operations

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【机器学习部署】Machine Learning Operations(MLOps) --1(利用fastapi部署yolov3模型)

建立一个文件夹用来返回图片预测的结果import osdir_name = "images_uploaded"if not os.path.exists(dir_name): os.mkdir(dir_name)接下来实现部署模型代码import ioimport uvicornimport numpy as npimport nest_asynciofrom enum import Enumfrom fastapi import FastAPI, U

ABC261

Intersection Tournament Result NewFolder(1) Flipping and Bonus Many Operations Sorting Color Balls Replace Game on Graph

Many Operations

Problem Statement We have a variable $X$ and $N$ kinds of operations that change the value of $X$. Operation $i$ is represented as a pair of integers $(T_i,A_i)$, and is the following operation: if $T_i=1$, it replaces the value o

InnoDB handles SELECT COUNT(*) and SELECT COUNT(1) operations in the same way. There is no performan

MySQL :: MySQL 8.0 Reference Manual :: 12.20.1 Aggregate Function Descriptions https://dev.mysql.com/doc/refman/8.0/en/aggregate-functions.html#function_count COUNT(expr) [over_clause] Returns a count of the number of non-NULL values of expr in the row

CF1708A Difference Operations

cf题链 luogu题链这个A题相对较难分析。 Description 目标结果：经过若干次 $a_i=a_i-a_{i-1},i\in\left[2,n\right]\cap\mathbb{Z}$ 的差分操作，使 $\forall a_i=0,i\in\left[2,n\right]\cap\mathbb{Z}$。答案：返回能否达成目标结果。 Analysis 我们分数列中的各个数分析。

CF1708A Difference Operations 题解

这道题只要想到了思路就很简单啦！具体思路就是：让 $i$ 从 $2$ 一直枚举到 $n$，假如所有的 $a_i$ 能整除 $a_1$，就输出 yes，否则输出 no。思路清晰了，代码就非常简单写了。代码 // Author: CrazyWolf #include <bits/stdc++.h> using namespace std; const int maxn = 1e2

题解 P8271 [USACO22OPEN] COW Operations S

面对这种题目要大胆赋值。考虑序列肯定可以不停操作，最后剩下可能是：空、C、O、W。我们把空赋值为 $0$，COW 三个字母分别是 $1,2,3$。操作 $1$ 对应的是两个相等的数运算后得 $0$。操作 $2$ 则是两个不相等的数运算后得到第三个数。还有一点是 $0$ 和任何数运算得 \(0

Leetcode2011. 执行操作后的变量值(可以,一次过)

存在一种仅支持 4 种操作和 1 个变量 X 的编程语言： ++X 和 X++ 使变量 X 的值加 1 --X 和 X-- 使变量 X 的值减 1 最初，X 的值是 0 给你一个字符串数组 operations ，这是由操作组成的一个列表，返回执行所有操作后， X 的最终值。示例 1：输入：operations = ["--X","X++","X++"]

spring-data-redis 事务操作

方法一： @Autowired private RedisTemplate redisTemplate; /** * 事务方式一 */ @Test void testTransaction(){ //开启事务支持 //记得开启事务支持，但是不知道为何，我已经开启了，结果还是出现“ ERR EXEC without MULTI”,找不到原

AtCoder Beginner Contest 253 F - Operations on a Matrix // 树状数组

题目传送门：F - Operations on a Matrix (atcoder.jp) 题意：给一个N*M大小的零矩阵，以及Q次操作。操作1（l，r，x）：对于 [l，r] 区间内的每列都加上x；操作2（i，x）：对于第 i 行，替换为x；操作3（i，j）：查询矩阵第 i 行，第 j 列元素的值。 N、M、Q大小均为2E5. 思路：树状数组首先考虑没有操作2的情况，那么很

chaos

Chaos is the science of surprises, of the nonlinear and the unpredictable. It teaches us to expect the unexpected. While most traditional science deals with supposedly predictable phenomena like gravity, electricity, or chemical reactions, Chaos Theory de

swagger提示No operations defined in spec!的解决方案

swagger提示No operations defined in spec!的解决方案: 检查swagger的配置文件（我这里用的是java配置文件SwaggerConfig.java)： Docket docket = new Docket(DocumentationType.SWAGGER_2)//指定文档风格 .apiInfo(apiInfo) .s

【CF339D Xenia and Bit Operations】题解

题目链接题目 Xenia the beginner programmer has a sequence $ a $ , consisting of $ 2^{n} $ non-negative integers: $ a_{1},a_{2},...,a_{2^{n}} $ . Xenia is currently studying bit operations. To better understand how they work, Xenia decided to calculate some v

LeetCode 0075 Sort Colors

原题传送门 1. 题目描述 2. Solution 1 1、思路分析类似快排思想，把所有的0放到数组头部，把所有2放到数组尾部，这样1全部留在中间。 2、代码实现 package Q0099.Q0075SortColors; public class Solution { /* The idea is to sweep all 0s to the left and all 2s to t

codeforce C. Boboniu and Bit Operations

#include<stdio.h> #include<math.h> #include<string.h> #include<ctype.h> #include<iostream> #include<algorithm> #include<queue> #include<bitset> typedef long long ll; using namespace std; int a,b; int aa[2

AtCoder Beginner Contest 248 F - Ignore Operations // 贪心 + 大根堆

传送门：F - Keep Connect (atcoder.jp) 题意：给定长度为N的操作（ti，yi）。给定初值为0的x，对其进行操作：当t为1时，将x替换为y；当t为2时，将x加上y。最多可以跳过k步，求最终x的最大值。思路：注意到，当t为1时，进行替换操作，那么该位置前面的操作是不会对后面产生任何影响的，也就不会消耗k。

【分块构造】1654E - Arithmetic Operations

好久没有单独为一道题出题解，但是这道题得确有意义单出一篇题解来做为笔记记录 E. Arithmetic Operations time limit per test 5 seconds memory limit per test 1024 megabytes input standard input output standard output You are given an array of integers

E. Arithmetic Operations（根号分治，dp）

E. Arithmetic Operations Tag dp 根号分治 rating2300 题目来源 Codeforces Round #778 (Div. 1 + Div. 2, based on Technocup 2022 Final Round) 题目大意给定一个数列a，每次操作可以任意变动a的其中一个数，求最少的操作次数将a变成等差数列解题思路我们与其去计算最小的变

OAM

OAM abbr. 操作与维护(Operations and Maintenance); 操作、管理与维护(Operations, Administration and Maintenance); 操作员援助菜单(Operator Assistance Menu); [例句]Security management is an important function of OAM system. 其中，安全管理是O

【Python百日进阶-数据分析】Day107 - Pandas V1.4.0 用户手册(七)索引对象

文章目录六、Index objects 索引对象 6.1 Index 索引 6.1.1 Properties 属性 6.1.2 Modifying and computations 修改和计算 6.1.3 Compatibility with MultiIndex 与 MultiIndex 的兼容性 6.1.4 Missing values 缺失值 6.1.5 Conversion 转换 6.1.6 Sorting 排序 6.

Appendix D: Graphics matrix operations

Great Microprocessors of the Past and Present (V 13.4.0) (cpushack.com) 3-D points are generally stored in four element vectors, defined as:[X, Y, Z, W]...where X, Y, and Z are the point 3-D coordinates, and W is the 'weight', and is used to nor

Codeforces 1638 E. Colorful Operations —— 线段树+暴力，有丶东西

This way 题意：给你一个长度为n的数组a，一开始所有位置的颜色都为1，值都为0.每次有三种操作： Color l r x:把[l,r]中的所有位置颜色变成x Add c:把颜色为c的所有位置的值加上c Query i:问i位置的值是多少。题解：这个感觉还是有点绕的，我也是想了比较长的一段时间。最后在通过某

DB as a Service complete with Day 2 Operations

OOTB Shared DB as a Service complete with Day 2 Operations Published: August 21, 2017 Automation Cloud-Services How-to vRealize-Automation vRealize-Suite Purpose: The first part of the post dealt with "Out of the box Shared DB as a Service". In

文件operations

在RTDM中 rtdm_open static inline int rtdm_open(const char *path, int oflag, ...) { return __rtdm_dev_open(path, oflag); } __rtdm_dev_open() int __rtdm_dev_open(const char *path, int oflag) { struct rtdm_dev_context *context; struct rtdm_device *dev;

Power Automate and its connector for Microsoft Dynamics 365 Finance and Operations apps

In this third and final post about Power Automate and its connector for Microsoft Dynamics 365 Finance and Operations apps, we are going to talk about the different write operations that exist in it. These operations are the following: Get record Update