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  • best way to find similar items in python

    - by user230911
    I have 1M numbers:N[], and 1 single number n, now I want to find in those 1M numbers that are similar to that single number, say an area of [n-10, n+10]. what's the best way in python to do this? Do I have to sort the 1M number and do an iteration? Thanks

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  • How to restrict operations to certain lines?

    - by Ayman
    I have to work on some relatively huge code files in vim. How do I restrict some operations like find-next normal-n and others to a certain function / block? How would I visually know if I'm within that block or outside it? Looking and line numbers seems awkward, specially that the line numbers I need to work with are generally 5 digits long!

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  • What is the best algorithm for this problem?

    - by mark
    What is the most efficient algorithm to solve the following problem? Given 6 arrays, D1,D2,D3,D4,D5 and D6 each containing 6 numbers like: D1[0] = number D2[0] = number ...... D6[0] = number D1[1] = another number D2[1] = another number .... ..... .... ...... .... D1[5] = yet another number .... ...... .... Given a second array ST1, containing 1 number: ST1[0] = 6 Given a third array ans, containing 6 numbers: ans[0] = 3, ans[1] = 4, ans[2] = 5, ......ans[5] = 8 Using as index for the arrays D1,D2,D3,D4,D5 and D6, the number that goes from 0, to the number stored in ST1[0] minus one, in this example 6, so from 0 to 6-1, compare each res array against each D array My algorithm so far is: I tried to keep everything unlooped as much as possible. EML := ST1[0] //number contained in ST1[0] EML1 := 0 //start index for the arrays D While EML1 < EML if D1[ELM1] = ans[0] goto two if D2[ELM1] = ans[0] goto two if D3[ELM1] = ans[0] goto two if D4[ELM1] = ans[0] goto two if D5[ELM1] = ans[0] goto two if D6[ELM1] = ans[0] goto two ELM1 = ELM1 + 1 return 0 //bad row of numbers, if while ends two: EML1 := 0 start index for arrays Ds While EML1 < EML if D1[ELM1] = ans[1] goto two if D2[ELM1] = ans[1] goto two if D3[ELM1] = ans[1] goto two if D4[ELM1] = ans[1] goto two if D5[ELM1] = ans[1] goto two if D6[ELM1] = ans[1] goto two ELM1 = ELM1 + 1 return 0 three: EML1 := 0 start index for arrays Ds While EML1 < EML if D1[ELM1] = ans[2] goto two if D2[ELM1] = ans[2] goto two if D3[ELM1] = ans[2] goto two if D4[ELM1] = ans[2] goto two if D5[ELM1] = ans[2] goto two if D6[ELM1] = ans[2] goto two ELM1 = ELM1 + 1 return 0 four: EML1 := 0 start index for arrays Ds While EML1 < EML if D1[ELM1] = ans[3] goto two if D2[ELM1] = ans[3] goto two if D3[ELM1] = ans[3] goto two if D4[ELM1] = ans[3] goto two if D5[ELM1] = ans[3] goto two if D6[ELM1] = ans[3] goto two ELM1 = ELM1 + 1 return 0 five: EML1 := 0 start index for arrays Ds While EML1 < EML if D1[ELM1] = ans[4] goto two if D2[ELM1] = ans[4] goto two if D3[ELM1] = ans[4] goto two if D4[ELM1] = ans[4] goto two if D5[ELM1] = ans[4] goto two if D6[ELM1] = ans[4] goto two ELM1 = ELM1 + 1 return 0 six: EML1 := 0 start index for arrays Ds While EML1 < EML if D1[ELM1] = ans[0] return 1 //good row of numbers if D2[ELM1] = ans[0] return 1 if D3[ELM1] = ans[0] return 1 if D4[ELM1] = ans[0] return 1 if D5[ELM1] = ans[0] return 1 if D6[ELM1] = ans[0] return 1 ELM1 = ELM1 + 1 return 0 As language of choice, it would be pure c

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  • How to take URL and split/string to get URL variables in Flash AS3

    - by Leon
    So I have a URL that I need my Flash movie to extract variables from: example link: http://www.example.com/example_xml.php?aID=1234&bID=5678 I need to get the aID and the bID numbers. I'm able to get the full URL into a String via ExternalInterface var url:String = ExternalInterface.call("window.location.href.toString"); if (url) testField.text = url; Just unsure as how to manipulate the String to just get the 1234 and 5678 numbers. Appreciate any tips, links or help with this!

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  • Finding what makes strings unique in a list, can you improve on brute force?

    - by Ed Guiness
    Suppose I have a list of strings where each string is exactly 4 characters long and unique within the list. For each of these strings I want to identify the position of the characters within the string that make the string unique. So for a list of three strings abcd abcc bbcb For the first string I want to identify the character in 4th position d since d does not appear in the 4th position in any other string. For the second string I want to identify the character in 4th position c. For the third string it I want to identify the character in 1st position b AND the character in 4th position, also b. This could be concisely represented as abcd -> ...d abcc -> ...c bbcb -> b..b If you consider the same problem but with a list of binary numbers 0101 0011 1111 Then the result I want would be 0101 -> ..0. 0011 -> .0.. 1111 -> 1... Staying with the binary theme I can use XOR to identify which bits are unique within two binary numbers since 0101 ^ 0011 = 0110 which I can interpret as meaning that in this case the 2nd and 3rd bits (reading left to right) are unique between these two binary numbers. This technique might be a red herring unless somehow it can be extended to the larger list. A brute-force approach would be to look at each string in turn, and for each string to iterate through vertical slices of the remainder of the strings in the list. So for the list abcd abcc bbcb I would start with abcd and iterate through vertical slices of abcc bbcb where these vertical slices would be a | b | c | c b | b | c | b or in list form, "ab", "bb", "cc", "cb". This would result in four comparisons a : ab -> . (a is not unique) b : bb -> . (b is not unique) c : cc -> . (c is not unique) d : cb -> d (d is unique) or concisely abcd -> ...d Maybe it's wishful thinking, but I have a feeling that there should be an elegant and general solution that would apply to an arbitrarily large list of strings (or binary numbers). But if there is I haven't yet been able to see it. I hope to use this algorithm to to derive minimal signatures from a collection of unique images (bitmaps) in order to efficiently identify those images at a future time. If future efficiency wasn't a concern I would use a simple hash of each image. Can you improve on brute force?

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  • Writing number keypad

    - by theojero
    I am writing a program that prints out telephone numbers that are to be entered by users. On the west side, I need a picture icon of 3 people. In the center, I need the keypads. On the North side, I need the numbers to be displayed. Can someone help with the layout of the buttons?

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  • Interview question: f(f(n)) == -n

    - by Hrvoje Prgeša
    A question I got on my last interview: Design a function f, such that: f(f(n)) == -n Where n is a 32 bit signed integer; you can't use complex numbers arithmetic. If you can't design such a function for the whole range of numbers, design it for the largest range possible. Any ideas?

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  • Generating a readable colour from RGB?

    - by Joe Simpson
    Hi, I'm putting in a function which will allow a user to input a color (eg: purple) and it will change the look of their profile to be purple. It's interpreted from text into a 'Color' class which stores them inside itself as RGB numbers (int for red, one for green and other for blue). What i don't know how to do is logically turn these three numbers into another 3 which will make a readable colour. Can anyone help me on how to do this? Joe

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  • A[i] * A[j] = k in O(nlog(n))

    - by gleb-pendler
    A is an Array of n positive int numbers k given int Algorithm should find if there is a pair of numbers which product gives the result a. A[i] * A[j] = k b. A[i] = A[j] + k if there is such a couple the algorithm should return thier index. thanks in advance.

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  • Working with PHP Octals and String conversions

    - by krio
    I'm working with a database that has a bunch of serial numbers that are prefixed with leading 0's. So a serial number can look like 00032432 or 56332432. Problem is with PHP I don't understand how the conversion system with octals works. A specific example is that I'm trying to convert and compare all of these integer based numbers with strings. Is it possible to convert an octal, such as 00234 to a string like "00234" so that I can compare it?

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  • Rails validates_format_of

    - by squids
    Hi, I want to use validates_format_of to validate a comma separated string with only letters (small and caps), and numbers. So. example1, example2, 22example44, ex24 not: ^&*, < , asfasfsdafas<#%$# Basically I want to have users enter comma separated words(incl numbers) without special characters. I'll use it to validate tags from acts_as_taggable_on. (i don't want to be a valid tag for example. Thanks in advance.

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  • containsObject - why doesen't this work?

    - by Emil
    Hi. I have an array that I am trying to check wether or not an indexPath(.row) exists in. I use this code: if ([array containsObject:[NSNumber numberWithInt:indexPath.row]]){ NSLog(@"Yep, it exists in there."); } the array consist of the numbers 3, 8 and 2. The index path loads numbers fromm 0 to 8 in a loop. Can anybody see why this doesen't work?

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  • RegEx for Dynamic URL Goals settings in Google Analytics

    - by gaaustralia
    Hi, I have tried to work this regex to set up a goal in GA for 2 days, but I cannot get my head around it... The url format is like this: /purchase.php?cDd=1&transaction_id=xxxxxxx&verify=xxxxxxxxxxxxxxxx=&method=creditcard&type=purchase transaction_id= is populated with a sept of numbers verify= is populated by a string of numbers, letters in both caps and lower case Basically I would like to only match URLs which finish by "&method=creditcard&type=purchase" I have tried to just put &method=creditcard&type=purchase but it does retrieve other URLs too Would anyone has any ideas

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  • F# - Facebook Hacker Cup - Double Squares

    - by Jacob
    I'm working on strengthening my F#-fu and decided to tackle the Facebook Hacker Cup Double Squares problem. I'm having some problems with the run-time and was wondering if anyone could help me figure out why it is so much slower than my C# equivalent. There's a good description from another post; Source: Facebook Hacker Cup Qualification Round 2011 A double-square number is an integer X which can be expressed as the sum of two perfect squares. For example, 10 is a double-square because 10 = 3^2 + 1^2. Given X, how can we determine the number of ways in which it can be written as the sum of two squares? For example, 10 can only be written as 3^2 + 1^2 (we don't count 1^2 + 3^2 as being different). On the other hand, 25 can be written as 5^2 + 0^2 or as 4^2 + 3^2. You need to solve this problem for 0 = X = 2,147,483,647. Examples: 10 = 1 25 = 2 3 = 0 0 = 1 1 = 1 My basic strategy (which I'm open to critique on) is to; Create a dictionary (for memoize) of the input numbers initialzed to 0 Get the largest number (LN) and pass it to count/memo function Get the LN square root as int Calculate squares for all numbers 0 to LN and store in dict Sum squares for non repeat combinations of numbers from 0 to LN If sum is in memo dict, add 1 to memo Finally, output the counts of the original numbers. Here is the F# code (See code changes at bottom) I've written that I believe corresponds to this strategy (Runtime: ~8:10); open System open System.Collections.Generic open System.IO /// Get a sequence of values let rec range min max = seq { for num in [min .. max] do yield num } /// Get a sequence starting from 0 and going to max let rec zeroRange max = range 0 max /// Find the maximum number in a list with a starting accumulator (acc) let rec maxNum acc = function | [] -> acc | p::tail when p > acc -> maxNum p tail | p::tail -> maxNum acc tail /// A helper for finding max that sets the accumulator to 0 let rec findMax nums = maxNum 0 nums /// Build a collection of combinations; ie [1,2,3] = (1,1), (1,2), (1,3), (2,2), (2,3), (3,3) let rec combos range = seq { let count = ref 0 for inner in range do for outer in Seq.skip !count range do yield (inner, outer) count := !count + 1 } let rec squares nums = let dict = new Dictionary<int, int>() for s in nums do dict.[s] <- (s * s) dict /// Counts the number of possible double squares for a given number and keeps track of other counts that are provided in the memo dict. let rec countDoubleSquares (num: int) (memo: Dictionary<int, int>) = // The highest relevent square is the square root because it squared plus 0 squared is the top most possibility let maxSquare = System.Math.Sqrt((float)num) // Our relevant squares are 0 to the highest possible square; note the cast to int which shouldn't hurt. let relSquares = range 0 ((int)maxSquare) // calculate the squares up front; let calcSquares = squares relSquares // Build up our square combinations; ie [1,2,3] = (1,1), (1,2), (1,3), (2,2), (2,3), (3,3) for (sq1, sq2) in combos relSquares do let v = calcSquares.[sq1] + calcSquares.[sq2] // Memoize our relevant results if memo.ContainsKey(v) then memo.[v] <- memo.[v] + 1 // return our count for the num passed in memo.[num] // Read our numbers from file. //let lines = File.ReadAllLines("test2.txt") //let nums = [ for line in Seq.skip 1 lines -> Int32.Parse(line) ] // Optionally, read them from straight array let nums = [1740798996; 1257431873; 2147483643; 602519112; 858320077; 1048039120; 415485223; 874566596; 1022907856; 65; 421330820; 1041493518; 5; 1328649093; 1941554117; 4225; 2082925; 0; 1; 3] // Initialize our memoize dictionary let memo = new Dictionary<int, int>() for num in nums do memo.[num] <- 0 // Get the largest number in our set, all other numbers will be memoized along the way let maxN = findMax nums // Do the memoize let maxCount = countDoubleSquares maxN memo // Output our results. for num in nums do printfn "%i" memo.[num] // Have a little pause for when we debug let line = Console.Read() And here is my version in C# (Runtime: ~1:40: using System; using System.Collections.Generic; using System.Diagnostics; using System.IO; using System.Linq; using System.Text; namespace FBHack_DoubleSquares { public class TestInput { public int NumCases { get; set; } public List<int> Nums { get; set; } public TestInput() { Nums = new List<int>(); } public int MaxNum() { return Nums.Max(); } } class Program { static void Main(string[] args) { // Read input from file. //TestInput input = ReadTestInput("live.txt"); // As example, load straight. TestInput input = new TestInput { NumCases = 20, Nums = new List<int> { 1740798996, 1257431873, 2147483643, 602519112, 858320077, 1048039120, 415485223, 874566596, 1022907856, 65, 421330820, 1041493518, 5, 1328649093, 1941554117, 4225, 2082925, 0, 1, 3, } }; var maxNum = input.MaxNum(); Dictionary<int, int> memo = new Dictionary<int, int>(); foreach (var num in input.Nums) { if (!memo.ContainsKey(num)) memo.Add(num, 0); } DoMemoize(maxNum, memo); StringBuilder sb = new StringBuilder(); foreach (var num in input.Nums) { //Console.WriteLine(memo[num]); sb.AppendLine(memo[num].ToString()); } Console.Write(sb.ToString()); var blah = Console.Read(); //File.WriteAllText("out.txt", sb.ToString()); } private static int DoMemoize(int num, Dictionary<int, int> memo) { var highSquare = (int)Math.Floor(Math.Sqrt(num)); var squares = CreateSquareLookup(highSquare); var relSquares = squares.Keys.ToList(); Debug.WriteLine("Starting - " + num.ToString()); Debug.WriteLine("RelSquares.Count = {0}", relSquares.Count); int sum = 0; var index = 0; foreach (var square in relSquares) { foreach (var inner in relSquares.Skip(index)) { sum = squares[square] + squares[inner]; if (memo.ContainsKey(sum)) memo[sum]++; } index++; } if (memo.ContainsKey(num)) return memo[num]; return 0; } private static TestInput ReadTestInput(string fileName) { var lines = File.ReadAllLines(fileName); var input = new TestInput(); input.NumCases = int.Parse(lines[0]); foreach (var lin in lines.Skip(1)) { input.Nums.Add(int.Parse(lin)); } return input; } public static Dictionary<int, int> CreateSquareLookup(int maxNum) { var dict = new Dictionary<int, int>(); int square; foreach (var num in Enumerable.Range(0, maxNum)) { square = num * num; dict[num] = square; } return dict; } } } Thanks for taking a look. UPDATE Changing the combos function slightly will result in a pretty big performance boost (from 8 min to 3:45): /// Old and Busted... let rec combosOld range = seq { let rangeCache = Seq.cache range let count = ref 0 for inner in rangeCache do for outer in Seq.skip !count rangeCache do yield (inner, outer) count := !count + 1 } /// The New Hotness... let rec combos maxNum = seq { for i in 0..maxNum do for j in i..maxNum do yield i,j }

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  • Maintaing list order in binary tree.

    - by TheBigO
    Given a sequence of numbers, I want to insert the numbers into a balanced binary tree such that when I do a preorder traversal on the tree, it gives me the sequence back. How can I construct the insert method corresponding to this requirement? Remember that the tree must be balanced, so there isn't a completely trivial solution. I was trying to do this with a modified version of an AVL tree, but I'm not sure if this can work out.

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  • How to increment counters based on a printed array

    - by Sam Liew
    I managed to developed a simple board of 5x5 using random numbers and array. Big achievement for someone like me! :) Now I have to increment the counters depending on the frequency of the numbers. If the value within 0-49 is printed..then nCounter++ If the value within 50-75 is printed..then pCounter++ something like that. The problem is that I don't know how to increase the counters based on the printed board. Here is the code: #include <stdio.h> #include <stdlib.h> #include <time.h> int main() { //Initialize Variables int randomNumber; int rows; int columns; int hdCounter =0; int hCounter = 0; int cCounter = 0; int pCounter = 0; int nCounter = 0; //Declare board size int board[5][5]; //size of board is 5 x 5 //Create the random number generator seed srand(time(NULL)); //Assign the random numbers from 1 - 25 into variable randomNumber //Create the rows for the board for ( rows = 1; rows <= 5 ; rows++ ) { //Create the colimns for the board for ( columns = 1; columns <= 5 ; columns++ ) { //Assign variable randomNumber into variable board randomNumber = rand() %100 + 1; board[rows][columns] = randomNumber; //print the board printf("%d\t", board[rows][columns]); //calculate the frequency of numbers on the printed board if (randomNumber >= 85 && randomNumber <= 100 ) hdCounter++; else if ( randomNumber >= 75 ) hCounter++; else if ( randomNumber >= 65 ) cCounter++; else if ( randomNumber >= 50 ) pCounter++; else if ( randomNumber >= 1 ) nCounter++; else continue; } //Newline after the end of 5th column. printf("\n\n"); } printf( "N \t P \t C \t H \t HD\n\n" ); printf("%d \t %d \t %d \t %d \t %d \t", nCounter, pCounter, cCounter, hCounter, hdCounter); }//end main I tried replacing randomNumber in the if-statement with board[rows][columns] but I seem to get the same undesired results.

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  • plot matrix missing points in different color using gnuplot

    - by kitt
    I have a file 'matrix.dat': 1 2 3 4 5 5 - 3 4 5 - 4 5 B - 1 B 2 B 3 - 3 2 - 3 I want to plot numbers using palette, '-' using white color and 'B' using black color. In gnuplot, I use this palette (blue - cyan - green - orange - red): set palette model HSV functions 0.666*(1-gray), 1, 1 And set '-' as missing data: set datafile missing "-" plot 'matrix.dat' matrix with image Now I can only plot numbers and '-' in correct colors.

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  • sanitation script in php for login credentials...

    - by Matt
    What I am looking for currently is a simple, basic, login credentials sanitation script. I understand that I make a function to do so and I have one...but all it does right now is strip tags... am I doomed to use replace? or is there a way i can just remove all special characters and spaces and limit it to only letters and numbers...then as for the password limit it to only letters and numbers exclimation points, periods, and other special chars that cannot affect my SQL query. Please help :/ Thanks, Matt

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  • How to block calls in android

    - by Anurag Uniyal
    I want to block calls from few numbers, for that I want to write a app of my own. So what are the APIs which I should be using? Basically I want to get notified when a call comes, i want to compare numbers if it is what i want to block, i want to cut the call or mute it or if possible mute it and record it.

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  • Probelm with String.split() in java

    - by Matt
    What I am trying to do is read a .java file, and pick out all of the identifiers and store them in a list. My problem is with the .split() method. If you run this code the way it is, you will get ArrayOutOfBounds, but if you change the delimiter from "." to anything else, the code works. But I need to lines parsed by "." so is there another way I could accomplish this? import java.io.BufferedReader; import java.io.FileNotFoundException; import java.io.FileReader; import java.io.IOException; import java.util.*; public class MyHash { private static String[] reserved = new String[100]; private static List list = new LinkedList(); private static List list2 = new LinkedList(); public static void main (String args[]){ Hashtable hashtable = new Hashtable(997); makeReserved(); readFile(); String line; ListIterator itr = list.listIterator(); int listIndex = 0; while (listIndex < list.size()) { if (itr.hasNext()){ line = itr.next().toString(); //PROBLEM IS HERE!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! String[] words = line.split("."); //CHANGE THIS AND IT WILL WORK System.out.println(words[0]); //TESTING TO SEE IF IT WORKED } listIndex++; } } public static void readFile() { String text; String[] words; BufferedReader in = null; try { in = new BufferedReader(new FileReader("MyHash.java")); //NAME OF INPUT FILE } catch (FileNotFoundException ex) { Logger.getLogger(MyHash.class.getName()).log(Level.SEVERE, null, ex); } try { while ((text = in.readLine()) != null){ text = text.trim(); words = text.split("\\s+"); for (int i = 0; i < words.length; i++){ list.add(words[i]); } for (int j = 0; j < reserved.length; j++){ if (list.contains(reserved[j])){ list.remove(reserved[j]); } } } } catch (IOException ex) { Logger.getLogger(MyHash.class.getName()).log(Level.SEVERE, null, ex); } try { in.close(); } catch (IOException ex) { Logger.getLogger(MyHash.class.getName()).log(Level.SEVERE, null, ex); } } public static int keyIt (int x) { int key = x % 997; return key; } public static int horner (String word){ int length = word.length(); char[] letters = new char[length]; for (int i = 0; i < length; i++){ letters[i]=word.charAt(i); } char[] alphabet = new char[26]; String abc = "abcdefghijklmnopqrstuvwxyz"; for (int i = 0; i < 26; i++){ alphabet[i]=abc.charAt(i); } int[] numbers = new int[length]; int place = 0; for (int i = 0; i < length; i++){ for (int j = 0; j < 26; j++){ if (alphabet[j]==letters[i]){ numbers[place]=j+1; place++; } } } int hornered = numbers[0] * 32; for (int i = 1; i < numbers.length; i++){ hornered += numbers[i]; if (i == numbers.length -1){ return hornered; } hornered = hornered % 997; hornered *= 32; } return hornered; } public static String[] makeReserved (){ reserved[0] = "abstract"; reserved[1] = "assert"; reserved[2] = "boolean"; reserved[3] = "break"; reserved[4] = "byte"; reserved[5] = "case"; reserved[6] = "catch"; reserved[7] = "char"; reserved[8] = "class"; reserved[9] = "const"; reserved[10] = "continue"; reserved[11] = "default"; reserved[12] = "do"; reserved[13] = "double"; reserved[14] = "else"; reserved[15] = "enum"; reserved[16] = "extends"; reserved[17] = "false"; reserved[18] = "final"; reserved[19] = "finally"; reserved[20] = "float"; reserved[21] = "for"; reserved[22] = "goto"; reserved[23] = "if"; reserved[24] = "implements"; reserved[25] = "import"; reserved[26] = "instanceof"; reserved[27] = "int"; reserved[28] = "interface"; reserved[29] = "long"; reserved[30] = "native"; reserved[31] = "new"; reserved[32] = "null"; reserved[33] = "package"; reserved[34] = "private"; reserved[35] = "protected"; reserved[36] = "public"; reserved[37] = "return"; reserved[38] = "short"; reserved[39] = "static"; reserved[40] = "strictfp"; reserved[41] = "super"; reserved[42] = "switch"; reserved[43] = "synchronize"; reserved[44] = "this"; reserved[45] = "throw"; reserved[46] = "throws"; reserved[47] = "trasient"; reserved[48] = "true"; reserved[49] = "try"; reserved[50] = "void"; reserved[51] = "volatile"; reserved[52] = "while"; reserved[53] = "="; reserved[54] = "=="; reserved[55] = "!="; reserved[56] = "+"; reserved[57] = "-"; reserved[58] = "*"; reserved[59] = "/"; reserved[60] = "{"; reserved[61] = "}"; return reserved; } }

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