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独立成分分析
Independent Components Analysis 目录Independent Components AnalysisAmbiguityDensities and Linear TransformationsICA Ambiguity ICA is ambiguous to scaling and permutation. but usually it doesn't matter. As long as the data is not Gaussian, it's pos[Oracle] LeetCode 1802 Maximum Value at a Given Index in a Bounded Array
You are given three positive integers: n, index, and maxSum. You want to construct an array nums (0-indexed) that satisfies the following conditions: nums.length == n nums[i] is a positive integer where 0 <= i < n. abs(nums[i] - nums[i+1]) <= 1[Google] LeetCode 2115 Find All Possible Recipes from Given Supplies
You have information about n different recipes. You are given a string array recipes and a 2D string array ingredients. The \(i\)-th recipe has the name recipes[i], and you can create it if you have all the needed ingredients from ingredients[i]. Ingredcartopy五个常用模块
cartopy五个常用模块 1、GeoAxes.set_extent() 对图形范围（经纬度）进行设置的函数 set_extent(self, extents, crs=None) Set the extent (x0, x1, y0, y1) of the map in the given coordinate system. If no crs is given, the extents' coordinate systLeetCode 828. Count Unique Characters of All Substrings of a Given String
原题链接在这里：https://leetcode.com/problems/count-unique-characters-of-all-substrings-of-a-given-string/ 题目： Let's define a function countUniqueChars(s) that returns the number of unique characters on s. For example, calling countUniqueChars(s) if s =CF1202C You Are Given a WASD-string...
调试题！ *2100 考虑答案由 \(\max(x),\min(x),\max(y),\min(y)\) 决定。 于是枚举一个位置，就拿 \(x\) 举例吧。 那么就要考虑 \([1,pos]\) 的累加最值，枚举当前 \(+-1\)，到当前枚举的位置的值，后缀的累加最值加上当前枚举的贡献，三者取个最值，那么就得到了。 有点难写和难调，注意 \((0,0)[LeetCode] 828. Count Unique Characters of All Substrings of a Given String
Let's define a function countUniqueChars(s) that returns the number of unique characters on s. For example, calling countUniqueChars(s) if s = "LEETCODE" then "L", "T", "C", "O", "D" are th[LeetCode]828. Count Unique Characters of All Substrings of a Given String 动态规划转移方程详解
题目描述 LeetCode原题链接：828. Count Unique Characters of All Substrings of a Given String Let's define a function countUniqueChars(s) that returns the number of unique characters on s. For example, calling countUniqueChars(s) if s = "LEETCODE" th多校联训 DS 专题
CF1039D You Are Given a Tree CF983E NN country [AGC001F] Wide Swap [AGC015E] Mr.Aoki Incubator [AGC007E] Shik and Travel CF1446D2 Frequency Problem (Hard Version) CF765F Souvenirs CF1458D Flip and ReversePHP接口报错：Unable to init from given binary data
前因： 事情是这样的，前几天不是使用Laravel做了一个图片比对的功能么，因为需要安装Composer扩展，并且这个扩展的使用，需要开启PHP的GD库的扩展支持。 所以本地以及都调试好了，于是今天就上线。然后问题就来了，上线后，请求测试方法，接口直接报 500了。 后果： 因为是线上环境，所以springboot 连接Oracle数据库报ORA-12505, TNS:listener does not currently know of SID given in connect descr
springboot 连接Oracle数据库报ORA-12505, TNS:listener does not currently know of SID given in connect descriptor 处理方式 将jdbc:oracle:thin:@127.0.0.1:1522:KFPTDB 改为 jdbc:oracle:thin:@127.0.0.1:1522/KFPTDB 两种连接方式不一样LeetCode常见题型——排序算法
1. 算法思想 【1】排序算法总结_7-SEVENS-CSDN博客 【2】算法：排序算法之堆排序_7-SEVENS-CSDN博客 【3】算法：排序算法之计数排序_7-SEVENS-CSDN博客 【4】算法：排序算法之基数排序_7-SEVENS-CSDN博客 【5】算法：排序算法之桶排序_7-SEVENS-CSDN博客_桶排序 【6】算法：排序算法之归代码测试工具
Hypothesis 初识 Hypothesis是一个Python库，用于创建单元测试，该单元测试编写起来更简单，运行时功能更强大，可以在您不需要的代码中查找极端情况。它稳定，强大且易于添加到任何现有测试套件中。 它的工作原理是让您编写断言每种情况都应该正确的测试，而不仅仅是您偶然想到的那些。英语题目翻译
题目地址： https://acs.jxnu.edu.cn/problem/NOIOPJENGLISH13 Same Remainder 同样的余数 描述： Given A and B. Find the smallest X that X is greater than 1 and A modulo X equals to B modulo X. 给出A和B。找到最小的X，X是大于1且A模X等于B模X。 输入： Two positive integers社会科学问题研究的计算实践——1、社会网络基础（聚集系数与嵌入性、友谊悖论的验证）
学习资源来自，一个哲学学生的计算机作业 (karenlyu21.github.io) 1、背景问题 “网络”由节点组成，节点之间可能有边相连。网络常常是对社会的一种有效抽象，节点代表社会中的行动者，边代表行动者之间的联系。我们可以用一个矩阵（称为邻接矩阵）来表示网络，在程序中一般就是对应一个二维数【leetcode】1646. Get Maximum in Generated Array
题目如下： You are given an integer n. A 0-indexed integer array nums of length n + 1 is generated in the following way: nums[0] = 0 nums[1] = 1 nums[2 * i] = nums[i] when 2 <= 2 * i <= n nums[2 * i + 1] = nums[i] + nums[i + 1] when 2 <= 2 * i + 1CF1039D You Are Given a Tree
更好的阅读体验 题意 给定一棵有 \(n\) 个节点的树. 对于满足 \(1\le k\le n\) 的每一个 \(k\)，把树分成若干条包含 \(k\) 个顶点的链，其中每个点最多属于一条链，问最多能分得几条链. \(n\le 10^5\) 题解 考虑 \(k\) 固定时怎么做 我们自下而上贪心，对于一个点，如果在它的子树内有一条10 Python 3 - Strings
Strings are amongst the most popular types in Python. We can create them simply by enclosing characters in quotes. Python treats single quotes the same as double quotes. Creating strings is as simple as assigning a value to a variable. For example − var1Goldbach Conjecture（翻译）
http://noi.openjudge.cn/english/11/ 描述 Given the sum of prime A and prime B, find A and B. 输入 One positive integer indicating the sum (<= 10000). 输出 Two integers A and B.英文翻译9
OpenJudge - 09:Least Common Multiple 描述 Given A and B. Find the least positive M which is a common multiple of both A and B. 输入 Two positive integers A and B (A,B <= 10000). 输出 One integer M. 翻译： 给出A和B，找到A和B的最小公倍数。 输入： 两个正整数A和B 输翻译练习 Day9
题目：Tic-tac-toe | JXNUOJ 翻译： Tic-tac-toe 1000ms 65536K 描述： Certainly, everyone is familiar with tic-tac-toe game. The rules are very simple indeed. Two players take turns marking the cells in a 3 × 3 grid (one player always draws crosses, the other — noGreatest Common Divisor（翻译）
http://noi.openjudge.cn/english/08/ 描述 Given A and B. Find the greatest D which is a common divisor of both A and B. 输入 Two positive integers A and B (A,B <= 10000). 输出 One integer D.线段树求和
P1003: A Simple Problem with Integers 题目描述 You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbe【无标题】
2022.1.21 题目网址： https://acs.jxnu.edu.cn/problem/CF3C 原题： Tic-tac-toe 1000ms 65536K 描述： Certainly, everyone is familiar with tic-tac-toe game. The rules are very simple indeed. Two players take turns marking the cells in a 3 × 3 grid (one play[LeetCode] 1282. Group the People Given the Group Size They Belong To 用户分组
There are n people that are split into some unknown number of groups. Each person is labeled with a unique ID from 0 to n - 1. You are given an integer array groupSizes, where groupSizes[i] is the size of the group that person i is in. For examp