<?xml version="1.0" encoding="utf-8"?><rss version="2.0" xmlns:dc="http://purl.org/dc/elements/1.1/" xmlns:sy="http://purl.org/rss/1.0/modules/syndication/"><channel><title>MagicLen</title><link>https://magiclen.org/2020/07/feed</link><description>多元化原創文章，內容包羅萬象，有程式語言、網站應用、軟體介紹、硬體介紹、作業系統、旅行遊記、專題採訪、益智問題、文學創作，以及作者們的知識分享和作品分享。</description><language>zh-TW</language><lastBuildDate>Sat, 27 Jun 2026 12:25:23 +0000</lastBuildDate><sy:updatePeriod>hourly</sy:updatePeriod><sy:updateFrequency>8</sy:updateFrequency><sy:updateBase>1970-01-01T00:00+00:00</sy:updateBase><item><title>台北大縱走 第三段和第二段：擎天崗─竹篙山─七星山─小油坑山─大屯山─清天宮</title><link>https://magiclen.org/taipei-mountaineering-section-3-2</link><description><![CDATA[    <img src="https://magiclen.org/wp-content/uploads/2020/03/P1150414-220x162.jpg" srcset="https://magiclen.org/wp-content/uploads/2020/03/P1150414-220x162.jpg 1x, https://magiclen.org/wp-content/uploads/2020/03/P1150414-440x324.jpg 2x, https://magiclen.org/wp-content/uploads/2020/03/P1150414-660x486.jpg 3x" alt="">
臺北市政府工務局大地工程處規劃了「台北大縱走」的活動，規劃了七段主要環繞台北市區、可縱走台北郊山的路程。這個活動雖然在走完全程後有送一些東西，但是我並不想要參加，只是想參考活動路程來爬山。台北大縱走的第三段位於陽明山主線步道上，就是在擎天崗附近，由於比較不會迷路、有補給而且很容易搭到公車，因此第三段(西半段甚至是全段)可以與第二段的大屯山縱走一起納入同一趟的縱走路線。台北大縱走第二段官方預估的路程...]]></description><category>徒步</category><category>七星山</category><category>冷水坑</category><category>台北</category><category>大屯山</category><category>小油坑</category><category>擎天崗</category><category>竹篙山</category><category>菁山吊橋</category><category>台北大縱走</category><category>大屯山連峰步道</category><category>清天宮</category><category>面天坪</category><category>小油坑山</category><category>竹子湖</category><category>冷擎步道</category><category>夢幻湖山</category><category>陽明溪畔步道</category><category>頂湖海芋步道</category><category>水尾巴拉卡步道</category><guid>https://magiclen.org/taipei-mountaineering-section-3-2</guid><pubDate>Thu, 30 Jul 2020 12:00:14 +0800</pubDate><dc:creator>Magic Len</dc:creator></item><item><title>用動態規劃解決問題：找零錢問題(Coin Change Problem)</title><link>https://magiclen.org/dynamic-programming-coin-change</link><description><![CDATA[    <img src="https://magiclen.org/wp-content/uploads/2020/04/technology-1283624_1920-220x162.jpg" srcset="https://magiclen.org/wp-content/uploads/2020/04/technology-1283624_1920-220x162.jpg 1x, https://magiclen.org/wp-content/uploads/2020/04/technology-1283624_1920-440x324.jpg 2x, https://magiclen.org/wp-content/uploads/2020/04/technology-1283624_1920-660x486.jpg 3x" alt="">
許多人認為身上如果帶太多的零錢會讓行動變得不方便，因此會希望商店店員在找零錢的時候能夠以最少的硬幣數來找，而不是全部都用1元塞給我們。]]></description><category>研究分享</category><category>Java</category><category>Rust</category><category>演算法</category><category>Go</category><category>JavaScript</category><category>Java</category><category>JavaScript</category><category>Rust</category><category>TypeScript</category><category>動態規劃</category><category>Go</category><guid>https://magiclen.org/dynamic-programming-coin-change</guid><pubDate>Tue, 28 Jul 2020 12:00:30 +0800</pubDate><dc:creator>Magic Len</dc:creator></item><item><title>台北大縱走 第一段：關渡─光武山─忠義山(嘎嘮別山)─向天池─向天山─面天山─二子坪</title><link>https://magiclen.org/taipei-mountaineering-section-1</link><description><![CDATA[    <img src="https://magiclen.org/wp-content/uploads/2020/03/P1120892-220x162.jpg" srcset="https://magiclen.org/wp-content/uploads/2020/03/P1120892-220x162.jpg 1x, https://magiclen.org/wp-content/uploads/2020/03/P1120892-440x324.jpg 2x, https://magiclen.org/wp-content/uploads/2020/03/P1120892-660x486.jpg 3x" alt="">
臺北市政府工務局大地工程處規劃了「台北大縱走」的活動，規劃了七段主要環繞台北市區、可縱走台北郊山的路程。這個活動雖然在走完全程後有送一些東西，但是我並不想要參加，只是想參考活動路程來爬山。台北大縱走的第一段官方的原路線只會登到一座忠義山(嘎嘮別山)，我覺得有點無趣，就小改了一下路線，多爬幾座山。原路線之官方預估的路程時間為7到8小時，沿途沒什麼美景可看。]]></description><category>徒步</category><category>台北</category><category>台北大縱走</category><category>光武山</category><category>忠義山</category><category>嘎嘮別山</category><category>向天池</category><category>向天山</category><category>面天山</category><category>二子坪</category><category>下青礐步道</category><category>面天山步道</category><category>大屯山連峰步道</category><category>貴子坑步道</category><guid>https://magiclen.org/taipei-mountaineering-section-1</guid><pubDate>Fri, 24 Jul 2020 12:00:38 +0800</pubDate><dc:creator>Magic Len</dc:creator></item><item><title>用動態規劃解決問題：基本觀念(有重疊子問題的問題)</title><link>https://magiclen.org/dynamic-programming-basic</link><description><![CDATA[    <img src="https://magiclen.org/wp-content/uploads/2020/04/technology-1283624_1920-220x162.jpg" srcset="https://magiclen.org/wp-content/uploads/2020/04/technology-1283624_1920-220x162.jpg 1x, https://magiclen.org/wp-content/uploads/2020/04/technology-1283624_1920-440x324.jpg 2x, https://magiclen.org/wp-content/uploads/2020/04/technology-1283624_1920-660x486.jpg 3x" alt="">
動態規劃(Dynamic Programming，簡稱DP)是一種解決問題的技巧，主要被用來優化那些「記不住自己過去曾解出來的答案所以只好重複再解」的演算法，讓它們可以「記憶」已經找出來的答案，從而不斷利用，以大大降低時間複雜度(從指數級降到線性)。]]></description><category>研究分享</category><category>Java</category><category>Rust</category><category>演算法</category><category>Go</category><category>JavaScript</category><category>Java</category><category>JavaScript</category><category>Rust</category><category>TypeScript</category><category>費氏數列</category><category>動態規劃</category><category>Go</category><guid>https://magiclen.org/dynamic-programming-basic</guid><pubDate>Tue, 21 Jul 2020 12:00:16 +0800</pubDate><dc:creator>Magic Len</dc:creator></item><item><title>Linux Mint 20 Cinnamon 安裝教學</title><link>https://magiclen.org/linux-mint-20-cinnamon-install</link><description><![CDATA[    <img src="https://magiclen.org/wp-content/uploads/2020/07/shot-52-220x162.png" srcset="https://magiclen.org/wp-content/uploads/2020/07/shot-52-220x162.png 1x, https://magiclen.org/wp-content/uploads/2020/07/shot-52-440x324.png 2x, https://magiclen.org/wp-content/uploads/2020/07/shot-52-660x486.png 3x" alt="">
Linux Mint是基於Ubuntu的衍生發行版，通常在Ubuntu新版本釋出後一個月左右，Linux Mint也會釋出相對應的新版本。版本號碼的話，Linux Mint 15對應到Ubuntu 13.04、Linux Mint 16對應到Ubuntu 13.10，依此類推。但在Linux Mint 17之後，大版本號是跟著Ubuntu的LTS(長期支援)版本，有5年的更新支援，Linux Mi...]]></description><category>Linux</category><category>Cinnamon</category><category>Linux</category><category>Linux Mint</category><guid>https://magiclen.org/linux-mint-20-cinnamon-install</guid><pubDate>Thu, 16 Jul 2020 12:00:17 +0800</pubDate><dc:creator>Magic Len</dc:creator></item><item><title>認識數字系統，進行任意進制的超長整數運算(加、減、乘、除)</title><link>https://magiclen.org/numeral-system-big-integer</link><description><![CDATA[    <img src="https://magiclen.org/wp-content/uploads/2020/05/big-number-220x162.jpg" srcset="https://magiclen.org/wp-content/uploads/2020/05/big-number-220x162.jpg 1x, https://magiclen.org/wp-content/uploads/2020/05/big-number-440x324.jpg 2x, https://magiclen.org/wp-content/uploads/2020/05/big-number-660x486.jpg 3x" alt="">
電腦程式在進行整數數值計算的時候，會受到變數型別的記憶體空間使用大小而有數值表示範圍的限制，所以當遇到過長的數值時，就無法正常地運算出結果。]]></description><category>研究分享</category><category>Java</category><category>Rust</category><category>數學邏輯</category><category>Go</category><category>JavaScript</category><category>Java</category><category>JavaScript</category><category>Rust</category><category>大數運算</category><category>TypeScript</category><category>Go</category><guid>https://magiclen.org/numeral-system-big-integer</guid><pubDate>Tue, 14 Jul 2020 12:00:42 +0800</pubDate><dc:creator>Magic Len</dc:creator></item><item><title>如何用程式進行質因數分解和尋找最大公因數與最小公倍數？</title><link>https://magiclen.org/prime-factorization-gcd-lcm</link><description><![CDATA[    <img src="https://magiclen.org/wp-content/uploads/2018/10/pay-819587_1920-220x162.jpg" srcset="https://magiclen.org/wp-content/uploads/2018/10/pay-819587_1920-220x162.jpg 1x, https://magiclen.org/wp-content/uploads/2018/10/pay-819587_1920-440x324.jpg 2x, https://magiclen.org/wp-content/uploads/2018/10/pay-819587_1920-660x486.jpg 3x" alt="">
若正整數a除以正整數b可以整除，則稱b為a的因數(Factor)，a為b的倍數(Multiple)，1是所有正整數最小的因數，任意正整數最大的因數就是該正整數本身。若a同時是x和y的因數，則稱a是x和y的公因數(Common Divisor)，如果a是x和y的公因數中最大的一個，則稱a是x和y的最大公因數(Greatest Common Divisor，簡稱GCD)。若a同時是x和y的倍數，則稱a...]]></description><category>研究分享</category><category>Java</category><category>Rust</category><category>數學邏輯</category><category>演算法</category><category>Go</category><category>JavaScript</category><category>Java</category><category>JavaScript</category><category>Rust</category><category>TypeScript</category><category>質因數分解</category><category>最大公因數</category><category>最小公倍數</category><category>輾轉相除法</category><category>Go</category><guid>https://magiclen.org/prime-factorization-gcd-lcm</guid><pubDate>Thu, 9 Jul 2020 12:00:35 +0800</pubDate><dc:creator>Magic Len</dc:creator></item><item><title>如何用FFmpeg合併影片和字幕？</title><link>https://magiclen.org/ffmpeg-subtitle</link><description><![CDATA[    <img src="https://magiclen.org/wp-content/uploads/2016/01/shot-06-220x162.png" srcset="https://magiclen.org/wp-content/uploads/2016/01/shot-06-220x162.png 1x, https://magiclen.org/wp-content/uploads/2016/01/shot-06-440x324.png 2x, https://magiclen.org/wp-content/uploads/2016/01/shot-06-660x486.png 3x" alt="">
FFmpeg全名是Fast Forward MPEG(Moving Picture Experts Group)，為開源的影音多媒體處理框架，可以進行影音的解碼、編碼、編碼轉換、混合、抽取、串流和濾鏡，無論影音格式是從哪個地方出來的，從過去到現在的影音格式它幾乎都能夠支援。當然，我們也可以用它來合併影片與字幕。]]></description><category>研究分享</category><category>軟體介紹</category><category>FFmpeg</category><category>字幕</category><category>SRT</category><category>ASS</category><guid>https://magiclen.org/ffmpeg-subtitle</guid><pubDate>Tue, 7 Jul 2020 12:00:52 +0800</pubDate><dc:creator>Magic Len</dc:creator></item><item><title>如何使用Rust程式語言處理寬度(位元組數量)浮動的UTF-8字串？</title><link>https://magiclen.org/utf8-width</link><description><![CDATA[    <img src="https://magiclen.org/wp-content/uploads/2018/04/rust-220x162.png" srcset="https://magiclen.org/wp-content/uploads/2018/04/rust-220x162.png 1x, https://magiclen.org/wp-content/uploads/2018/04/rust-440x324.png 2x, https://magiclen.org/wp-content/uploads/2018/04/rust-660x486.png 3x" alt="">
Rust程式語言的字串是使用UTF-8編碼的，這種編碼方式讓每個字元有四種不同的寬度，例如「4」會被編碼成「[52]」；「ß」會被編碼成「[195, 159]」；「中」會被編碼成「[228, 184, 173]」；「𩸽」(ㄌㄨㄥˇ)會被編碼成「[240, 169, 184, 189]」。再加上Rust是標榜安全的程式語言，因此在切割字串的時候會去做索引位置的檢查，避免取到不正確的資料範圍，不過這也...]]></description><category>研究分享</category><category>Rust</category><category>Rust</category><category>utf8-width</category><guid>https://magiclen.org/utf8-width</guid><pubDate>Fri, 3 Jul 2020 06:00:47 +0800</pubDate><dc:creator>Magic Len</dc:creator></item></channel></rss>