1970. Last Day Where You Can Still Cross

There is a 1-based binary matrix where 0 represents land and 1 represents water. You are given integers row and col representing the number of rows and columns in the matrix, respectively. Initially on day 0, the entire matrix is land. However, each day a new cell becomes flooded with water. You are given a 1-based 2D array cells, where cells[i] = [ri, ci] represents that on the ith day, the cell on the rith row and cith column (1-based coordinates) will be covered with water (i.e., changed to 1). You want to find the last day that it is possible to walk from the top to the bottom by only walking on land cells. You can start from any cell in the top row and end at any cell in the bottom row. You can only travel in the four cardinal directions (left, right, up, and down). Return the last day where it is possible to walk from the top to the bottom by only walking on land cells.
/**
 * @param {number} row
 * @param {number} col
 * @param {number[][]} cells
 * @return {number}
 */
var latestDayToCross = function(row, col, cells) {
    // Helper: check if it's possible to cross on a given day
    function canCross(day) {
        // Build a grid where 1 = water, 0 = land
        const grid = Array.from({ length: row }, () => Array(col).fill(0));

        // Flood the first `day` cells
        for (let i = 0; i < day; i++) {
            const r = cells[i][0] - 1; // convert

how to refresh list of available packages in apt? 

apt update

TotalCMD替代资源管理器 

;date:2014-07-06
;用TC替换资源管理器及恢复资源管理器
;主要来自http://blog.xiazhiri.com/tags/totalcmd/的文章，
;又根据http://qing.blog.sina.com.cn/2002017477/77545cc533002ie4.html的文章加了右键用explorer打开
;另外参考了http://blog.csdn.net/lord_is_layuping/article/details/7435989的文章
RegRead, IsExp, HKEY_LOCAL_MACHINE, SOFTWARE\Classes\Folder\shell\open\command, DelegateExecute
If(IsExp="{11dbb47c-a525-400b-9e80-a54615a090c0}")
{
    RegDelete HKEY_LOCAL_MACHINE, SOFTWARE\Classes\Folder\shell\open\command, DelegateExecute
;Direct

ShortCuts快捷方式同步 

' 修正后的脚本（重点调整字符串连接格式）
Set WshShell = CreateObject("WScript.Shell")

' 重构命令构建方式（移除行尾注释干扰）
RoboCommand = "robocopy " & _
  Chr(34) & "D:\Settings\ShortCuts" & Chr(34) & " " & _
  Chr(34) & "C:\ProgramData\Microsoft\Windows\Start Menu\Programs\ShortCuts" & Chr(34) & " " & _
  "/purge /e /copy:DAT /mt /z /xc /xn /xo /mot:1"  ' 参数统一在行尾注释

' 验证命令完整性（调试时可取消注释查看）
 'MsgBox RoboCommand

' 执行命令
ReturnCode = WshShell.Run(RoboCommand, 0, True)
Set WshShell = Nothing

修改文件夹显示名称 

@echo off
set /p var=请输入需要显示的名称：
if exist 1%\desktop.ini (
del /a /f 1%\desktop.ini
)
(echo [.ShellClassInfo]
		echo LocalizedResourceName=%var%
		)>"%1\desktop.ini"
attrib +s +h %1\desktop.ini
attrib +r /s /d "%1"
taskkill /f /im explorer.exe
start explorer

Rclone同步数据到云盘(使用cd2或者openlist作为中介) 

/*
    AutoRclone Pro V3.0 (双模式+排除优化版)
    更新日志：
    - V3.0: 支持 Sync/Backup 双模式；支持数组形式配置排除文件
    - V2.2: 启动自检
    - V2.0: WinAPI 监控内核
*/

#Requires AutoHotkey v2.0
#SingleInstance Force
Persistent

; ===================== 核心配置区域 =====================
Global RclonePath := "D:\CloudSoft\CMDTools\Tools\rclone.exe"  ; ★ 修改为你的 Rclone 路径
Global LogFile    := "D:\UsersData\Zakary\Documents\Rclone\autoRclone.log" ; ★ 日志路径

/* 
   配置说明：
   - Mode: "Sync" (镜像，本地删网盘也删) | "Backup" (备份，本地删网

wget 

wget \
  --recursive \
  --spider \
  --no-verbose \
  --reject "*.*" \
  up.top


  tail -f /vhosts/multisite/logs/nginx/nginx.access.log

NS Task 

||*Domain*||*NS*||
|ug.com|are.com|

Load CSS file in editor 

<?php 

// https://developer.wordpress.org/news/2025/07/you-dont-need-theme-json-for-block-theme-styles/
add_action( 'after_setup_theme', 'themeslug_add_editor_styles' );

function themeslug_add_editor_styles() {
	add_editor_style( get_stylesheet_uri() );
}

Claude Code - Workflows Studio 

# Veille Technologique : Claude Code Workflow Studio

**Date :** 30 décembre 2025  
**Sujet :** Extension VS Code pour la création visuelle de workflows Claude Code  
**Éditeur :** breaking-brake (projet communautaire, non officiel Anthropic)  
**Version actuelle :** 3.10.0

---

## Résumé exécutif

Claude Code Workflow Studio est une extension VS Code communautaire permettant de créer visuellement des workflows d'automatisation pour Claude Code. Via une interface drag-and-drop inspirée de Dify,

how can I see all users in a Linux system? 

cat /etc/passwd

How can I check the list of group that a user belong? 

groups username

# for example
groups pcattaneo

General - Packaging Linux 

# Taxonomie des solutions de packaging Linux

## Niveau 1 : Gestionnaires de paquets système

**Objectif** : gérer l'OS, les librairies système, les outils CLI.

| Outil | Distribution | Philosophie |
|-------|--------------|-------------|
| `apt` | Debian/Ubuntu | Stabilité, paquets validés |
| `dnf` | Fedora/RHEL | Cycle plus rapide qu'apt |
| `pacman` | Arch | Rolling release, dernières versions |

**Caractéristique commune** : les paquets partagent les dépendances système. Une mise à jour pe

age - encrypt / decrypt files with password 

# Encrypt with password
age -p -o .env.age .env

# Decrypt
age -d -o .env .env.age


# To download age:
# https://github.com/FiloSottile/age

756. Pyramid Transition Matrix

You are stacking blocks to form a pyramid. Each block has a color, which is represented by a single letter. Each row of blocks contains one less block than the row beneath it and is centered on top. To make the pyramid aesthetically pleasing, there are only specific triangular patterns that are allowed. A triangular pattern consists of a single block stacked on top of two blocks. The patterns are given as a list of three-letter strings allowed, where the first two characters of a pattern represent the left and right bottom blocks respectively, and the third character is the top block. For example, "ABC" represents a triangular pattern with a 'C' block stacked on top of an 'A' (left) and 'B' (right) block. Note that this is different from "BAC" where 'B' is on the left bottom and 'A' is on the right bottom. You start with a bottom row of blocks bottom, given as a single string, that you must use as the base of the pyramid. Given bottom and allowed, return true if you can build the pyramid all the way to the top such that every triangular pattern in the pyramid is in allowed, or false otherwise.
/**
 * @param {string} bottom
 * @param {string[]} allowed
 * @return {boolean}
 */
var pyramidTransition = function(bottom, allowed) {
    // STEP 1: Build a mapping from (left,right) -> list of possible tops
    // Example: "ABC" means A+B -> C
    const map = new Map();
    for (let pattern of allowed) {
        const left = pattern[0];
        const right = pattern[1];
        const top = pattern[2];
        const key = left + right;

        if (!map.has(key)) {
            map.set(key, [])

dataframe 

import pandas as pd

# Sample data based on your image
data = {
    'Division': ['BED, BARGARH'] * 5,
    'Sub-Division': ['ATABIRA', 'BARGARH-1', 'BARGARH-2', 'BHATLI', 'BHEDEN'],
    'Total Complaint Count (A+B+C)': [10] * 5,
    'Complain Count (A)': [2] * 5,
    'Hour <2 (A)': [3] * 5,
    'Hour 2<4 (A)': [1] * 5,
    'Hour 4<8 (A)': [4] * 5,
    'Complain Count (B)': [3] * 5,
    'Hour <2 (B)': [3] * 5,
    'Hour 2<4 (B)': [1] * 5,
    'Hour 4<8 (B)': [4] * 5,
    'Closed Out