Search Results

Search found 5655 results on 227 pages for 'stl algorithm'.

facebook twitter digg google delicious technorati stumbleupon myspace wordpress linkedin gmail igoogle windows live tumblr viadeo yahoo buzz yahoo mail yahoo bookmarks favorites email print

Page 116/227 | < Previous Page | 112 113 114 115 116 117 118 119 120 121 122 123  | Next Page >

  • All possibilities in 2d array

    - by valli-R
    I have this array: $array = array ( array('1', '2', '3'), array('!', '@'), array('a', 'b', 'c', 'd'), ); And I want to know all character combination of sub arrays.. for example : 1!a 1!b 1!c 1!d 1@a 1@b 1@c 1@d 2!a 2!b 2!c 2!d 2@a 2@b ... Currently I am having this code : for($i = 0; $i < count($array[0]); $i++) { for($j = 0; $j < count($array[1]); $j++) { for($k = 0; $k < count($array[2]); $k++) { echo $array[0][$i].$array[1][$j].$array[2][$k].'<br/>'; } } } It works, but I think it is ugly, and when I am adding more arrays, I have to add more for. I am pretty sure there is a way to do this recursively, but I don't know how to start/how to do it. A little help could be nice! Thanks you!

    Read the article

  • Interview Q: given an array of numbers, return array of products of all other numbers (no division)

    - by polygenelubricants
    I was asked this question in a job interview, and I'd like to know how others would solve it. I'm most comfortable with Java, but solutions in other languages are welcome. Given an array of numbers, nums, return an array of numbers products, where products[i] is the product of all nums[j], j != i. Input : [1, 2, 3, 4, 5] Output: [(2*3*4*5), (1*3*4*5), (1*2*4*5), (1*2*3*5), (1*2*3*4)] = [120, 60, 40, 30, 24] You must do this in O(N) without using division.

    Read the article

  • Minimizing distance to a weighted grid

    - by Andrew Tomazos - Fathomling
    Lets suppose you have a 1000x1000 grid of positive integer weights W. We want to find the cell that minimizes the average weighted distance.to each cell. The brute force way to do this would be to loop over each candidate cell and calculate the distance: int best_x, best_y, best_dist; for x0 = 1:1000, for y0 = 1:1000, int total_dist = 0; for x1 = 1:1000, for y1 = 1:1000, total_dist += W[x1,y1] * sqrt((x0-x1)^2 + (y0-y1)^2); if (total_dist < best_dist) best_x = x0; best_y = y0; best_dist = total_dist; This takes ~10^12 operations, which is too long. Is there a way to do this in or near ~10^8 or so operations?

    Read the article

  • Programmatically determine the relative "popularities" of a list of items (books, songs, movies, etc

    - by Horace Loeb
    Given a list of (say) songs, what's the best way to determine their relative "popularity"? My first thought is to use Google Trends. This list of songs: Subterranean Homesick Blues Empire State of Mind California Gurls produces the following Google Trends report: (to find out what's popular now, I restricted the report to the last 30 days) Empire State of Mind is marginally more popular than California Gurls, and Subterranean Homesick Blues is far less popular than either. So this works pretty well, but what happens when your list is 100 or 1000 songs long? Google Trends only allows you to compare 5 terms at once, so absent a huge round-robin, what's the right approach? Another option is to just do a Google Search for each song and see which has the most results, but this doesn't really measure the same thing

    Read the article

  • question about permut-by-sorting

    - by davit-datuashvili
    hi i have following question from book introduction in algorithms second edition there is such problem suppose we have some array A int a[]={1,2,3,4} and we have some random priorities array P={36,3,97,19} we shoud permut array a randomly using this priorities array here is pseudo code P ERMUTE -B Y-S ORTING ( A) 1 n ? length[A] 2 for i ? 1 to n do P[i] = R ANDOM(1, n 3 ) 3 4 sort A, using P as sort keys 5 return A and result will be permuted array B={2, 4, 1, 3}; please help any ideas i have done this code and need aideas how continue import java.util.*; public class Permut { public static void main(String[]args){ Random r=new Random(); int a[]=new int[]{1,2,3,4}; int n=a.length; int b[]=new int[a.length]; int p[]=new int[a.length]; for (int i=0;i<p.length;i++){ p[i]=r.nextInt(n*n*n)+1; } // for (int i=0;i<p.length;i++){ // System.out.println(p[i]); //} } } please help

    Read the article

  • question about Littles Law

    - by davit-datuashvili
    I know that Little's Law states (paraphrased): the average number of things in a system is the product of the average rate at which things leave the system and the average time each one spends in the system, or: n=x*(r+z); x-throughput r-response time z-think time r+z - average response time now i have question about a problem from programming pearls: Suppose that system makes 100 disk accesses to process a transaction (although some systems require fewer, some systems will require several hundred disk access per transaction). How many transactions per hour per disk can the system handle? please help

    Read the article

  • How can i optimize this python code

    - by RandomVector
    def maxVote(nLabels): count = {} maxList = [] maxCount = 0 for nLabel in nLabels: if nLabel in count: count[nLabel] += 1 else: count[nLabel] = 1 #Check if the count is max if count[nLabel] > maxCount: maxCount = count[nLabel] maxList = [nLabel,] elif count[nLabel]==maxCount: maxList.append(nLabel) return random.choice(maxList) nLabels contains a list of integers. The above function returns the integer with highest frequency, if more than one have same frequency then a randomly selected integer from them is returned. E.g. maxVote([1,3,4,5,5,5,3,12,11]) is 5

    Read the article

  • binary number comparison

    - by EquinoX
    If I have a 32 bit two's complement number and I want to know what is the easiest way to know of two numbers are equal... what would be the fastest bitwise operator to know this? I know xor'ing both numbers and check if the results are zero works well... any other one's? how about if a number is greater than 0?? I can check the 31'st bit to see if it's greater or equal to 0..but how about bgtz?

    Read the article

  • Calculating similarites between sentences

    - by codecreator
    I have datbase with thousands of rows of error logs and their description.This error log is for an application that running 24/7. I want to create a dashboard/UI to view the current common errors happening for prodcution support. The problem I am having is that even though there are lot of common errors, the error description differs by the transcation ID or user ID or things that are unique for that sigle prcoess. e.g Error trasaction XYz failed for user 233 e.g 2. Error trasaction XYz failed for user 567 I consider these two erros to be same. So I want to a program that will go through the new error logs and classify them into groups. I am trying to use "edit distance" but its very slow.Since I alraedy have old error logs, i am trying to think of solutions using that information too. Any thoughts?

    Read the article

  • Loop through different sets of unique permutations

    - by user558610
    Hi I'm having a hard time getting started to layout code for this problem. I have a fixed amount of random numbers, in this case 8 numbers. R[] = { 1, 2, 3, 4, 5, 6, 7, 8 }; That are going to be placed in 3 sets of numbers, with the only constraint that each set contain minimum one value, and each value can only be used once. For example: R1[] = { 1, 4 } R2[] = { 2, 8, 5, 6 } R3[] = { 7, 4 } I need to loop through all possible combinations of a set R1, R2, R3. Order is not important, so if the above example happened, I don't need R1[] = { 4, 1 } R2[] = { 2, 8, 5, 6 } R3[] = { 7, 4 } NOR R1[] = { 2, 8, 5, 6 } R2[] = { 7, 4 } R3[] = { 1, 4 } What is a good method?

    Read the article

  • question about random select

    - by davit-datuashvili
    here is code print number in decreasing order import java.util.*; public class select { public static void Select(int m,int n){ Random r=new Random(); if (m>0) if (r.nextInt(0x3fff8001) % n <m ){ System.out.println(n-1); Select(m-1,n-1); } else{ Select(m,n-1); } } public static void main(String[]args){ int m=35; int n=200; Select(m,n); } } and question is how to changes code such that print number in increasing order? please help

    Read the article

  • Data structure name: combination array/linked list

    - by me_and
    I have come up with a data structure that combines some of the advantages of linked lists with some of the advantages of fixed-size arrays. It seems very obvious to me, and so I'd expect someone to have thought of it and named it already. Does anyone know what this is called: Take a small fixed-size array. If the number of elements you want to put in your array is greater than the size of the array, add a new array and whatever pointers you like between the old and the new. Thus you have: Static array ————————————————————————— |1|2|3|4|5|6|7|8|9|a|b|c| ————————————————————————— Linked list ———— ———— ———— ———— ———— |1|*->|2|*->|3|*->|4|*->|5|*->NULL ———— ———— ———— ———— ———— My thing: ———————————— ———————————— |1|2|3|4|5|*->|6|7|8|9|a|*->NULL ———————————— ————————————

    Read the article

  • command&pointer&malloc [closed]

    - by gcc
    input 23 3 4 4 42 n 23 0 9 9 n n n 3 9 9 x //according to input,i should create int pointer arrays. pointer arrays starting from 1 (that is initial arrays is arrays[1].when program sees n ,it must be jumb to arrays 2 expected output arrays[1] 3 4 5 42 arrays[2] 23 0 9 9 arrays[5] 3 9 9 x is stopper n is comman to create new pointer array i am new in this site anyone help me how can i write

    Read the article

  • fill array with binary numbers

    - by davit-datuashvili
    hi, first of all this is not homework!! my question is from book: Algorithms in C++ third edition by robert sedgewick question is: there is given array of size n by 2^n(two dimensional) and we should fill it by binary numbers with bits size exactly n or for example n=5 so result will be 00001 00010 00011 00100 00101 00110 00111 and so on we should put this sequence of bits into arrays please help me

    Read the article

  • Common strategies to deal with rounding errors in currency-intensive soft?

    - by Max
    What is your advice on: compensation of accumulated error in bulk math operations on collections of Money objects. How is this implemented in your production code? (things like variable rounding, etc...) theory behind rounding in accountancy. any literature on topic. I currently read Fowler. He mentions Money type, but says nothing on strategies. Older posts on money-rounding (here, and here) do not provide a details and formality I need. Thanks for help.

    Read the article

  • Need Multiple Sudoku Solutions

    - by user1567909
    I'm trying to output multiple sudoku solutions in my program. For example, when You enter this as input: 8..6..9.5.............2.31...7318.6.24.....73...........279.1..5...8..36..3...... .'s denote blank spaces. Numbers represent already-filled spaces. The output should be a sudoku solution like so: 814637925325149687796825314957318462241956873638274591462793158579481236183562749 However, I want to output multiple solutions. This would be all the solutions that should be printed: 814637925325149687796825314957318462241956873638274591462793158579481236183562749 814637925325941687796825314957318462241569873638472591462793158579184236183256749 834671925125839647796425318957318462241956873368247591682793154579184236413562789 834671925125839647796524318957318462241956873368247591682793154519482736473165289 834671925125839647796524318957318462241965873368247591682793154519482736473156289 But my program only prints out one solution. Below is my recursive solution to solving a sudoku solution bool sodoku::testTheNumber(sodoku *arr[9][9], int row, int column) { if(column == 9) { column = 0; row++; if(row == 9) return true; } if(arr[row][column]->number != 0) { return testTheNumber(arr, row, column+1); } for(int k = 1; k < 10; k++) { if(k == 10) { arr[row][column]->number = 0; return false; } if(rowIsValid(arr, k, row) && columnIsValid(arr, k, column) && boxIsValid(arr, k, row, column)) { arr[row][column]->number = k; if(testTheNumber(arr, row, column+1)==true) { return true; } arr[row][column]->number = 0; } } return false; } Could anyone help me come up with a way to print out multiple solutions? Thanks.

    Read the article

facebook twitter digg google delicious technorati stumbleupon myspace wordpress linkedin gmail igoogle windows live tumblr viadeo yahoo buzz yahoo mail yahoo bookmarks favorites email print

< Previous Page | 112 113 114 115 116 117 118 119 120 121 122 123  | Next Page >