<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[HomePage的专栏]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/designer_</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; designer_]]></copyright><item><title><![CDATA[make: g++：命令未找到]]></title><link>https://blog.youkuaiyun.com/designer_/article/details/6080433</link><guid>https://blog.youkuaiyun.com/designer_/article/details/6080433</guid><author>designer_</author><pubDate>Thu, 16 Dec 2010 16:28:00 +0800</pubDate><description><![CDATA[<br />装好Ubuntu后，第一次编译时会出现错误：<br />make: g++：命令未找到
make:离开目录“/home/administrator/QtProject/HRD-build-desktop”
make: *** [main.o] 错误 127
The process "/usr/bin/make" exited with code %2.
Error while building project HRD (target: Desktop)
When executing b]]></description><category></category></item><item><title><![CDATA[在Qt里添加图片资源]]></title><link>https://blog.youkuaiyun.com/designer_/article/details/6080403</link><guid>https://blog.youkuaiyun.com/designer_/article/details/6080403</guid><author>designer_</author><pubDate>Thu, 16 Dec 2010 16:18:00 +0800</pubDate><description><![CDATA[在工程目录下创建一个文件夹（例如images），把图片放在里面；右击位于窗口左上角的工程名，选择 “Add New...”，在弹出的对话框中依次选择  “Qt” -> “Qt Resource file”，填上名字，                            选择好路径（*/*/*/images） ，接下来确定就行了；点击主窗口位于下方  “Add”  Button，选择 “Add Prefix”， 再点击 “Add”  Button，选择 “Add Files”， 进行选择图片。 <br />]]></description><category></category></item><item><title><![CDATA[恢复Unbuntu的启动项]]></title><link>https://blog.youkuaiyun.com/designer_/article/details/6078473</link><guid>https://blog.youkuaiyun.com/designer_/article/details/6078473</guid><author>designer_</author><pubDate>Wed, 15 Dec 2010 20:04:00 +0800</pubDate><description><![CDATA[<br />windows下用wubi安装的unbuntu系统，在windows系统还原后会丢失ubuntu的启动项，有一个解决办法，就是把C盘下的wubildr.mbr， wubildr ， boot.ini三个文件事先保存在其他盘里面，即使还原后丢失ubuntu的启动项，只要把三个文件拷过来就行了。<br /> <br />boot.ini 里的内容（可以用记事本打开）<br />[boot loader]
timeout=10
default=multi(0)disk(0)rdisk(0)part]]></description><category></category></item><item><title><![CDATA[添加模版到右键菜单中]]></title><link>https://blog.youkuaiyun.com/designer_/article/details/6078453</link><guid>https://blog.youkuaiyun.com/designer_/article/details/6078453</guid><author>designer_</author><pubDate>Wed, 15 Dec 2010 19:56:00 +0800</pubDate><description><![CDATA[<br />在windows下，邮件菜单中有新建Word 文档什么的，可是ubuntu下都没有，其实只要在windows下新建需要的模板（就是新建各种文件），然后拷贝到ubuntu下的模板（Templates）文件夹内，右键看效果。<br /> <br />貌似两个系统的读写方式不同（暂时先搁着）]]></description><category></category></item><item><title><![CDATA[安装Qt SDK]]></title><link>https://blog.youkuaiyun.com/designer_/article/details/6078426</link><guid>https://blog.youkuaiyun.com/designer_/article/details/6078426</guid><author>designer_</author><pubDate>Wed, 15 Dec 2010 19:45:00 +0800</pubDate><description><![CDATA[<br />从官网上下载：http://qt.nokia.com/<br /> <br />把安装文件放在administrator目录下<br />在终端输入以下两句：<br /> /*这个是从官网上看来的，就在下载页*/
/*On Linux/Unix, you need to make the file executable in order to run it. You can either do this with your desktop's file manager or, at the c]]></description><category></category></item><item><title><![CDATA[更新Ubuntu LTS版后进不了系统]]></title><link>https://blog.youkuaiyun.com/designer_/article/details/6078399</link><guid>https://blog.youkuaiyun.com/designer_/article/details/6078399</guid><author>designer_</author><pubDate>Wed, 15 Dec 2010 19:31:00 +0800</pubDate><description><![CDATA[<br />我是在XP下，用wubi安装的Ubuntu LTS版，上来就是好几百M的更新啊，Ubuntu下载的速度不知道为什么这么慢，总在6MB/S那边，真叫我纠结，从晚上0点下到上午快10点，后来一看停住了，跳出个对话框，说是更新Grub2还是什么，一堆的英文，用谷歌翻译了下，还是不能破解其中奥秘，于是能打勾的打勾，等中午吃好饭回来一重启，懵了。<br /> <br />error： no such device： ~！@#￥%……      <br />grub rescue><br /> <br />]]></description><category></category></item><item><title><![CDATA[安装Ubuntu]]></title><link>https://blog.youkuaiyun.com/designer_/article/details/6074203</link><guid>https://blog.youkuaiyun.com/designer_/article/details/6074203</guid><author>designer_</author><pubDate>Mon, 13 Dec 2010 22:02:00 +0800</pubDate><description><![CDATA[        一个多月前安装了桌面版的最新版Ubuntu(中文版), 由于一些原因, 今天重新安装 Ubuntu, 选择的是桌面版的LTS版Ubuntu( English US). 我用的是Wubi安装, 操作比较简单. 安装程序(从官网取得,http://www.ubuntu.com/ )是一个ios 文件, youshijianjiezhexie, haizaizhuangzhongwenshurufa]]></description><category></category></item><item><title><![CDATA[位操作应用——异或]]></title><link>https://blog.youkuaiyun.com/designer_/article/details/6008237</link><guid>https://blog.youkuaiyun.com/designer_/article/details/6008237</guid><author>designer_</author><pubDate>Sun, 14 Nov 2010 13:47:00 +0800</pubDate><description><![CDATA[<br />http://acm.zju.edu.cn/onlinejudge/showContestProblem.do?problemId=4095<br /> Sample Input2
aabcdef
bzyxwvu
bzyxwvu
4
aqwerty
eas fgh
aqwerty
easdfgh
easdfgh
aqwerty
aqwerty
2
0x0abcd
0ABCDEF
0x0abcd
Sample Outputaabcdef
eas fgh
0ABC]]></description><category></category></item><item><title><![CDATA[三分]]></title><link>https://blog.youkuaiyun.com/designer_/article/details/6008132</link><guid>https://blog.youkuaiyun.com/designer_/article/details/6008132</guid><author>designer_</author><pubDate>Sun, 14 Nov 2010 12:12:00 +0800</pubDate><description><![CDATA[拿这题来说事：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3421 ZOJ Problem Set - 3421Error CurvesTime Limit: 2 Seconds      Memory Limit: 65536 KB 
Josephina is a clever girl and addicted to Machine Learning recently. She pays much attention to ]]></description><category></category></item><item><title><![CDATA[位图去空白]]></title><link>https://blog.youkuaiyun.com/designer_/article/details/5863601</link><guid>https://blog.youkuaiyun.com/designer_/article/details/5863601</guid><author>designer_</author><pubDate>Sat, 04 Sep 2010 18:42:00 +0800</pubDate><description><![CDATA[<br />显示位图并去掉位图周围空白区域：<br />需要调用函数：TransparentBlt<br />具体参见：http://baike.baidu.com/view/1080409.htm?fr=ala0_1<br /> <br />　　函数功能：该函数对指定的源设备环境中的矩形区域像素的颜色数据进行位块（bit_block）转换，并将结果置于目标设备环境。 <br />　　函数原型：BOOL TransparentBlt(HDC hdcDest, int nXOriginDest, int nY]]></description><category></category></item><item><title><![CDATA[POJ 3292]]></title><link>https://blog.youkuaiyun.com/designer_/article/details/5832492</link><guid>https://blog.youkuaiyun.com/designer_/article/details/5832492</guid><author>designer_</author><pubDate>Mon, 23 Aug 2010 17:13:00 +0800</pubDate><description><![CDATA[<br />http://acm.pku.edu.cn/JudgeOnline/problem?id=3292<br /> <br />题目意思：在1，5，9，13，17。。。4*n+1中，<br />H-primes：如5，9，13，因素只有1和其本身。<br />H-composites：在4*n+1中，除1和H-primes以外的数（n≥0）<br />H-semi-prime：恰好只由两个H-primes组成，可以存在多对符合要求的H-primes组合，例如：441=21*21=9*49；<br /]]></description><category></category></item><item><title><![CDATA[过桥问题]]></title><link>https://blog.youkuaiyun.com/designer_/article/details/5786649</link><guid>https://blog.youkuaiyun.com/designer_/article/details/5786649</guid><author>designer_</author><pubDate>Tue, 03 Aug 2010 22:49:00 +0800</pubDate><description><![CDATA[<br />背景：在一个伸手不见五指的黑夜里，有一群人过一独木桥，只有一盏灯，每次只能过两人，于是没过去两个必须回来一个人。<br /> <br />分析：<br />N：总人数，按所花时间从小到大排序：A<B<......<Y<Z，分别对应所用时间a，b......y,z。<br /> <br />分情况讨论：<br />假设数据已经按以上规则排好序。<br /> <br />N=1时，taketime=a；<br /> <br />        A  ->  a;<br /> <br />N=2时，t]]></description><category></category></item><item><title><![CDATA[Dominos 2(DFS)(容器)]]></title><link>https://blog.youkuaiyun.com/designer_/article/details/5662232</link><guid>https://blog.youkuaiyun.com/designer_/article/details/5662232</guid><author>designer_</author><pubDate>Thu, 10 Jun 2010 20:10:00 +0800</pubDate><description><![CDATA[http://acm.hdu.edu.cn/showproblem.php?pid=2754大致意思：有一堆骨牌，码好（当然不只是单线程，最初考虑单线程的，一直wa）给你 n， m， l。n表示n张骨牌（从1-n标记），m表示 （x，y）对数，x倒了，y必倒，l 表示用手推倒的牌数，对应相应骨牌的标记号码。计算有几张骨牌倒了。第一次开了10001*10001的数组来存，明显超内存了；第二次用结构体来存（x，y）；
218MS 324K 793 B C++ 
46MS540K733 BC++#includ]]></description><category></category></item><item><title><![CDATA[Monthly Expense（二分）]]></title><link>https://blog.youkuaiyun.com/designer_/article/details/5632025</link><guid>https://blog.youkuaiyun.com/designer_/article/details/5632025</guid><author>designer_</author><pubDate>Fri, 28 May 2010 23:43:00 +0800</pubDate><description><![CDATA[http://acm.pku.edu.cn/JudgeOnline/problem?id=3273  把N天分为M段连续区域，求M段区域中的最小最大值。 其实自己还没怎么完全理解二分的实现，下面的代码是看别人写的。惭愧。  #includeint main(){	int N,M,max,min,mid,i,k,sum,money[100001];	w]]></description><category></category></item><item><title><![CDATA[Fibonacci 矩阵乘法]]></title><link>https://blog.youkuaiyun.com/designer_/article/details/5612371</link><guid>https://blog.youkuaiyun.com/designer_/article/details/5612371</guid><author>designer_</author><pubDate>Thu, 20 May 2010 21:34:00 +0800</pubDate><description><![CDATA[斐波那切数列.F(0) = 0, F(1) = 1, F(n) = F(n-1) + F(n-2) (n>=2) 每行输入一个数n和m（n 每行输出F(n)%m.  第 n 个数为：┌          ┐  ^n  ┌  ┐ |   0 1  |            0 |   1 1  |            1└          ┘         └ ]]></description><category></category></item><item><title><![CDATA[进制题]]></title><link>https://blog.youkuaiyun.com/designer_/article/details/5602356</link><guid>https://blog.youkuaiyun.com/designer_/article/details/5602356</guid><author>designer_</author><pubDate>Mon, 17 May 2010 23:37:00 +0800</pubDate><description><![CDATA[http://acm.pku.edu.cn/JudgeOnline/problem?id=3191其实这题是看别人的分析做的；那人很强悍，把此题看作 -2 进制来做；核心代码：while(num != 1){	if(abs(num)%2)	{		b[i++]=1;		num=(num-1)/-2;	}	else	{		b[i++]=0;]]></description><category></category></item><item><title><![CDATA[Treats for the Cows（dp）]]></title><link>https://blog.youkuaiyun.com/designer_/article/details/5602340</link><guid>https://blog.youkuaiyun.com/designer_/article/details/5602340</guid><author>designer_</author><pubDate>Mon, 17 May 2010 23:30:00 +0800</pubDate><description><![CDATA[http://acm.pku.edu.cn/JudgeOnline/problem?id=3186 有一些食物，放在一个两端开口的仓库里，每天只能从两端选择一端取出一件食物，并且食物的价值是随着天数逐天递增，第i天的价值 本来价值*i，求n天取出食物，使之取得最大价值；  状态转移方程：dp[i][j]=max(dp[i+1][j]+t[i]*(n-j+i), dp[i][]]></description><category></category></item><item><title><![CDATA[用strlen（）时的粗心]]></title><link>https://blog.youkuaiyun.com/designer_/article/details/5602315</link><guid>https://blog.youkuaiyun.com/designer_/article/details/5602315</guid><author>designer_</author><pubDate>Mon, 17 May 2010 23:23:00 +0800</pubDate><description><![CDATA[http://acm.pku.edu.cn/JudgeOnline/problem?id=3193 strlen（）放在for（）里面，反复进行多余的调用，导致超时 正确代码：while(m--){		gets(str);		len=strlen(str);		for(i=0; i<n; i++)		{			if(leng[i] < len)	conti]]></description><category></category></item><item><title><![CDATA[射线与圆的位置关系]]></title><link>https://blog.youkuaiyun.com/designer_/article/details/5598199</link><guid>https://blog.youkuaiyun.com/designer_/article/details/5598199</guid><author>designer_</author><pubDate>Sun, 16 May 2010 20:21:00 +0800</pubDate><description><![CDATA[http://acm.nit.net.cn/showproblem.jsp?pid=1553 设圆心（a，b），r起点（ox，oy），增量（dx，dy）； 先判断圆半径 和 圆心到直线的距离（dis = fabs(dy*(a-ox)+dx*(oy-b))/sqrt(dx*dx+dy*dy) - r;）关系；if(dis > 0) printf("yes/n");el]]></description><category></category></item><item><title><![CDATA[Cow Picnic]]></title><link>https://blog.youkuaiyun.com/designer_/article/details/5596038</link><guid>https://blog.youkuaiyun.com/designer_/article/details/5596038</guid><author>designer_</author><pubDate>Sat, 15 May 2010 21:08:00 +0800</pubDate><description><![CDATA[http://acm.pku.edu.cn/JudgeOnline/problem?id=3256 题意：有k头牛，n个牧场，m条路（每条路相连两个牧场且单向），求全部牛都能到达的牧场有几个。 分析：用DFS，从每头牛所在牧场开始，走遍该牛能到的牧场，对应能到达的牧场记录来过的牛数 心得：原先我是对每个牧场作为起点，进行DFS，当然条件是有路并且对应牧场有牛。不过比较牧场数]]></description><category></category></item></channel></rss>