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  • 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.

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  • priority queue implementation

    - by davit-datuashvili
    i have implemented priority queue and i am interested if it is correct public class priqueue { private int n,maxsize; int x[]; void swap(int i,int j){ int t=x[i]; x[i]=x[j]; x[j]=t; } public priqueue(int m){ maxsize=m; x=new int [maxsize+1]; n=0; } void insert(int t){ int i,p; x[++n]=t; for (i=n;i>1 && x[p=i/2] >x[i];i=p) swap(p,i); } public int extramin(){ int i,c; int t=x[1]; x[1]=x[n--]; for (i=1;(c=2*i)<=n;i=c){ if (c+1<=n && x[c+1]<x[c]) c++; if (x[i]<=x[c]) break; swap(c,i); } return t; } public void display(){ for (int j=0;j<x.length;j++){ System.out.println(x[j]); } } } public class priorityqueue { public static void main(String[] args) { priqueue pr=new priqueue(12); pr.insert(20); pr.insert(12); pr.insert(22); pr.insert(15); pr.insert(35); pr.insert(17); pr.insert(40); pr.insert(51); pr.insert(26); pr.insert(19); pr.insert(29); pr.insert(23); pr.extramin(); pr.display(); } } //result: 0 12 15 17 20 19 22 40 51 26 35 29 23

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  • "Teach" a computer how to do addition?

    - by ffar
    The problem is to learn computer to do addition. Computer have as input a knowladge of a numbers: he "knows" that after 1 goes 2, after 2 goes 3 and so on... Having that data computer can easyly get next number. Next, computer have knowlandge as input that x+0=x and x+(y+1)=(x+1)+y. This axioms let computer to do addition. For example, to add 5 and 3, computer makes following: 5+3 = 5+(2+1) = (5+1)+2 = 6+2 = 6+(1+1) = (6+1)+1 = 7+1 = 8. But this is too long to add numbers in such way. The problem is to develop program which can improve this way of addition using the rules of mathemetics and logic. The goal addition must executed in O(log(N)) time, not O(N) time, N is magnitude of added numbers. Have this program any science value? Is any program can do such things?

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  • 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

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  • 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

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  • longest increasing subsequent

    - by davit-datuashvili
    i have write this code is it correct? public class subsequent{ public static void main(String[] args){ int a[]=new int[]{0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15}; int a_b[]=new int[a.length]; a_b[0]=1; int int_max=0; int int_index=0; for (int i=0;i<a.length;i++){ for (int j=0;j<i;j++){ if (a[i]>a[j] && a_b[i]<(a_b[j]+1)){ a_b[i]=a_b[j]+1; } } if (a_b[i]>int_max){ int_max=a_b[i]; int_index=i; } } int k=int_max+1; int list[]=new int[k]; for (int i=int_index;i>0;i--){ if (a_b[i]==k-1){ list[k-1]=a[i]; k=a_b[i]; } } for (int g=0;g<list.length;g++){ System.out.println(list[g]); } } }

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  • Linear color interpolation?

    - by user146780
    If I have a straight line that mesures from 0 to 1, then I have colorA(255,0,0) at 0 on the line, then at 0.3 I have colorB(20,160,0) then at 1 on the line I have colorC(0,0,0). How could I find the color at 0.7? Thanks

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  • Calculating color shades

    - by matejv
    I have the next problem. I have a base color with couple of different shades of that color. Example: Base color: #4085c5 Shade: #005cb1 Now, I have a different color (let's say #d60620), but no shades of it. From the color I would like to calculate shades, that have similar difference as colors mentioned in first paragraph. First I tried calculating difference of RGB elements and applying them to second color, but the result was not like I expected to be. Than I tried with converting color to HSV, reading saturation value and applying the difference to second color, but again the resulting color was still weird. The formula was something like: (HSV(BaseColor)[S] - HSV(Shade)[S]) + HSV(SecondColor)[H] Does anyone know how this problem could be solved? I know I am doing something wrong, but I don't know what. :)

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  • Need to sync two lists with atrribute time, but times aren't equal

    - by virgula24
    I gonna try to describe my problem the best i can. I have two lists, one with audio frames and other with color frames (not relevant). Both of them have timestamps, they were captured at the same moment but at different instants. So, i have like this: index COLOR AUDIO 0 841 846 1 873 897 2 905 948 3 940 1000 the frames start at high numbers because they were captured and then trimmed to specific parts, but im shot, frame 0 is synced with only 5ms apart(timestamp in ms). On every case i have, the audio frames count is less than the color count. I need to make them have the same count. The stating frames may be coloraudio, color

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  • fastest way to perform string search in general and in python

    - by Rkz
    My task is to search for a string or a pattern in a list of documents that are very short (say 200 characters long). However, say there are 1 million documents of such time. What is the most efficient way to perform this search?. I was thinking of tokenizing each document and putting the words in hashtable with words as key and document number as value, there by creating a bag of words. Then perform the word search and retrieve the list of documents that contained this word. From what I can see is this operation will take O(n) operations. Is there any other way? may be without using hash-tables?. Also, is there a python library or third party package that can perform efficient searches?

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  • programming books [closed]

    - by davit-datuashvili
    I have only one dream -- to buy these two books: Introduction to Algorithms, Third Edition Concrete Mathematics: A Foundation for Computer Science Sorry for my dream, this is a site for posting programming questions, but unfortunately everything happened

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  • Merging elements in a scala list

    - by scompt.com
    I'm trying to port the following Java snippet to Scala. It takes a list of MyColor objects and merges all of the ones that are within a delta of each other. It seems like a problem that could be solved elegantly using some of Scala's functional bits. Any tips? List<MyColor> mergedColors = ...; MyColor lastColor = null; for(Color aColor : lotsOfColors) { if(lastColor != null) { if(lastColor.diff(aColor) < delta) { lastColor.merge(aColor); continue; } } lastColor = aColor; mergedColors.add(aColor); }

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  • Recursive solution to finding patterns

    - by user2997162
    I was solving a problem on recursion which is to count the total number of consecutive 8's in a number. For example: input: 8801 output: 2 input: 801 output: 0 input: 888 output: 3 input: 88088018 output:4 I am unable to figure out the logic of passing the information to the next recursive call about whether the previous digit was an 8. I do not want the code but I need help with the logic. For an iterative solution, I could have used a flag variable, but in recursion how do I do the work which flag variable does in an iterative solution. Also, it is not a part of any assignment. This just came to my mind because I am trying to practice coding using recursion.

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  • yet another logic.

    - by Sunil
    I'm working on a research problem out of curiosity and I don't know how to program the logic that I've in mind. Let me explain it to you : I've 4 vectors say for example, v1 = 1 1 1 1 v2 = 2 2 2 2 v3 = 3 3 3 3 v4 = 4 4 4 4 Now what I want to do is to add them combination-wise. i.e v12 = v1+v2 v13 = v1+v3 v14 = v1+v4 v23 = v2+v3 v24 = v2+v4 v34 = v3+v4 Till this step it is just fine. The problem/trick is now, at the end of each iteration I give the obtained vectors into a black box function and it returns only few of the vectors say v12, v13 and v34. Now, I want to add each of these vectors one vector from v1,v2,v3,v4 which it hasn't added before. For example v3 and v4 hasn't been added to v12 so I want to create v123 and v124. similarly for all the vectors like, v12 should become : v123 = v12+v3 v124 = v12+v4 v13 should become : v132 // this should not occur because I already have v123 v134 = v13+v4; v14,v23 and v24 cannot be considered because it was deleted in the black box function so all we have in our hands to work with is v12,v13 and v34. v34 should become : v341 // cannot occur because we have 134 v342 = v34+v2 It is important that I do not do all at one step at the start like for example I can do (4 choose 3) 4C3 and finish it off but I want to do it step by step at each iteration. I've asked a modified version of this question before (without including the black box function) and got answers here. Can anybody tell me how to do it when the black box function is included ? A modification of the previous answer would also be great. Thanks in advance.

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  • Why is i-- faster than i++ in loops? [closed]

    - by Afshin Mehrabani
    Possible Duplicate: JavaScript - Are loops really faster in reverse…? I don't know if this question is valid in other languages or not, but I'm asking this specifically for JavaScript. I see in some articles and questions that the fastest loop in JavaScript is something like: for(var i = array.length; i--; ) Also in Sublime Text 2, when you try to write a loop, it suggests: for (var i = Things.length - 1; i >= 0; i--) { Things[i] }; I want to know, why is i-- faster than i++ in loops?

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  • Python: Determine whether list of lists contains a defined sequence

    - by duhaime
    I have a list of sublists, and I want to see if any of the integer values from the first sublist plus one are contained in the second sublist. For all such values, I want to see if that value plus one is contained in the third sublist, and so on, proceeding in this fashion across all sublists. If there is a way of proceeding in this fashion from the first sublist to the last sublist, I wish to return True; otherwise I wish to return False. In other words, for each value in sublist one, for each "step" in a "walk" across all sublists read left to right, if that value + n (where n = number of steps taken) is contained in the current sublist, the function should return True; otherwise it should return False. (Sorry for the clumsy phrasing--I'm not sure how to clean up my language without using many more words.) Here's what I wrote. a = [ [1,3],[2,4],[3,5],[6],[7] ] def find_list_traversing_walk(l): for i in l[0]: index_position = 0 first_pass = 1 walking_current_path = 1 while walking_current_path == 1: if first_pass == 1: first_pass = 0 walking_value = i if walking_value+1 in l[index_position + 1]: index_position += 1 walking_value += 1 if index_position+1 == len(l): print "There is a walk across the sublists for initial value ", walking_value - index_position return True else: walking_current_path = 0 return False print find_list_traversing_walk(a) My question is: Have I overlooked something simple here, or will this function return True for all true positives and False for all true negatives? Are there easier ways to accomplish the intended task? I would be grateful for any feedback others can offer!

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  • Multiply without multiplication, division and bitwise operators, and no loops. Recursion

    - by lxx22
    public class MultiplyViaRecursion{ public static void main(String[] args){ System.out.println("8 * 9 == " + multiply(8, 9)); System.out.println("6 * 0 == " + multiply(6, 0)); System.out.println("0 * 6 == " + multiply(0, 6)); System.out.println("7 * -6 == " + multiply(7, -6)); } public static int multiply(int x, int y){ int result = 0; if(y > 0) return result = (x + multiply(x, (y-1))); if(y == 0) return result; if(y < 0) return result = -multiply(x, -y); return result; } } My question is very simple and basic, why after each "if" the "return" still cannot pass the compilation, error shows missing return.

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