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  • 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

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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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  • 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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  • 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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  • Sort vector<int>(n) in O(n) time using O(m) space?

    - by Adam
    I have a vector<unsigned int> vec of size n. Each element in vec is in the range [0, m], no duplicates, and I want to sort vec. Is it possible to do better than O(n log n) time if you're allowed to use O(m) space? In the average case m is much larger than n, in the worst case m == n. Ideally I want something O(n). I get the feeling that there's a bucket sort-ish way to do this: unsigned int aux[m]; aux[vec[i]] = i; Somehow extract the permutation and permute vec. I'm stuck on how to do 3. In my application m is on the order of 16k. However this sort is in the inner loops and accounts for a significant portion of my runtime.

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  • select k th mimimum from array a[0..n-1]

    - by davit-datuashvili
    i have done folloing code from progrmming pearls here is code import java.util.*; public class select { public static int select1(int x[],int l,int u,int k){ //pre l<=k<=u //post x[l..k-1]<=x[k]<=x[k+1..u] Random r=new Random(); int t=r.nextInt(u-1-l)+l; if (l>=u) return -1 ; swap(l,t); int s=x[l]; int i=l; int j=u+1; while (true){ do { i++; }while (i<=u && x[i]<t); do { j--; }while (x[j]>t); if (i>j) break; int temp=x[i]; x[i]=x[j];x[j]=t; swap(l,j); if (j<k){ return select1(x,j+1,u,k); } } return select1(x,l,j-1,k); } public static void main(String[] args) { int x[]=new int[]{4,7,9,3,2,12,13,10,20}; select1(x,0,x.length-1,5); } public static void swap(int i,int j){ int c=i; i=j; j=c; } } but here is mistake Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: -1 at select.select1(select.java:21) at select.main(select.java:36) Java Result: 1 please help

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  • Number of the different elements in an array.

    - by AB
    Is it possible to compute the number of the different elements in an array in linear time and constant space? Let us say it's an array of long integers, and you can not allocate an array of length sizeof(long). P.S. Not homework, just curious. I've got a book that sort of implies that it is possible.

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  • it is very important for me this problem [closed]

    - by davit-datuashvili
    please help this is very important problem for me i am going to get job and need such kind of practise implement heaps priortiy queue and so on what is wrong in my java code please tell i want insert number with heap property and return minimum element what is wrong explain please look http://stackoverflow.com/questions/2902781/priority-queue-implementation/2903288#2903288

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  • How to map a long integer number to a N-dimensional vector of smaller integers (and fast inverse)?

    - by psihodelia
    Given a N-dimensional vector of small integers is there any simple way to map it with one-to-one correspondence to a large integer number? Say, we have N=3 vector space. Can we represent a vector X=[(int32)x1,(int32)x2,(int32)x3] using an integer (int48)y? The obvious answer is "Yes, we can". But the question is: "What is the fastest way to do this and its inverse operation?"

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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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  • 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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  • 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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  • Enumerate all k-partitions of 1d array with N elements?

    - by user301217
    This seems like a simple request, but google is not my friend because "partition" scores a bunch of hits in database and filesystem space. I need to enumerate all partitions of an array of N values (N is constant) into k sub-arrays. The sub-arrays are just that - a starting index and ending index. The overall order of the original array will be preserved. For example, with N=4 and k=2: [ | a b c d ] (0, 4) [ a | b c d ] (1, 3) [ a b | c d ] (2, 2) [ a b c | d ] (3, 1) [ a b c d | ] (4, 0) I'm pretty sure this isn't an original problem (and no, it's not homework), but I'd like to do it for every k <= N, and it'd be great if the later passes (as k grows) took advantage of earlier results. If you've got a link, please share.

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