Time, Delays, and Deferred Work
Dealing with time involves the following tasks, in order of increasing complexity:
? Measuring time lapses and comparing times
? Knowing the cur...
分类:
其他好文 时间:
2014-08-12 00:41:13
阅读次数:
484
把denty改成allowhttpd.conf文件中。 AllowOverride None Options None Order allow,denyDeny from all -->将所有的Deny改成Allow
分类:
数据库 时间:
2014-08-11 20:47:53
阅读次数:
247
B - Symmetric Order(3.3.1)
Time Limit:1000MS Memory Limit:30000KB 64bit IO Format:%I64d
& %I64u
Submit Status Practice POJ
2013
Description
In your job at Albatross Circus ...
分类:
其他好文 时间:
2014-08-11 17:54:22
阅读次数:
202
孤狼前不久写了个存储过程,大概内容就是查询每个人的订单总额,用的是表关联,执行速度还可以。如果有哪位大神有更好的建议或者有什么疑问,欢迎能够多多指出。 有下面两张表 UserInfo表Order表 如果要查询每个人的订单总额,方法一就是用循环,但是此方法不但复杂,而且执行速度很慢,孤狼...
分类:
其他好文 时间:
2014-08-11 11:44:02
阅读次数:
218
Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.
You may assume no duplicates in the array....
分类:
其他好文 时间:
2014-08-10 18:48:50
阅读次数:
203
Consider integer numbers from 1 to n. Let us call the sum of digits of an integer number its weight. Denote the weight of the number x as w(x).
Now let us order the numbers using so called graduated ...
分类:
其他好文 时间:
2014-08-10 18:47:20
阅读次数:
406
题目:
You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Outpu...
分类:
其他好文 时间:
2014-08-10 13:08:30
阅读次数:
296
在执行update命令之后,出现系统校验和不符,网上找了一些方法,最后在大神的帮助下终于解决了!!
1.更改
/etc/apt/apt.conf.d/00aptitude 文件,在最后一行加入:Acquire::CompressionTypes::Order "gz";
然后执行 update就可以了。 (我是用这个方法解决的)
2.在软件更新管理器中点击右下角的“设置”,在其它软件中...
分类:
其他好文 时间:
2014-08-10 13:07:30
阅读次数:
219
Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).
zigzag层序遍历树
For example:
Given binary...
分类:
其他好文 时间:
2014-08-09 23:19:59
阅读次数:
363
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:
2014-08-09 21:34:09
阅读次数:
365