I hereby claim:
- I am arknave on github.
- I am arknave (https://keybase.io/arknave) on keybase.
- I have a public key whose fingerprint is 86D1 F11A 1C62 FD23 EAFD E7E2 A108 24C9 4330 0001
To claim this, I am signing this object:
# Maintainer: Thorsten Wißmann <edu@thorsten-wissmann.de> | |
pkgname=kattis-problemtools | |
pkgver=v1.20201230 | |
pkgrel=1 | |
pkgdesc="Tools to manage problem packages using the Kattis problem package format" | |
arch=('i686' 'arm' 'x86_64') | |
url="https://github.com/Kattis/problemtools" | |
license=('MIT') | |
depends=('python' 'gmp' 'java-environment' 'plastex' 'python-yaml') | |
makedepends=('python-setuptools' 'boost') |
// Say you have an integer array of length n. Then this monstrosity sorts it, slowly. | |
void cursed_sort(int* a, int n) { | |
int*q,i; | |
for(i=0;i<n;++i) | |
for(q=a;q+1<a+n;++q) | |
if(*q>*(q+1)) | |
*q^=*(q+1)^=*q^=*(q+1); | |
} |
Hello Challenge! |
import Data.List.Split | |
valid :: String -> Bool | |
valid "" = False | |
valid s = not $ foldr (\a b -> b || a `elem` ["::", "import", "->", "module" ]) False $ words s | |
count :: String -> Int | |
count s = if valid s then length . concat $ words s else 0 | |
main = do |
public class Trie { | |
final static int ALPHABET_SIZE = 26; | |
Trie[] children; | |
public Trie() { | |
this.children = new Trie[ALPHABET_SIZE]; | |
} | |
// return true if the word added is new. | |
public boolean insert(String s) { |
#include <iostream> | |
#include <cstring> | |
#define MAX_LETTERS 600005 | |
#define ALPHABET_SIZE 26 | |
using namespace std; | |
// These two structures should be intiialized to all 0 | |
// (0 is the root and should never be the child of another node) | |
int trie[MAX_LETTERS][ALPHABET_SIZE]; |
N = length(perm); | |
do { | |
print(perm); | |
} while (next_permutation(perm, perm + N)); |
for (int i = 0; i < N; i++) { | |
dist[i] = INFINITY; | |
} | |
dist[start] = 0; | |
for (int k = 0; k < N - 1; k++) { | |
for (int i = 0; i < N; i++) { | |
for (int e = 0; e < adj[i].size(); e++) { | |
int j = adj[i][e].vertex; |
// Assume N is the number of vertices | |
for (int k = 0; k < N; k++) { | |
for (int i = 0; i < N; i++) { | |
for (int j = 0; j < N; j++) { | |
dist[i][j] = min(dist[i][j], | |
dist[i][k] + dist[k][j]); | |
} | |
} | |
} |
I hereby claim:
To claim this, I am signing this object: