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HDU 4923 Room and Moor
Problem Description PM Room defines a sequence A = {A1, A2,..., AN}, each of which is either 0 or 1. In order to beat him, programmer Moor has to construct another sequence B = {B1, B2,... , BN} of...
分类:其他好文   时间:2014-08-07 23:18:54    阅读次数:308
How To mount/Browse Windows Shares【在linux{centos}上挂载、浏览window共享】
How to mount remote Windows sharesContentsRequired packagesBasic methodBetter MethodEven-better methodYet Another Even-better methodOK, we live in the...
分类:Windows程序   时间:2014-08-07 22:05:17    阅读次数:444
【12C考题精解】OCP 1z0-060 QUESTION 3: Managing the SYSAUX Tablespace
QUESTION3Whichtwostatementsaretrueabouttheuseoftheprocedureslistedinthev$sysaux_occupants.move_procedurecolumn?A.TheproceduremaybeusedforsomecomponentstorelocatecomponentdatatotheSYSAUXtablespacefromitscurrenttablespace.B.Theproceduremaybeusedforsomecompone..
分类:其他好文   时间:2014-08-07 19:22:41    阅读次数:295
Transposed Matrix
Transposed MatrixIn linear algebra, the transpose of a matrixAis another matrixAT(also writtenA′,Atr,tAorAt) created by any one of the following equiv...
分类:其他好文   时间:2014-08-07 18:52:40    阅读次数:242
nl2br() 函数
nl2br() 函数在字符串中的每个新行 (\n) 之前插入 HTML 换行符 ()。经常用于在线编辑的内容,以便显示。用法:nl2br(string)参数描述string必需。规定要检查的字符串。例子输出:One line.Another line.
分类:其他好文   时间:2014-08-06 18:54:17    阅读次数:272
LeetCode "Longest Valid Parentheses"
Your intuition would tell you that there's a O(n) solution. Actually it is another stack-based problem to solve.class Solution {public: struct Rec ...
分类:其他好文   时间:2014-08-06 14:37:41    阅读次数:186
LeetCode "Reverse Nodes in k-Group"
Another list manipulation problem.class Solution {public: ListNode *reverseKGroup(ListNode *head, int k) { if (!head) return head; if...
分类:其他好文   时间:2014-08-06 06:14:10    阅读次数:232
升级ubuntu中的gcc和g++版本
在利用张乐博士的最大熵模型工具包(Maximum Entropy Modeling Toolkit for Python and C++)和条件随机场的经典工具包CRF++(CRF++: Yet Another CRF toolkit)进行分词的时候,发现工具包不能正常安装,从报出的错误推测是gcc的版本较低,上述工具包发布于2011年,推测使用了较新的C++11标准。我们知道C++11标准开始支...
分类:其他好文   时间:2014-08-06 01:59:30    阅读次数:151
Light OJ Dynamic Programming
1068 - Investigation 数位dp dp[i][j][k] 到第i位每位数字之和的余数为j 当前数字余数为k 1079 - Just another Robbery 01背包 所有钱之和为背包体积 不被抓的概率为物品价值 1140 数位dp dp[i][j][k] 到第i为出现j个0出现过0以外的数(k==true)...
分类:其他好文   时间:2014-08-05 15:55:19    阅读次数:262
Lesson 41-42 How would you respond?
一 Words 1 think through 充分考虑 schoolwork 功课,学业 2 only to 结果却 eg. Or we will solve one problem only to have to deal with another instead. 3 tutor 导师,家庭教...
分类:其他好文   时间:2014-08-05 09:24:48    阅读次数:277
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