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<rss xmlns:dc="http://purl.org/dc/elements/1.1/" xmlns:atom="http://www.w3.org/2005/Atom" xmlns:media="http://search.yahoo.com/mrss/" version="2.0"><channel><title>飞行雪绒的博客</title><link>https://homeofmjr.top</link><atom:link href="https://homeofmjr.top/rss.xml" rel="self" type="application/rss+xml"/><description>飞行雪绒的博客</description><generator>Halo v2.23.2</generator><language>zh-cn</language><image><url>https://123.homeofmjr.top/blog/avatar.webp</url><title>飞行雪绒的博客</title><link>https://homeofmjr.top</link></image><lastBuildDate>Sun, 19 Apr 2026 10:26:46 GMT</lastBuildDate><item><title><![CDATA[leetcode-1979. 找出数组的最大公约数]]></title><link>https://homeofmjr.top/archives/leetcode-1979.-zhao-chu-shu-zu-de-zui-da-gong-yue-shu</link><description><![CDATA[<img src="https://homeofmjr.top/plugins/feed/assets/telemetry.gif?title=leetcode-1979.%20%E6%89%BE%E5%87%BA%E6%95%B0%E7%BB%84%E7%9A%84%E6%9C%80%E5%A4%A7%E5%85%AC%E7%BA%A6%E6%95%B0&amp;url=/archives/leetcode-1979.-zhao-chu-shu-zu-de-zui-da-gong-yue-shu" width="1" height="1" alt="" style="opacity:0;">题目链接:1979. 找出数组的最大公约数 这个题还是非常简单的,返回数组最大数和最小数的最大公约数,我们要找最大数和最小数,直接排个序不就行了. Python代码如下: import math class Solution: def findGCD(self, nums: List[int]]></description><guid isPermaLink="false">/archives/leetcode-1979.-zhao-chu-shu-zu-de-zui-da-gong-yue-shu</guid><dc:creator>Administrator</dc:creator><category>算法题</category><pubDate>Tue, 7 Apr 2026 05:35:30 GMT</pubDate></item><item><title><![CDATA[洛谷-P1029 [NOIP 2001 普及组] 最大公约数和最小公倍数问题]]></title><link>https://homeofmjr.top/archives/wei-ming-ming-wen-zhang-LyiaJVaS</link><description><![CDATA[<img src="https://homeofmjr.top/plugins/feed/assets/telemetry.gif?title=%E6%B4%9B%E8%B0%B7-P1029%20%5BNOIP%202001%20%E6%99%AE%E5%8F%8A%E7%BB%84%5D%20%E6%9C%80%E5%A4%A7%E5%85%AC%E7%BA%A6%E6%95%B0%E5%92%8C%E6%9C%80%E5%B0%8F%E5%85%AC%E5%80%8D%E6%95%B0%E9%97%AE%E9%A2%98&amp;url=/archives/wei-ming-ming-wen-zhang-LyiaJVaS" width="1" height="1" alt="" style="opacity:0;">题目链接:P1029 [NOIP 2001 普及组] 最大公约数和最小公倍数问题 这个题可以先读题,先输入x0和y0,然后需要求P,Q,这两个数的最大公约数是x0,最大公倍数是y0.求有多少个这样的P,Q 我们简单的想到穷举.因为x0是两个数字的公约数,那这个数一定能被x0整除,所以我们从x0穷举,]]></description><guid isPermaLink="false">/archives/wei-ming-ming-wen-zhang-LyiaJVaS</guid><dc:creator>Administrator</dc:creator><category>算法题</category><pubDate>Tue, 7 Apr 2026 05:20:58 GMT</pubDate></item><item><title><![CDATA[洛谷-B3634 最大公约数和最小公倍数]]></title><link>https://homeofmjr.top/archives/luo-gu-b3634-zui-da-gong-yue-shu-he-zui-xiao-gong-bei-shu</link><description><![CDATA[<img src="https://homeofmjr.top/plugins/feed/assets/telemetry.gif?title=%E6%B4%9B%E8%B0%B7-B3634%20%E6%9C%80%E5%A4%A7%E5%85%AC%E7%BA%A6%E6%95%B0%E5%92%8C%E6%9C%80%E5%B0%8F%E5%85%AC%E5%80%8D%E6%95%B0&amp;url=/archives/luo-gu-b3634-zui-da-gong-yue-shu-he-zui-xiao-gong-bei-shu" width="1" height="1" alt="" style="opacity:0;">题目链接:B3634 最大公约数和最小公倍数 求最大公约数需要用辗转相除法.例如找24和16的最大公约数 16%24=0......16 24%16=1......8 16%8=2......0 8%0=? 就是拿一个数当除数,一个数当被除数,谁在前都无所谓,如果前面的小,第二轮也会替换位置的,然后]]></description><guid isPermaLink="false">/archives/luo-gu-b3634-zui-da-gong-yue-shu-he-zui-xiao-gong-bei-shu</guid><dc:creator>Administrator</dc:creator><category>算法题</category><pubDate>Mon, 6 Apr 2026 12:46:17 GMT</pubDate></item><item><title><![CDATA[洛谷-P10424 [蓝桥杯 2024 省 B] 好数]]></title><link>https://homeofmjr.top/archives/luo-gu-p10424-lan-qiao-bei-2024-sheng-b-hao-shu</link><description><![CDATA[<img src="https://homeofmjr.top/plugins/feed/assets/telemetry.gif?title=%E6%B4%9B%E8%B0%B7-P10424%20%5B%E8%93%9D%E6%A1%A5%E6%9D%AF%202024%20%E7%9C%81%20B%5D%20%E5%A5%BD%E6%95%B0&amp;url=/archives/luo-gu-p10424-lan-qiao-bei-2024-sheng-b-hao-shu" width="1" height="1" alt="" style="opacity:0;">题目链接:P10424 [蓝桥杯 2024 省 B] 好数 这个题求1-N之间的好数,也就是一个数字从低位到高位,奇数位是奇数,偶数位是偶数,就是一个好数. 直接for循环从1遍历到n.然后分别判断每个数是不是好数即可. cpp代码如下: #include&lt;iostream&gt; using names]]></description><guid isPermaLink="false">/archives/luo-gu-p10424-lan-qiao-bei-2024-sheng-b-hao-shu</guid><dc:creator>Administrator</dc:creator><category>算法题</category><pubDate>Mon, 6 Apr 2026 05:29:17 GMT</pubDate></item><item><title><![CDATA[洛谷-P8716 [蓝桥杯 2020 省 AB2] 回文日期]]></title><link>https://homeofmjr.top/archives/luo-gu-p8716-lan-qiao-bei-2020-sheng-ab2-hui-wen-ri-qi</link><description><![CDATA[<img src="https://homeofmjr.top/plugins/feed/assets/telemetry.gif?title=%E6%B4%9B%E8%B0%B7-P8716%20%5B%E8%93%9D%E6%A1%A5%E6%9D%AF%202020%20%E7%9C%81%20AB2%5D%20%E5%9B%9E%E6%96%87%E6%97%A5%E6%9C%9F&amp;url=/archives/luo-gu-p8716-lan-qiao-bei-2020-sheng-ab2-hui-wen-ri-qi" width="1" height="1" alt="" style="opacity:0;">题目链接:P8716 [蓝桥杯 2020 省 AB2] 回文日期 这个题我一开始的想法是直接从输入的数字遍历到99999999,然后判断每一个数是不是回文数.但是很明显超时了.虽然是O(n)复杂度,但是几千万次循环也是很多的. 然后我们可以想,对于一个确定的年份,比如2012,那它对应的回文日期肯定]]></description><guid isPermaLink="false">/archives/luo-gu-p8716-lan-qiao-bei-2020-sheng-ab2-hui-wen-ri-qi</guid><dc:creator>Administrator</dc:creator><category>算法题</category><pubDate>Sun, 5 Apr 2026 12:55:38 GMT</pubDate></item><item><title><![CDATA[leetcode-2226. 每个小孩最多能分到多少糖果]]></title><link>https://homeofmjr.top/archives/leetcode-2226.-mei-ge-xiao-hai-zui-duo-neng-fen-dao-duo-shao-tang-guo</link><description><![CDATA[<img src="https://homeofmjr.top/plugins/feed/assets/telemetry.gif?title=leetcode-2226.%20%E6%AF%8F%E4%B8%AA%E5%B0%8F%E5%AD%A9%E6%9C%80%E5%A4%9A%E8%83%BD%E5%88%86%E5%88%B0%E5%A4%9A%E5%B0%91%E7%B3%96%E6%9E%9C&amp;url=/archives/leetcode-2226.-mei-ge-xiao-hai-zui-duo-neng-fen-dao-duo-shao-tang-guo" width="1" height="1" alt="" style="opacity:0;">题目链接: 2226. 每个小孩最多能分到多少糖果 这个题我们先看一下样例1.[5,8,6],将8和6进行拆分,变成[5,5,3,5,1]然后三个孩子分别拿走3个5,每个孩子最多能分到的糖果就是5个.然后看题意,每个孩子只能拿一堆.多余的堆是可以舍弃不要的. 然后样例2的话,糖果是[2,5],加起来]]></description><guid isPermaLink="false">/archives/leetcode-2226.-mei-ge-xiao-hai-zui-duo-neng-fen-dao-duo-shao-tang-guo</guid><dc:creator>Administrator</dc:creator><category>算法题</category><pubDate>Sat, 4 Apr 2026 06:49:17 GMT</pubDate></item><item><title><![CDATA[Dr.com校园网自动登录方案]]></title><link>https://homeofmjr.top/archives/dr.comxiao-yuan-wang-zi-dong-deng-lu-fang-an</link><description><![CDATA[<img src="https://homeofmjr.top/plugins/feed/assets/telemetry.gif?title=Dr.com%E6%A0%A1%E5%9B%AD%E7%BD%91%E8%87%AA%E5%8A%A8%E7%99%BB%E5%BD%95%E6%96%B9%E6%A1%88&amp;url=/archives/dr.comxiao-yuan-wang-zi-dong-deng-lu-fang-an" width="1" height="1" alt="" style="opacity:0;">Dr.COM校园网自动登录认证方案: 1. 连接giwifi校园网后进入这个网站http://172.25.251.2/a79.htm 2. 然后打开F12开发者模式.勾选保留日志. 3. 执行一次登录操作.在网络哪里会看到一条get请求]]></description><guid isPermaLink="false">/archives/dr.comxiao-yuan-wang-zi-dong-deng-lu-fang-an</guid><dc:creator>Administrator</dc:creator><category>自动化脚本</category><pubDate>Sat, 4 Apr 2026 03:16:12 GMT</pubDate></item><item><title><![CDATA[leetcode-2563. 统计公平数对的数目]]></title><link>https://homeofmjr.top/archives/leetcode-2563.-tong-ji-gong-ping-shu-dui-de-shu-mu</link><description><![CDATA[<img src="https://homeofmjr.top/plugins/feed/assets/telemetry.gif?title=leetcode-2563.%20%E7%BB%9F%E8%AE%A1%E5%85%AC%E5%B9%B3%E6%95%B0%E5%AF%B9%E7%9A%84%E6%95%B0%E7%9B%AE&amp;url=/archives/leetcode-2563.-tong-ji-gong-ping-shu-dui-de-shu-mu" width="1" height="1" alt="" style="opacity:0;">题目链接:2563. 统计公平数对的数目 这个题求公平对的和在[lower,upper]内.就是数字两两组合,看在不在这个区间,所以答案和数据的顺序就无关,我们可以先进行一个排序. 然后对于$lower\le x+y\le upper$,可以变形为$lower-x\le y \le upper-x$]]></description><guid isPermaLink="false">/archives/leetcode-2563.-tong-ji-gong-ping-shu-dui-de-shu-mu</guid><dc:creator>Administrator</dc:creator><category>算法题</category><pubDate>Sat, 4 Apr 2026 01:57:45 GMT</pubDate></item><item><title><![CDATA[洛谷-P2249 【深基13.例1】查找]]></title><link>https://homeofmjr.top/archives/luo-gu-p2249-shen-ji-13.li-1-cha-zhao</link><description><![CDATA[<img src="https://homeofmjr.top/plugins/feed/assets/telemetry.gif?title=%E6%B4%9B%E8%B0%B7-P2249%20%E3%80%90%E6%B7%B1%E5%9F%BA13.%E4%BE%8B1%E3%80%91%E6%9F%A5%E6%89%BE&amp;url=/archives/luo-gu-p2249-shen-ji-13.li-1-cha-zhao" width="1" height="1" alt="" style="opacity:0;">题目链接:P2249 【深基13.例1】查找 这个题目当我们看到单调不减.以及查找一个数字的时候就应该想到二分查找了.但是我们可以看到题目最后的说明,数据量是比较大的,并且也提示我们使用较快的IO方式.所以就不只是简单的二分查找,我们要通过一些手段来优化时间复杂度. Python代码的话,我们可以先]]></description><guid isPermaLink="false">/archives/luo-gu-p2249-shen-ji-13.li-1-cha-zhao</guid><dc:creator>Administrator</dc:creator><category>算法题</category><pubDate>Sat, 4 Apr 2026 01:25:04 GMT</pubDate></item><item><title><![CDATA[洛谷-P8218 【深进1.例1】求区间和]]></title><link>https://homeofmjr.top/archives/luo-gu-p8218-shen-jin-1.li-1-qiu-qu-jian-he</link><description><![CDATA[<img src="https://homeofmjr.top/plugins/feed/assets/telemetry.gif?title=%E6%B4%9B%E8%B0%B7-P8218%20%E3%80%90%E6%B7%B1%E8%BF%9B1.%E4%BE%8B1%E3%80%91%E6%B1%82%E5%8C%BA%E9%97%B4%E5%92%8C&amp;url=/archives/luo-gu-p8218-shen-jin-1.li-1-qiu-qu-jian-he" width="1" height="1" alt="" style="opacity:0;">题目链接: 求区间和 如果就按题意,我们简单的想的话,那就是这个题读取完数据,然后再读区间,确定起始索引和终止索引.然后用for循环对这个区间求和就可以了. python代码如下: import sys input = lambda: sys.stdin.readline().strip() n]]></description><guid isPermaLink="false">/archives/luo-gu-p8218-shen-jin-1.li-1-qiu-qu-jian-he</guid><dc:creator>Administrator</dc:creator><category>算法题</category><pubDate>Fri, 3 Apr 2026 09:45:56 GMT</pubDate></item><item><title><![CDATA[leetcode-1.两数之和]]></title><link>https://homeofmjr.top/archives/wei-ming-ming-wen-zhang</link><description><![CDATA[<img src="https://homeofmjr.top/plugins/feed/assets/telemetry.gif?title=leetcode-1.%E4%B8%A4%E6%95%B0%E4%B9%8B%E5%92%8C&amp;url=/archives/wei-ming-ming-wen-zhang" width="1" height="1" alt="" style="opacity:0;">题目链接: 两数之和 这个题我们可以简单的想到暴力枚举的方案,让所有的数字两两组合,如果等于目标值.就返回对应的索引. python代码如下: class Solution: def twoSum(self, nums: List[int], target: int) -&gt; List[int]:]]></description><guid isPermaLink="false">/archives/wei-ming-ming-wen-zhang</guid><dc:creator>Administrator</dc:creator><category>算法题</category><pubDate>Fri, 3 Apr 2026 08:42:23 GMT</pubDate></item></channel></rss>