<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[buzzaz的博客]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/buzzaz</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; buzzaz]]></copyright><item><title><![CDATA[Springboot项目用docker容器部署]]></title><link>https://blog.youkuaiyun.com/buzzaz/article/details/135130067</link><guid>https://blog.youkuaiyun.com/buzzaz/article/details/135130067</guid><author>buzzaz</author><pubDate>Fri, 22 Dec 2023 18:25:57 +0800</pubDate><description><![CDATA[整个Springboot项目都容器化，部署到云服务器。]]></description><category></category></item><item><title><![CDATA[Springboot开发遇到的问题]]></title><link>https://blog.youkuaiyun.com/buzzaz/article/details/134907087</link><guid>https://blog.youkuaiyun.com/buzzaz/article/details/134907087</guid><author>buzzaz</author><pubDate>Wed, 13 Dec 2023 18:03:04 +0800</pubDate><description><![CDATA[Springboot开发后端，前端JS及其框架。]]></description><category></category></item><item><title><![CDATA[Vue3 +Vite项目部署遇见的问题]]></title><link>https://blog.youkuaiyun.com/buzzaz/article/details/134771586</link><guid>https://blog.youkuaiyun.com/buzzaz/article/details/134771586</guid><author>buzzaz</author><pubDate>Sun, 03 Dec 2023 22:14:41 +0800</pubDate><description><![CDATA[Vue3 + Vite部署单页。]]></description><category></category></item><item><title><![CDATA[云计算：概念、技术、架构 学习]]></title><link>https://blog.youkuaiyun.com/buzzaz/article/details/134677295</link><guid>https://blog.youkuaiyun.com/buzzaz/article/details/134677295</guid><author>buzzaz</author><pubDate>Tue, 28 Nov 2023 23:42:48 +0800</pubDate><description><![CDATA[读研的研究课题是边缘计算，先学习云计算基础。云计算是分布式计算的一种特殊形式，它引入效用模型来远程供给可扩展和可测量的资源。]]></description><category></category></item><item><title><![CDATA[模拟退火+粒子群算法寻找最小值点]]></title><link>https://blog.youkuaiyun.com/buzzaz/article/details/125975603</link><guid>https://blog.youkuaiyun.com/buzzaz/article/details/125975603</guid><author>buzzaz</author><pubDate>Mon, 25 Jul 2022 16:03:55 +0800</pubDate><description><![CDATA[模拟退火+粒子群算法]]></description><category></category></item><item><title><![CDATA[POJ - 3468区间修改查询]]></title><link>https://blog.youkuaiyun.com/buzzaz/article/details/115599234</link><guid>https://blog.youkuaiyun.com/buzzaz/article/details/115599234</guid><author>buzzaz</author><pubDate>Sun, 11 Apr 2021 15:13:01 +0800</pubDate><description><![CDATA[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 numbers in a given interval.
Input
The first line c]]></description><category></category></item><item><title><![CDATA[AtCoder - abc194_c 数学公式]]></title><link>https://blog.youkuaiyun.com/buzzaz/article/details/115443986</link><guid>https://blog.youkuaiyun.com/buzzaz/article/details/115443986</guid><author>buzzaz</author><pubDate>Mon, 05 Apr 2021 16:32:29 +0800</pubDate><description><![CDATA[题目：题目
题意：求出∑i=2N∑j=1i−1(ai−aj)2\sum_{i=2}^N\sum_{j=1}^{i-1}(a_i-a_j)^2∑i=2N​∑j=1i−1​(ai​−aj​)2
思路：暴力求解是O(n2)O(n^2)O(n2)的时间复杂度，必然TLE，需要对式子进行化简。
∑i=2N∑j=1i−1(ai−aj)2\sum_{i=2}^N\sum_{j=1}^{i-1}(a_i-a_j)^2∑i=2N​∑j=1i−1​(ai​−aj​)2
=(N−1)∑i=1Nai2−∑i=2N∑j=1i−12a]]></description><category></category></item><item><title><![CDATA[AtCoder - abc194_e Mex Min]]></title><link>https://blog.youkuaiyun.com/buzzaz/article/details/115441999</link><guid>https://blog.youkuaiyun.com/buzzaz/article/details/115441999</guid><author>buzzaz</author><pubDate>Mon, 05 Apr 2021 14:50:04 +0800</pubDate><description><![CDATA[题目：AtCoder - abc194_e 
题意：我们要找到长度为 n 数组 A 的所有的长度为 m 的切片 si 最小的 mex(si) 值。
思路：
可以发现，如果记录切片中每一个数字出现的次数，找到第一个出现次数为0的就是答案。
但是我们不能每次切片改变都去遍历一次，这样会TLE，因此可以用一个数组M维护区间（i,i+m）数字出现的次数。
切片的末尾指向了 Ai 的时候，那么对应的 Ma[i]M_{a[i]}Ma[i]​就会加1；而Ma[i−m]M_{a[i-m]}Ma[i−m]​ 就会减1。
接下]]></description><category></category></item><item><title><![CDATA[AtCoder - arc114_a 质因数]]></title><link>https://blog.youkuaiyun.com/buzzaz/article/details/115308391</link><guid>https://blog.youkuaiyun.com/buzzaz/article/details/115308391</guid><author>buzzaz</author><pubDate>Mon, 29 Mar 2021 19:05:23 +0800</pubDate><description><![CDATA[AtCoder - arc114_a
题目
题意：找出一个数Y使得它与数组a内所有的数a[i]都不互质，即gcd(Y,a[i])!=1
思路：大部分题目都是要求互质的数，本题反其道而行，而且求不互质的数并不容易。
gcd(Y,a[i])!=1意味着什么？即a中每个数都与Y要有一个公因子，可以证明，公因子d必然为质数。
另外，本题的a[i]&lt;=50，因此可以暴力枚举2~50所有的质数，利用二进制枚举模拟所有质数选择的情况，必然能找到一个答案。
code：
#include &lt;bits/stdc++]]></description><category></category></item><item><title><![CDATA[CodeForces - 1473B]]></title><link>https://blog.youkuaiyun.com/buzzaz/article/details/115308195</link><guid>https://blog.youkuaiyun.com/buzzaz/article/details/115308195</guid><author>buzzaz</author><pubDate>Mon, 29 Mar 2021 18:52:23 +0800</pubDate><description><![CDATA[CodeForces - 1473B
题目
Let’s define a multiplication operation between a string a and a positive integer x: a⋅x is the string that is a result of writing x copies of a one after another. For example, “abc” ⋅ 2 = “abcabc”, “a” ⋅ 5 = “aaaaa”.
A string a is di]]></description><category></category></item><item><title><![CDATA[AtCoder - abc193_c 思维，枚举]]></title><link>https://blog.youkuaiyun.com/buzzaz/article/details/115307944</link><guid>https://blog.youkuaiyun.com/buzzaz/article/details/115307944</guid><author>buzzaz</author><pubDate>Mon, 29 Mar 2021 18:38:17 +0800</pubDate><description><![CDATA[AtCoder - abc193_c
题目
Problem Statement
Given is an integer
N
. How many integers between
1
and
N
(inclusive) are unrepresentable as
a
b
, where
a
and
b
are integers not less than
2
?
Constraints
N
is an integer.
1
≤
N
≤
10
10
Input
Input is given from Sta]]></description><category></category></item><item><title><![CDATA[POJ - 3264]]></title><link>https://blog.youkuaiyun.com/buzzaz/article/details/115273031</link><guid>https://blog.youkuaiyun.com/buzzaz/article/details/115273031</guid><author>buzzaz</author><pubDate>Sat, 27 Mar 2021 22:38:11 +0800</pubDate><description><![CDATA[POJ - 3264
For the daily milking, Farmer John’s N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer John decides to organize a game of Ultimate Frisbee with some of the cows. To keep things simple, he will take a contiguous range of co]]></description><category></category></item><item><title><![CDATA[CodeForces - 1486A 思维]]></title><link>https://blog.youkuaiyun.com/buzzaz/article/details/115272920</link><guid>https://blog.youkuaiyun.com/buzzaz/article/details/115272920</guid><author>buzzaz</author><pubDate>Sat, 27 Mar 2021 22:18:39 +0800</pubDate><description><![CDATA[CodeForces - 1486A
CodeForces - 1486A
You have n stacks of blocks. The i-th stack contains hi blocks and it’s height is the number of blocks in it. In one move you can take a block from the i-th stack (if there is at least one block) and put it to the i+1-]]></description><category></category></item><item><title><![CDATA[Div2. Find Maximum]]></title><link>https://blog.youkuaiyun.com/buzzaz/article/details/115056848</link><guid>https://blog.youkuaiyun.com/buzzaz/article/details/115056848</guid><author>buzzaz</author><pubDate>Sun, 21 Mar 2021 22:08:15 +0800</pubDate><description><![CDATA[Find Maximum （二进制）
CodeForces - 353C [GO](https://codeforces.com/problemset/problem/353/C)

Valera has array a, consisting of n integers a0, a1, …, an - 1, and function f(x), taking an integer from 0 to 2n - 1 as its single argument. Value f(x) is calculat]]></description><category></category></item></channel></rss>