[ad_1] Even if I agree with Get Off My Lawn, you could have a solution that wok for both, with simple use of modulo. x = x – (x%10000) + (x%1000) //29492 => 29492 – 9492 + 492 [ad_2] solved Replace thousand number with 0 on an integer [closed]
[ad_1] Searching for a single item in an array is O(N) Searching for a longest run of increasing numbers in an array is O(N^2) if you use a straightforward algorithm with two nested loops* Note: This is not Java-specific. * A faster algorithm exists to do this search. [ad_2] solved Java Search algorithm [closed]
[ad_1] After you try to find the tile in the board in the move function, you should check if the obtained array indices are valid or not. Because, in case user inputs number outside the permitted values, then some garbage value is used for further condition checking. So make the _i and _j variables -1 … Read more
[ad_1] Its a modified version of jasons: import java.util.concurrent.Executor; import java.util.concurrent.Executors; import java.util.concurrent.atomic.AtomicInteger; public class Test { public static void main(String[] args){ final int max = 100; final AtomicInteger i = new AtomicInteger(0); Executor dd = Executors.newFixedThreadPool(2); final Object lock = new Object(); dd.execute(new Runnable() { @Override public void run() { while (i.get() < max) … Read more
Introduction [ad_1] The Executor framework in Java is a powerful tool for managing multiple threads. It allows you to easily create and manage threads, and to execute tasks in parallel. In this tutorial, we will learn how to use the Executor framework to print odd and even numbers from two separate threads. We will also … Read more
[ad_1] double is not enough for the precision you should use high-precision to solve it 6 [ad_2] solved Online Judgement System – Why am I getting Wrong Answer for this thread? [closed]
[ad_1] If I understand correctly, you want to find the first positive integer number that is not in the list. (since you’re using “the” first number, I assume you don’t need an array of numbers with all holes.) The way you do this is (assuming the array is sorted like in your examples) is by … Read more
[ad_1] Notice the extra { void union{(int a, int b,int *p) ^ NOTE: This answer was posted before the OP edited the question to remove the typo (mentioned in this answer). Check the edits for clarity. ANOTHER NOTE: Use the standard definition of main() int main(void) //if no command line arguments. 10 [ad_2] solved Getting … Read more
[ad_1] Your first implementation worked but had degenerate behavior with specific datasets. The pivot in your original working code was *(last – 1) so the simplest fix would be to swap a random element with *(last – 1). The rest of your original partition code would work unchanged. 2 [ad_2] solved C++ QuickSort Isn’t Working … Read more
[ad_1] The question is a bit complicated and hard to follow.. Anyways I tried writing some code to the best of my understanding of the problem. Here is the part that finds the possible combinations: % input matrix A=[1 2 0 1 2 0 0 0 2 1 1 1 0 2 2 0 3 … Read more
[ad_1] You can try this: for (int i=6; i>1; i–) { for (int j=0; j<i; j++) cout<<“*”; cout<<endl; } Output: ****** ***** **** *** ** 2 [ad_2] solved Print 6,5,4,3,2 characters separated by a new line in C++?
[ad_1] The binary heap data structure is usually drawn as a binary tree with certain properties, but is typically implemented using a plain array. The reason for this is that the array version uses less memory than the explicit binary tree and is a lot easier to manipulate. Consequently, the array A that you’re seeing … Read more
[ad_1] The correct code is: #include <stdlib.h> #include <stdio.h> void main() { int pn = 0; //Processes Number int CPU = 0; //CPU Current time int allTime = 0; // Time neded to finish all processes printf(“Enrer Processes Count: “); scanf(“%d”,&pn); int AT[pn]; int ATt[pn]; int NoP = pn; int PT[pn]; //Processes Time int PP[pn]; … Read more
[ad_1] You can implement adjacency list with vectors like this, it is much easier than using pointers. Check the code, it is also much easier to understand how it works. #include <bits/stdc++.h> using namespace std; vector<int> edges[5]; bool visited[5]; void dfs(int x) { visited[x] = true; for(int i=0; i < edges[x].size(); i++) if(!visited[edges[x][i]]) dfs(edges[x][i]); } … Read more
[ad_1] Given that i^j=N, you can solve the equation for j by taking the log of both sides:j log(i) = log(N) or j = log(N) / log(i). So the algorithm becomes for i=2 to N { j = log(N) / log(i) if((Power(i,j)==N) print(i,j) } Note that due to rounding errors with floating point calculations, you … Read more