=sort_by_number) But some strings do not have numbers and this leads to an errors. Thanks.

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  • MySQL: Digits (1.75) to Friendly Time (1 hour and 45 minutes)

    - by Nick
    In my MySQL database, I have a float field named "HoursSpent" with values like "0.25", "1.75", "2.5", etc. Is there a way that I can have my SELECT statement format those values in a friendly format like this?: 0.25 = 15 minutes 1.75 = 1 hour and 45 minutes 2.5 = 2 hours and 30 minutes The "HoursSpent" field is supposed to only have values in 0.25 increments, but if somebody were to put something random like 0.16, it would be nice if the SELECT statement handled that by rounding it up to the nearest 0.25 (so in this case 0.16 would become 0.25, or 15 minutes).

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  • Is there a version of the Arial or Tahoma font with monospaced digits and spaces?

    - by rossmcm
    The digits in the Arial font supplied with Windows are monospaced, in that they each take up the same horizontal space, but they seem to have neglected to provide a "monospaced" version of the space character. This means that you can't format a column of digits right-justified in (say) 12 spaces and have the right-hand edge be aligned. For example: 1 12 123 1234 12345 1234567 12345678 123456789 1234567890 works because the font used for code examples has spaces the same width as digits. This however doesn't work if the same text is displayed in Arial (I can't demonstrate because I can't figure out how to defeat SU's reformatting at the moment!). It just so happens that with Tahoma 8 point you can cheat because a space is exactly half the number of pixels as a digit, but that is messy and very specific.

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  • How to round a number to n decimal places in Java

    - by Alex Spurling
    What I'd like is a method to convert a double to a string which rounds using the half-up method. I.e. if the decimal to be rounded is a 5, it always rounds up the previous number. This is the standard method of rounding most people expect in most situations. I also would like only significant digits to be displayed. That is there should not be any trailing zeroes. I know one method of doing this is to use the String.format method: String.format("%.5g%n", 0.912385); returns: 0.91239 which is great, however it always displays numbers with 5 decimal places even if they are not significant: String.format("%.5g%n", 0.912300); returns: 0.91230 Another method is to use the DecimalFormatter: DecimalFormat df = new DecimalFormat("#.#####"); df.format(0.912385); returns: 0.91238 However as you can see this uses half-even rounding. That is it will round down if the previous digit is even. What I'd like is this: 0.912385 -> 0.91239 0.912300 -> 0.9123 What is the best way to achieve this in Java?

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  • Define a positive number to be isolated if none of the digits in its square are in its cube [closed]

    - by proglaxmi
    Define a positive number to be isolated if none of the digits in its square are in its cube. For example 163 is n isolated number because 69*69 = 26569 and 69*69*69 = 4330747 and the square does not contain any of the digits 0, 3, 4 and 7 which are the digits used in the cube. On the other hand 162 is not an isolated number because 162*162=26244 and 162*162*162 = 4251528 and the digits 2 and 4 which appear in the square are also in the cube. Write a function named isIsolated that returns 1 if its argument is an isolated number, it returns 0 if its not an isolated number and it returns -1 if it cannot determine whether it is isolated or not (see the note below). The function signature is: int isIsolated(long n) Note that the type of the input parameter is long. The maximum positive number that can be represented as a long is 63 bits long. This allows us to test numbers up to 2,097,151 because the cube of 2,097,151 can be represented as a long. However, the cube of 2,097,152 requires more than 63 bits to represent it and hence cannot be computed without extra effort. Therefore, your function should test if n is larger than 2,097,151 and return -1 if it is. If n is less than 1 your function should also return -1. Hint: n % 10 is the rightmost digit of n, n = n/10 shifts the digits of n one place to the right. The first 10 isolated numbers are N n*n n*n*n 2 4 8 3 9 27 8 64 512 9 81 729 14 196 2744 24 576 13824 28 784 21952 34 1156 39304 58 3364 195112 63 3969 250047

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  • Is your Credit Card Number valid?

    - by Rekha
    The credit card numbers may look like some random unique 16 digits number but those digits inform more than what we think it could be. The first digit of the card is the Major Industry Identifier: 1 and 2 -  Airlines 3  – Travel and Entertainment 4 and 5 -  Banking and Financial 6 – Merchandizing and Banking 7 – Petroleum 8 – Telecommunications 9 – National assignment The first 6 digits represent the Issuer Identification Number: Visa – 4xxxxx Master Card – 51xxxx & 55xxxx The 7th and following digits, excluding the last digit, are the person’s account number which leads to trillion possible combinations if the maximum of 12 digits is used. Many cards only use 9 digits. The final digit is the checksum or check digit. It is used to validate the card number using Luhn algorithm. How To Validate Credit Card Number? Take any credit card number, for example 5588 3201 2345 6789. Step 1: Double every other digit from the right: 5*2      8*2      3*2      0*2      2*2      4*2      6*2      8*2 ————————————————————————- 10        16        6          0          4          8      12        16 Step 2: Add these new digits to undoubled digits. All double digit numbers are added as a sum of their digits, so 16 becomes 1+6 = 7: Undoubled digits:       5          8          2          1          3          5          7          9 Doubled Digits:          10       16         6          0          4          8         12         16 Sum:  5+1+0+8+1+6+2+6+1+0+3+4+5+8+7+1+2+9+1+6 = 76 If the final sum is divisible by 10, then the Credit Card number is valid, if not, the number is invalid or fake!!! Hence the example is a fake number? via mint  cc and image credit This article titled,Is your Credit Card Number valid?, was originally published at Tech Dreams. Grab our rss feed or fan us on Facebook to get updates from us.

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  • Fastest way to find the largest power of 10 smaller than x

    - by peoro
    Is there any fast way to find the largest power of 10 smaller than a given number? I'm using this algorithm, at the moment, but something inside myself dies anytime I see it: 10**( int( math.log10(x) ) ) # python pow( 10, (int) log10(x) ) // C I could implement simple log10 and pow functions for my problems with one loop each, but still I'm wondering if there is some bit magic for decimal numbers.

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  • SQL Server: Must numbers all be specified with latin numeral digits?

    - by Ian Boyd
    Does SQL server expect numbers to be specified with digits from the latin alphabet, e.g.: 0123456789 Is it valid to give SQL Server digits in other alphabets? Rosetta Stone: Latin: 01234567890 Arabic: ?????????? Bengali: ?????????? i know that the client (ADO) will convert 8-bit strings to 16-bit unicode strings using the current culture. But the client is also converting numbers to strings using their current culture, e.g.: SELECT * FROM Inventory WHERE Quantity > ???,?? Which throws SQL Server for fits. i know that the server/database has it's defined code page and locale, but that is for strings. Will SQL Server interpret numbers using the active (or per-login specified) locale, or must all numeric values be specifid with latin numeral digits?

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  • C Program not running as intended, hangs after input

    - by user41419
    The program I am writing to take a number and display that number as a calculator would display it (shown below) is compiling with no issues, but when I try to run it, I am able to input my number, but nothing happens. It seems like it is "hanging", since no further output is shown as I would have expected. Might anyone know what the problem is? #include <stdio.h> #define MAX_DIGITS 20 char segments[10][7] = /* seven segment array */ {{'1','1','1','1','1','1','0'}, /* zero */ {'0','1','1','0','0','0','0'}, /* one */ {'1','1','0','1','1','0','1'}, /* two */ {'1','1','1','1','0','0','1'}, /* three */ {'0','1','1','0','0','1','1'}, /* four */ {'1','0','1','1','0','1','1'}, /* five */ {'1','0','1','1','1','1','1'}, /* six */ {'1','1','1','0','0','0','0'}, /* seven */ {'1','1','1','1','1','1','1'}, /* eight */ {'1','1','1','0','0','1','1'}};/* nine */ char digits[3][MAX_DIGITS * 4]; /* digits array */ int i, j; /* count variables */ int adjust; /* output formatting */ int main(void) { clear_digits_array(); int digit[20]; for (i = 0; i < 20; i++) { digit[i] = 0; } int count = 20; int position = 0; printf("Enter a number: "); int number = scanf("%d%d%d%d%d%d%d%d%d%d%d%d%d%d%d%d%d%d%d%d", &digit[0], &digit[1], &digit[2], &digit[3], &digit[4], &digit[5], &digit[6], &digit[7], &digit[8], &digit[9], &digit[10], &digit[11], &digit[12], &digit[13], &digit[14], &digit[15], &digit[16], &digit[17], &digit[18], &digit[19]); //NOTHING HAPPENS AFTER HERE printf("Got input, number is %d", number); while (count > 0) { printf("Reading digits, count is %d", count); process_digit(digit[20 - count], position); position++; count--; } print_digits_array(); printf("\n"); return 0; } void clear_digits_array(void) { /* fill all positions in digits array with blank spaces */ for (i = 0; i < 3; i++) { for (j = 0; j < (MAX_DIGITS * 4); j++) { digits[i][j] = ' '; } } } void process_digit(int digit, int position) { /* check each segment to see if segment should be filled in for given digit */ for (i = 0; i < 7; i++) { printf("Processing digit %d at position %d, i is %d", digit, position, i); if (segments[digit][i] == 1) { switch (i) { case 0: digits[0][(position * 4) + 1] = '_'; break; case 1: digits[1][(position * 4) + 2] = '|'; break; case 2: digits[2][(position * 4) + 2] = '|'; break; case 3: digits[2][(position * 4) + 1] = '_'; break; case 4: digits[2][(position * 4) + 0] = '|'; break; case 5: digits[1][(position * 4) + 0] = '|'; break; case 6: digits[1][(position * 4) + 1] = '_'; break; } } } } void print_digits_array(void) { /* print each character in digits array */ for (i = 0; i < 3; i++) { for (j = 0; j < (MAX_DIGITS * 4); j++) { printf("%c", digits[i][j]); } printf("/n"); } }

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  • How to print all possibilities of a number that occur interchanging the digits in a number?

    - by RADHA KRISHNA LOLLA
    I want to print that the all possibilities of a number that occurs interchanging the digits in a number? I want to print that the all possibilities of a number that occurs interchanging the digits in a number? Suppose a number 1234 it have 24 possibilities 4!=24 123 have 6 possibilities 3!=6 123 132 213 231 312 321 i want these numbers as output please tell me the structure of this program in the python

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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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  • Is there a way to sort digits within a cell?

    - by Iszi
    I'd like to know if there's an Excel Function, or a formula, that can be used to sort digits within a cell. For example, if the value of A1 is 6193254807 and B1 has our formula with parameters set to sort ascending then B1 should equal 0123456789 (leading zero included). Of course, if the parameters were set so that it would sort descending then B1 should equal 9876543210. Is there a way to do this, without wandering into VBScript, or macros, or something similarly exotic?

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  • [AJAX Numeric Updown Control] Microsoft JScript runtime error: The number of fractional digits is out of range

    - by Jenson
    If you have using Ajax control toolkits a lot (which I will skip the parts on where to download and how to configure it in Visual Studio 2010), you might have encountered some bugs or limitations of the controls, or rather, some weird behaviours. I would call them weird behaviours though. Recently, I've been working on a Ajax numeric updown control, which i remember clearly it was working fine without problems. In fact, I use 2 numeric updown control this time. So I went on to configure it to be as simple as possible and I will just use the default up and down buttons provided by it (so that I won't need to design my own). I have two textbox controls to display the value controlled by the updown control. One for month, and another for year. <asp:TextBox ID="txtMonth" runat="server" CssClass="txtNumeric" ReadOnly="True" Width="150px" /> <asp:TextBox ID="txtYear" runat="server" CssClass="txtNumeric" ReadOnly="True" Width="150px" /> So I will now drop 1 numeric updown control for each of the textboxes. <asp:NumericUpDownExtender ID="txtMonth_NumericUpDownExtender"     runat="server" TargetControlID="txtMonth" Maximum="12" Minimum="1" Width="152"> </asp:NumericUpDownExtender>                          <asp:NumericUpDownExtender ID="txtYear_NumericUpDownExtender"     runat="server" TargetControlID="txtYear" Width="152"> </asp:NumericUpDownExtender>                                                  You noticed that I configure the Maximum and Minimum value for the first numericupdown control, but I never did the same for the second one (for txtYear). That's because it won't work, well, at least for me. So I remove the Minimum="2000" and Maximum="2099" from there. Then I would configure the initial value to the the current year, and let the year to flow up and down freely. If you want, you want write the codes to restrict it. Here are the codes I used on PageLoad:     Protected Sub Page_Load(ByVal sender As Object, ByVal e As System.EventArgs) Handles Me.Load         If Not Page.IsPostBack Then             If Trim(txtMonth.Text) = "" Then                 Me.txtMonth.Text = System.DateTime.Today.Month             End If             If Trim(txtYear.Text) = "" Then                 Me.txtYear.Text = System.DateTime.Today.Year             End If         End If     End Sub   Enjoy!

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  • How to get last 12 digits from a sting in MySQL?

    - by Nick Gorbikoff
    Hello. How would I get last 12 digits of a string using mysql? Let's say I have a varchar field with a tracking number, that may be anywhere from 5 to 20 varchars long. But I only need to select last 12 digits or less if there are less. so in a field = 12345678123456789012 I would only need to get what's in brackets field = 12345678[123456789012] I saw a few examples using mid, etc, but they dont' produce the desired result or I can't find an example that makes sense :-( Thank you.

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