Verificar Stock 

SELECT * 
FROM SUCURSAL_STOCK SS
WHERE SS.cveProducto = 'VS-3245-R'
	AND SS.cveSistema = '0'
	AND SS.cveSucursal = '001'

1477. Find Two Non-overlapping Sub-arrays Each With Target Sum

You are given an array of integers arr and an integer target. You have to find two non-overlapping sub-arrays of arr each with a sum equal target. There can be multiple answers so you have to find an answer where the sum of the lengths of the two sub-arrays is minimum. Return the minimum sum of the lengths of the two required sub-arrays, or return -1 if you cannot find such two sub-arrays.
/**
 * @param {number[]} arr
 * @param {number} target
 * @return {number}
 */
var minSumOfLengths = function(arr, target) {
    const n = arr.length;
    // minLen[i] stores the minimum length of a valid sub-array ending at or before index i
    const minLen = new Array(n).fill(Infinity);
    
    let left = 0;
    let currentSum = 0;
    let minWindowLen = Infinity;
    let ans = Infinity;
    
    for (let right = 0; right < n; right++) {
        currentSum += arr[right];
        
        //

POP Pull

A better advect by volume wrangle than the default node. Replicates the "pull" style of force injection seen in the volume source node and has a minimum threshold so that particles outside the velocit yvolume bounds don't get stuck.
// Sampling
vector curr_v = v@v;
vector vol_v = volumesamplev(1, "vel", v@P);
vol_v *= chf("scale_force");

// Speeds
float curr_speed = length(curr_v);
float vol_speed = length(vol_v);

// Directional vectors
vector curr_dir = normalize(curr_v);
vector vol_dir = normalize(vol_v);

// Weights
float weight_v = 1;
float weight_v_dir = weight_v;

float weight_vol = 0;
if (curr_speed < vol_speed) weight_vol = chf("acceleration");
if (curr_speed > vol_speed) weight_vol = chf("decel

JSON CPP Example 

// const auto hwRev = readHwRevConfig(platformData, hwRevMapPath);
const std::string hwRevStr = R"(
    {
        "description": "2 drivers xxxxx | xxxxx lighting",
        "maxIspGainDay": 8192,
        "backlightDrivers": {
            "driverVer": 1,
            "selectorGpio": {
                "switch": { "lineNum": 110 }
            },
            "pwm": {
                "irGroup": { "chip": 5, "line": 0, "periodUs": 250, "dutyCycleLowLimit": 50000, "dutyCycleHighLimit": 90000 },

1621. Number of Sets of K Non-Overlapping Line Segments

Given n points on a 1-D plane, where the ith point (from 0 to n-1) is at x = i, find the number of ways we can draw exactly k non-overlapping line segments such that each segment covers two or more points. The endpoints of each segment must have integral coordinates. The k line segments do not have to cover all n points, and they are allowed to share endpoints. Return the number of ways we can draw k non-overlapping line segments. Since this number can be huge, return it modulo 109 + 7.
/**
 * @param {number} n
 * @param {number} k
 * @return {number}
 */
const MOD = 1_000_000_007;

// Fast exponentiation: computes (a^b) % MOD
function modPow(a, b) {
    let res = 1n;
    let x = BigInt(a);
    let exp = BigInt(b);

    while (exp > 0n) {
        // If lowest bit is 1, multiply result by current base
        if (exp & 1n) {
            res = (res * x) % BigInt(MOD);
        }
        // Square the base each step
        x = (x * x) % BigInt(MOD);
        exp >>= 1n; // shift ri

Cancel Cacher User's Membership 

1.  Impersonate user: 
[https://snippets.cacher.io/snippet/b61256b4da1f7d0c502e](https://snippets.cacher.io/snippet/b61256b4da1f7d0c502e)
2.  Actually cancelling the membership:
[https://snippets.cacher.io/snippet/564dd9780d6096e06cba](https://snippets.cacher.io/snippet/564dd9780d6096e06cba)

2472. Maximum Number of Non-overlapping Palindrome Substrings

You are given a string s and a positive integer k. Select a set of non-overlapping substrings from the string s that satisfy the following conditions: The length of each substring is at least k. Each substring is a palindrome. Return the maximum number of substrings in an optimal selection. A substring is a contiguous sequence of characters within a string.
/**
 * @param {string} s
 * @param {number} k
 * @return {number}
 */
var maxPalindromes = function(s, k) {
    const n = s.length;
    const intervals = [];

    const tryCenter = (l, r) => {
        while (l >= 0 && r < n && s[l] === s[r]) {
            if (r - l + 1 >= k) {
                intervals.push([l, r]);
                return; // <-- CRITICAL: stop after first valid palindrome
            }
            l--; r++;
        }
    };

    for (let i = 0; i < n; i++) {
        tryCenter(i

IELTS_5_Practice_test_Set_4_Test_18_Listening 

Watson
16
1996
Dental assistant
388
fight bars
website
10:00
induction
personal trainer
A
C
F
H
J
B
F
I
J
M
secondary visit
source
flow levels
archives 
rolle
A
B
A
C
B
fear



habitants
disturbance
dominance

gram
captivity

836. Rectangle Overlap

An axis-aligned rectangle is represented as a list [x1, y1, x2, y2], where (x1, y1) is the coordinate of its bottom-left corner, and (x2, y2) is the coordinate of its top-right corner. Its top and bottom edges are parallel to the X-axis, and its left and right edges are parallel to the Y-axis. Two rectangles overlap if the area of their intersection is positive. To be clear, two rectangles that only touch at the corner or edges do not overlap. Given two axis-aligned rectangles rec1 and rec2, return true if they overlap, otherwise return false.
/**
 * @param {number[]} rec1
 * @param {number[]} rec2
 * @return {boolean}
 */
var isRectangleOverlap = function(rec1, rec2) {
    // If one rectangle is completely to one side of the other, no overlap
    if (rec1[2] <= rec2[0] ||  // rec1 right <= rec2 left
        rec1[0] >= rec2[2] ||  // rec1 left >= rec2 right
        rec1[3] <= rec2[1] ||  // rec1 top <= rec2 bottom
        rec1[1] >= rec2[3]) {  // rec1 bottom >= rec2 top
        return false;
    }
    return true;
};

Get Stats Info

Виводить вагу БД, кількість сайтів, постів і сторінок мультисайту
SET SESSION group_concat_max_len = 1000000;

SET @db = 'gambling1';

-- Генеруємо підзапити для всіх wp*_posts таблиць Multisite
SELECT GROUP_CONCAT(
               CONCAT(
                       'SELECT ',
                       'COUNT(CASE WHEN post_type = ''post'' THEN 1 END) AS posts, ',
                       'COUNT(CASE WHEN post_type = ''page'' THEN 1 END) AS pages ',
                       'FROM `', table_schema, '`.`', table_name, '` ',
                       'WHERE post_type IN (''post

835. Image Overlap

You are given two images, img1 and img2, represented as binary, square matrices of size n x n. A binary matrix has only 0s and 1s as values. We translate one image however we choose by sliding all the 1 bits left, right, up, and/or down any number of units. We then place it on top of the other image. We can then calculate the overlap by counting the number of positions that have a 1 in both images. Note also that a translation does not include any kind of rotation. Any 1 bits that are translated outside of the matrix borders are erased. Return the largest possible overlap.
/**
 * @param {number[][]} img1
 * @param {number[][]} img2
 * @return {number}
 */
var largestOverlap = function(img1, img2) {
    const n = img1.length;
    const A = [];
    const B = [];

    // Collect coordinates of 1s
    for (let r = 0; r < n; r++) {
        for (let c = 0; c < n; c++) {
            if (img1[r][c] === 1) A.push([r, c]);
            if (img2[r][c] === 1) B.push([r, c]);
        }
    }

    const map = new Map();
    let max = 0;

    // Count translation vectors
    for

3414. Maximum Score of Non-overlapping Intervals

You are given a 2D integer array intervals, where intervals[i] = [li, ri, weighti]. Interval i starts at position li and ends at ri, and has a weight of weighti. You can choose up to 4 non-overlapping intervals. The score of the chosen intervals is defined as the total sum of their weights. Return the lexicographically smallest array of at most 4 indices from intervals with maximum score, representing your choice of non-overlapping intervals. Two intervals are said to be non-overlapping if they do not share any points. In particular, intervals sharing a left or right boundary are considered overlapping.
/**
 * @param {number[][]} intervals
 * @return {number[]}
 */
var maximumWeight = function(intervals) {
    const n = intervals.length;

    // augment with original index
    let arr = intervals.map((it, i) => ({ l: it[0], r: it[1], w: it[2], idx: i }));

    // sort by end time, then start
    arr.sort((a, b) => a.r - b.r || a.l - b.l);

    // precompute prev[i]: last j with arr[j].r < arr[i].l
    const rights = arr.map(x => x.r);
    const prev = Array(n).fill(-1);
    for (let i = 0; i <

3483. Unique 3-Digit Even Numbers

You are given an array of digits called digits. Your task is to determine the number of distinct three-digit even numbers that can be formed using these digits. Note: Each copy of a digit can only be used once per number, and there may not be leading zeros.
/**
 * @param {number[]} digits
 * @return {number}
 */
var totalNumbers = function(digits) {
    // Count how many times each digit appears in the input
    const freq = Array(10).fill(0);
    for (let d of digits) freq[d]++;

    const result = new Set(); // store distinct valid numbers

    // Loop through all 3‑digit numbers
    for (let num = 100; num <= 999; num++) {
        // Must be even
        if (num % 2 !== 0) continue;

        // Break number into digits
        const a = Math.flo

Username list 

Maximium465

2265. Count Nodes Equal to Average of Subtree

Given the root of a binary tree, return the number of nodes where the value of the node is equal to the average of the values in its subtree. Note: The average of n elements is the sum of the n elements divided by n and rounded down to the nearest integer. A subtree of root is a tree consisting of root and all of its descendants.
/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number}
 */
var averageOfSubtree = function(root) {
    let ans = 0;

    function dfs(node) {
        if (!node) return [0, 0]; // sum, count

        const [ls, lc] = dfs(node.left);
        const [rs, rc] = dfs

Show-Me 

可以。以下我建議你採用 **Windows + Codex App / Codex CLI 的全域安裝 SOP**。目標是讓所有專案都能使用 `$show-me`，並把 HumanLayer 裡偏 Claude Code/macOS 的部分改成 Codex-friendly。

Codex 官方目前會從 `$HOME/.agents/skills` 載入使用者層級 Skill；專案層則是 `<repo>/.agents/skills`。Skill 核心只需要目錄裡有 `SKILL.md`，且 frontmatter 至少包含 `name` 與 `description`。([OpenAI Developers][1])

# SOP：安裝 `humanlayer/show-me`

## 1. 先確認 Node.js

PowerShell：

```powershell
node -v
npx -v
```

目前 `skills` CLI 的套件要求 Node.js `>= 22.20.0`。([GitHub][2])

如果你的 Node 已符合