Lulzx

Hello

Lulzx

Keybase proof

I hereby claim:

• I am lulzx on github.
• I am lulzx (https://keybase.io/lulzx) on keybase.
• I have a public key ASDA9wesgUb092VtWiMCnPM7a9WSG3MLl4y9r-KoZSFEaAo

To claim this, I am signing this object:

Lulzx

<?php
$servername = "localhost";
$username   = "username";
$password   = "password";
$conn       = new mysqli($servername, $username, $password);
if ($conn->connect_error)
  {
    die("Connection failed: " . $conn->connect_error);
  }
echo "Connected successfully";

Lulzx

import os
import sys
import signal
import subprocess

currentPID = os.getpid()
if 'pid' not in os.listdir():
    with open('pid', mode='w') as f:
        print(str(currentPID), file=f)
else:

Lulzx

const puppeteer = require('puppeteer');
(async () => {
  const browser = await puppeteer.launch()
  const page = await browser.newPage()
  
  await page.waitForSelector('.main_wrapper > .form > #form1 > p > #username')
  await page.click('.main_wrapper > .form > #form1 > p > #username')
  
  await page.waitForSelector('.main_wrapper > .form > #form1 > p > #password')
  await page.click('.main_wrapper > .form > #form1 > p > #password')

Lulzx

sudo dd if=Solus-3.9999-Budgie.iso of=/dev/sdb bs=1M;sudo sync;sudo eject /dev/sdb

Lulzx

Solve ( d^2 y(x))/( dx^2) + y(x) = 0:

Assume a solution will be proportional to e^(λ x) for some constant λ. Substitute y(x) = e^(λ x) into the differential equation: ( d^2 )/( dx^2)(e^(λ x)) + e^(λ x) = 0

Substitute ( d^2 )/( dx^2)(e^(λ x)) = λ^2 e^(λ x): λ^2 e^(λ x) + e^(λ x) = 0

Factor out e^(λ x):

Lulzx

#include <stdio.h>

int count(int l, int r, int n, int m) 
{ 
    int count = 0;
    for (int i = l; i <= r; i++) {
        int sum = 0, x = i, t;
        t = (i % m); 
        while (x != 0) { 
            sum += x % 10; 

Lulzx

#include <stdio.h>

int main(){
	int i;
	scanf("%d", &i);
	while(i != 1){
		printf("%d → ", i);
		if(!(i % 2)){
			i /= 2;
		}

Lulzx

#include <stdio.h>

void check(int matrix, int rows, int columns){
	printf("%d", matrix[0][0]);
}

int main(){
	int rows, columns, n, matrix[1000][1000];
	scanf("%d%d", &rows, &columns);
	for(int i = 0; i < rows; i++){