<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[Wqa556的博客]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/Wqa556</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; Wqa556]]></copyright><item><title><![CDATA[解决yolov5 报错 ValueError: not enough values to unpack (expected 3, got 0)]]></title><link>https://blog.youkuaiyun.com/Wqa556/article/details/124075986</link><guid>https://blog.youkuaiyun.com/Wqa556/article/details/124075986</guid><author>Wqa556</author><pubDate>Sun, 10 Apr 2022 12:27:38 +0800</pubDate><description><![CDATA[解决yolov5 报错 ValueError: not enough values to unpack (expected 3, got 0)]]></description><category></category></item><item><title><![CDATA[杭电多校第二场补提]]></title><link>https://blog.youkuaiyun.com/Wqa556/article/details/119358958</link><guid>https://blog.youkuaiyun.com/Wqa556/article/details/119358958</guid><author>Wqa556</author><pubDate>Tue, 03 Aug 2021 20:58:49 +0800</pubDate><description><![CDATA[hdu第二场补提
A - I love cube
数学
数学题 ， 由给的样例一个边长为1的小正方体联想到一个去分割一个大的正方体 ， 由顶点必须是整数点，则在正方体的三个主边上找相应长度的起点即可
枚举每一个长度，得到正方体种类：是从1 的 3次方 加到 n 的3次方。
13+23+33+……+n3=[n(n+1)/2]2
1^3+2^3+3^3+……+n^3=[n(n+1)/2]^2 
13+23+33+……+n3=[n(n+1)/2]2
由于数据比较大，采取分而算之的策略
#pragma GCC op]]></description><category></category></item></channel></rss>