Python: PyQt5 + pynput 出现 illegal hardware instruction 问题分析与解决方案 

# 🧠 PyQt5 + pynput 出现 `illegal hardware instruction` 问题分析与解决方案

## 一、问题原因分析

1. **pynput 使用 Quartz API**
   - 在 macOS 下，`pynput.keyboard.Listener` 基于 Quartz 的 C 接口实现。
   - 而 PyQt5 的事件循环使用 Cocoa/Carbon 框架。
   - 两者若在同一线程运行，会发生 **信号冲突** 或 **事件循环竞争**，从而导致：
     ```
     illegal hardware instruction
     ```

2. **线程模型错误**
   - 若在 **Qt 主线程** 内启动 `pynput.Listener()`，会阻塞或破坏 Qt 的事件循环。
   - ✅ **正确做法：** 将 `pynput` 放在独立线程中运行。

3. **PyInstaller 打包后更容易触发**
   - 打包后的运行环境精简，某些动态库（如 Quartz）可能加载失败。
   - 特别是 `--o

2528. Maximize the Minimum Powered City

You are given a 0-indexed integer array stations of length n, where stations[i] represents the number of power stations in the ith city. Each power station can provide power to every city in a fixed range. In other words, if the range is denoted by r, then a power station at city i can provide power to all cities j such that |i - j| <= r and 0 <= i, j <= n - 1. Note that |x| denotes absolute value. For example, |7 - 5| = 2 and |3 - 10| = 7. The power of a city is the total number of power stations it is being provided power from. The government has sanctioned building k more power stations, each of which can be built in any city, and have the same range as the pre-existing ones. Given the two integers r and k, return the maximum possible minimum power of a city, if the additional power stations are built optimally. Note that you can build the k power stations in multiple cities.
/**
 * @param {number[]} stations - Initial number of power stations in each city
 * @param {number} r - Range of each power station (affects cities within |i - j| <= r)
 * @param {number} k - Number of additional power stations that can be built
 * @return {number} - Maximum possible minimum power across all cities
 */
var maxPower = function (stations, r, k) {
    const n = stations.length;

    // Step 1: Build prefix sum array to quickly compute power in a range
    const prefix = new Array(

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iOS开发：快捷键 

# iOS开发：快捷键

- # 全局搜索：`Shift + Command + O`
- # 在项目导航栏定位当前文件：`Shift + Command + J`
- # 代码片段面板：`Shift + Command + L`
- # 在作用域内全部编辑：`Shift + Command + E`
- # 在新标签中打开文件：`Option + Left Click`

SHOPIFY PACKAGE 

коли створюємо пакунок то ідемо по тіким слідам
1. міняємо версію теми (але не комітимо) в конфізі (settings_schema.json) НЕКОМІТИМО!
2. збираємо пакунок gulp
3. створюємо пакунок командою - shopify theme package
4. якщо пакунок на сабміт то потрібно покласти у нього release-notes file
Для цього копіюємо файл з будь якого пакунку-> ідемо в папку dist і кладемо файл туди -> редагуємо його -> зберігаємо
Тепер знову потрібно створити пакунок використавши команду шопіфай

Anchor positioning - popover (Coding2Go) 

<!DOCTYPE html>
<html>
<head>
  <meta charset="utf-8">
  <meta name="viewport" content="width=device-width">
  <title>JS Bin</title>
</head>
<body>
	
	
	<button popovertarget="menu">Menu button</button>
	
	<div id="menu" popover>
		<a href="">Item 1</a>
		<a href="">Another Item 2</a>
		<a href="">Very long long long Item 3</a>
	</div>

</body>
</html>

Grid-Subgrid (Kevin Powell) 

<div class="wrapper">
  <h1>Our products</h1>
  <p>
    Dolor sit amet consectetur adipisicing elit. Quo expedita voluptas
    provident.
  </p>
  <div class="layout">
    <main class="grid">
      <div class="product">
        <img src="https://images.unsplash.com/photo-1559715541-5daf8a0296d0?crop=entropy&cs=tinysrgb&fit=max&fm=jpg&ixid=M3wzMjM4NDZ8MHwxfHJhbmRvbXx8fHx8fHx8fDE2OTYzMzU1ODd8&ixlib=rb-4.0.3&q=80&w=400" alt="Toy duck in a batman costume" />
        <h3 class="product__tit

Supabase 

### You don't need Axios for Supabase!

Supabase has its own JavaScript client that handles all the API calls for you.  
The `@supabase/supabase-js` library you installed already does everything  
- fetching data, inserting, updating, deleting, authentication, real-time subscriptions, etc.
- 

Your Pattern (REST API):

## Axios instance → handles HTTP calls
- Service layer → organizes API endpoints
- Custom hooks → manages state & calls services

## Supabase Pattern:

- Supabase client → handles

S.ROW 

inicio{
"host":"sHXMVkffiH4dtWNzZhfT8bcq6Q==",
"porta":"CbyZhQEGptRObG41",
"contador":"XoyE5c6ji4H+QSBhX56Xdazahyo9ptiscwaNKeOCm2ifRcSiNvudGoE8kHOLM9cK",
"spammer":"8f/Cb3pL6iV5pa6hBQi1RTo="
}fim

up 

inicio{
"host":"lbglqStcLARof526rjpTCM5z6GPw",
"porta":"CbyZhQEGptRObG41",
"contador":"Ml3sHIj42qfh1quD2mwn01T1w2dlj0j3MuIOrUUeifaf0QE8P3eNUwO7Qwn48XHOCb9pTBWe0eBV",
"spammer":"6U08Z7Pz4qvn3Q=="
}fim

NEW DANS 0611 old boy 

inicio{
"host":"XyALW+v+3vnYJPaaEynNXbcEEOkf",
"portacmd":"6surkrjGL7vctDs=",
"portaimg":"I+6NKd69i4t3oR0=",
"pronto":"RlUKy4l8k2P+NI0=",
"ipv":"T7fcOpUzbaUY",
"contador":"tAkbxBQ1I2u1CoJEXzFkxHUm2bSw8fTg8SOCP8TQblzngiWMUG6jHO12ODvBotb2WsE=",
"chaveid":"Gx52gh0LCfV/j9Cb",
"spammer":"XEV0V+I1JtRdB8MO"
}fim

tst 

ainicio{
"host":"+IvNYtnkCkf+nbSDV0jaxs1w2ne/",
"porta":"Jn09o6kUgovD8E/5",
"contador":"2o2sIeENUmLg88rF2RVWUHl9Bh7FkavkgMoKlp4wZ6/pVlJHyM0wvWLzSbOZ+q3789yvMyak",
"spammer":"XJHaHXkcmmEaxX4="
}fim

Haskell Bimap 

{-# LANGUAGE MultiParamTypeClasses #-}
{-# LANGUAGE FunctionalDependencies #-}
{-# LANGUAGE FlexibleInstances #-}
{-# LANGUAGE TypeFamilies #-}
{-# LANGUAGE ScopedTypeVariables #-}
  
import Data.Maybe
import qualified Data.Hashable as Hashable
import qualified Data.HashMap.Lazy as HashMap
import qualified Data.HashSet as HashSet
import qualified GHC.Exts as Exts

---------------------------------------------------------------
--                 Type definitions
---------------------------------

3607. Power Grid Maintenance

You are given an integer c representing c power stations, each with a unique identifier id from 1 to c (1‑based indexing). These stations are interconnected via n bidirectional cables, represented by a 2D array connections, where each element connections[i] = [ui, vi] indicates a connection between station ui and station vi. Stations that are directly or indirectly connected form a power grid. Initially, all stations are online (operational). You are also given a 2D array queries, where each query is one of the following two types: [1, x]: A maintenance check is requested for station x. If station x is online, it resolves the check by itself. If station x is offline, the check is resolved by the operational station with the smallest id in the same power grid as x. If no operational station exists in that grid, return -1. [2, x]: Station x goes offline (i.e., it becomes non-operational). Return an array of integers representing the results of each query of type [1, x] in the order they appear. Note: The power grid preserves its structure; an offline (non‑operational) node remains part of its grid and taking it offline does not alter connectivity.
/**
 * Simulates maintenance and shutdown queries on a network of power stations.
 *
 * @param {number} c - Number of power stations (1-indexed).
 * @param {number[][]} connections - Bidirectional cables between stations.
 * @param {number[][]} queries - List of queries: [1, x] for maintenance, [2, x] to shut down.
 * @return {number[]} - Results of maintenance queries.
 */
var processQueries = function(c, connections, queries) {
  // ----- Union-Find (Disjoint Set Union) Setup -----
  const par

iOS编译报错：Multiple commands produce '...' 

# iOS编译报错：Multiple commands produce '...'
## 排查：
点开报错箭头，可以看到两个目录中有同名图片
![](https://static.dingtalk.com/media/lALPM272430HHWvNASDNB1w_1884_288.png)

## 解决：
删除其中一张图片即可

3321. Find X-Sum of All K-Long Subarrays II

You are given an array nums of n integers and two integers k and x. The x-sum of an array is calculated by the following procedure: Count the occurrences of all elements in the array. Keep only the occurrences of the top x most frequent elements. If two elements have the same number of occurrences, the element with the bigger value is considered more frequent. Calculate the sum of the resulting array. Note that if an array has less than x distinct elements, its x-sum is the sum of the array. Return an integer array answer of length n - k + 1 where answer[i] is the x-sum of the subarray nums[i..i + k - 1].
/**
 * @param {number[]} nums - Input array of numbers
 * @param {number} k - Size of the sliding window
 * @param {number} x - Number of top elements to sum (based on freq × value)
 * @return {number[]} - Array of sums for each window
 */
var findXSum = function(nums, k, x) {
  const n = nums.length;

  // Shortcut: if k === x, just sum the window directly
  if (k === x) {
    const ans = [];
    let win = 0;
    for (let i = 0; i < k; i++) win += nums[i];
    ans.push(win);
    for (let i = k;