<?xml version="1.0" encoding="utf-8" standalone="yes"?><rss version="2.0" xmlns:atom="http://www.w3.org/2005/Atom"><channel><title>Knapsack-Dp on</title><link>https://h30306.github.io/blog/zh-tw/tags/knapsack-dp/</link><description>Recent content in Knapsack-Dp on</description><generator>Hugo -- gohugo.io</generator><language>zh-tw</language><managingEditor>hwork0511@gmail.com (Wei Yi Chung)</managingEditor><webMaster>hwork0511@gmail.com (Wei Yi Chung)</webMaster><copyright>© 2026 Wei Yi Chung</copyright><lastBuildDate>Tue, 13 Oct 2026 00:00:00 +0000</lastBuildDate><atom:link href="https://h30306.github.io/blog/zh-tw/tags/knapsack-dp/index.xml" rel="self" type="application/rss+xml"/><item><title>LeetCode 4081: Prime Subset Selection II</title><link>https://h30306.github.io/blog/zh-tw/leetcode/leetcode-4081/</link><pubDate>Tue, 13 Oct 2026 00:00:00 +0000</pubDate><author>hwork0511@gmail.com (Wei Yi Chung)</author><guid>https://h30306.github.io/blog/zh-tw/leetcode/leetcode-4081/</guid><description>LeetCode 4081 解題筆記：篩出質數後做 0/1 背包，dp[j] 是湊出 j 的最少個數；質數由大到小放，個數一樣時新的組合開頭比較小，所以用 &amp;lt;= 更新就是字典序最小</description><media:content xmlns:media="http://search.yahoo.com/mrss/" url="https://h30306.github.io/blog/leetcode/leetcode-4081/featured.png"/></item></channel></rss>