Guan Hua leftrk
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| // ==UserScript== | |
| // @name liaoxuefeng.com 左右方向键翻页 | |
| // @description liaoxuefeng.com 教程 增加翻页功能 | |
| // @namespace Big Scripts | |
| // @match *://www.liaoxuefeng.com/wiki/* | |
| // @grant none | |
| // @require https://code.jquery.com/jquery-3.5.1.min.js | |
| // @version 0.0.1 | |
| // ==/UserScript== |
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| def downloadFILE(url, name): | |
| resp = requests.get(url=url, stream=True) | |
| content_size = int(resp.headers['Content-Length']) / 1024 | |
| with open(name, "wb") as f: | |
| print("File total size is:", content_size, 'k,start...') | |
| for data in tqdm(iterable=resp.iter_content(1024), total=content_size + 1, unit='k', desc=name): | |
| f.write(data) | |
| print(name + " download finished!") |
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| /* | |
| * 1. right = nums.size() | |
| * 2. left < right | |
| * 3. right = mid | |
| * 4. return right | |
| */ | |
| int find(vector<int> &nums, int target) { | |
| int left = 0, right = nums.size(); |
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| #include "leetcode.h" | |
| class Solution { | |
| public: | |
| // 关键点：把 nums 塞入 value 中 | |
| void build_segment_tree(vector<int> &value, // 线段数组 value， 存储区间 sum | |
| vector<int> &nums, // 原始数组 nums | |
| int pos, // 当前线段【节点】在线段数组 value 中的下标 | |
| int left, // 当前线段的左端点 |
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| #define CATCH_CONFIG_MAIN | |
| //#include "catch.hpp" | |
| /** | |
| * function bucket-sort(array, n) is | |
| * bucket <- new array of n empty lists | |
| * | |
| * for i = 0 to (length(array)-1) do | |
| * insert array[i] into buckets[msbits(array[i], )] |
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| template<typename Object> | |
| struct Node { | |
| Object data; | |
| Node *prev; | |
| Node *next; | |
| Node(const Object &d = Object{}, Node *p = nullptr, Node *n = nullptr) | |
| : data{d}, prev{p}, next{n} {} | |
| Node(Object &&d, Node *p = nullptr, Node *n = nullptr) |
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| template<typename Object> | |
| class Vector { | |
| private: | |
| int theSize; | |
| int theCapacity; | |
| Object *objects; | |
| static const int SPARE_CAPACITY = 16; | |
| public: |
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| long long pow(long long x, int n) { | |
| if (n == 0) | |
| return 1; | |
| if (n == 1) | |
| return x; | |
| if (n & 0x1) | |
| return pow(x * x, n / 2) * x; | |
| else | |
| return pow(x * x, n / 2); | |
| } |
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| long long gcd(long long m, long long n) { | |
| while (n != 0) { | |
| long long rem = m % n; | |
| m = n; | |
| n = rem; | |
| } | |
| return m; | |
| } |
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| /** | |
| * @param a | |
| * @param left | |
| * @param right | |
| * @return maxSum between left and right | |
| * @brief 相连最大子序列和的递归算法 | |
| * @note 找出生成 [left ... right] 的子数组的最大和，不试图保留具体的最佳序列 | |
| */ | |
| int maxSumRec(const vector<int> &a, int left, int right) { | |
| if (left == right) // 基准情形 |
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