对 HTTP 304 的理解

蓝飞 蓝飞 | 时间:2012-08-01, Wed | 4,467 views
编程算法 

304 的标准解释是:Not Modified 客户端有缓冲的文档并发出了一个条件性的请求(一般是提供If-Modified-Since头表示客户只想比指定日期更新的文档)。服务器告诉客户,原来缓冲的文档还可以继续使用。

如果客户端在请求一个文件的时候,发现自己缓存的文件有 Last Modified ,那么在请求中会包含 If Modified Since ,这个时间就是缓存文件的 Last Modified 。因此,如果请求中包含 If Modified Since,就说明已经有缓存在客户端。只要判断这个时间和当前请求的文件的修改时间就可以确定是返回 304 还是 200 。对于静态文件,例如:CSS、图片,服务器会自动完成 Last Modified 和 If Modified Since 的比较,完成缓存或者更新。但是对于动态页面,就是动态产生的页面,往往没有包含 Last Modified 信息,这样浏览器、网关等都不会做缓存,也就是在每次请求的时候都完成一个 200 的请求。

因此,对于动态页面做缓存加速,首先要在 Response 的 HTTP Header 中增加 Last Modified 定义,其次根据 Request 中的 If Modified Since 和被请求内容的更新时间来返回 200 或者 304 。虽然在返回 304 的时候已经做了一次数据库查询,但是可以避免接下来更多的数据库查询,并且没有返回页面内容而只是一个 HTTP Header,从而大大的降低带宽的消耗,对于用户的感觉也是提高。

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[ACM_HDU_3177]Crixalis's Equipment

蓝飞 蓝飞 | 时间:2012-07-25, Wed | 16,891 views
编程算法 

Crixalis's Equipment

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1350 Accepted Submission(s): 543

Description

Crixalis's EquipmentCrixalis - Sand King used to be a giant scorpion(蝎子) in the deserts of Kalimdor. Though he's a guardian of Lich King now, he keeps the living habit of a scorpion like living underground and digging holes.

Someday Crixalis decides to move to another nice place and build a new house for himself (Actually it's just a new hole). As he collected a lot of equipment, he needs to dig a hole beside his new house to store them. This hole has a volume of V units, and Crixalis has N equipment, each of them needs Ai units of space. When dragging his equipment into the hole, Crixalis finds that he needs more space to ensure everything is placed well. Actually, the ith equipment needs Bi units of space during the moving. More precisely Crixalis can not move equipment into the hole unless there are Bi units of space left. After it moved in, the volume of the hole will decrease by Ai. Crixalis wonders if he can move all his equipment into the new hole and he turns to you for help.

Input

The first line contains an integer T, indicating the number of test cases. Then follows T cases, each one contains N + 1 lines. The first line contains 2 integers: V, volume of a hole and N, number of equipment respectively. The next N lines contain N pairs of integers: Ai and Bi.
0

Output

For each case output "Yes" if Crixalis can move all his equipment into the new hole or else output "No".

Sample Input

2

20 3
10 20
3 10
1 7

10 2
1 10
2 11

Sample Output

Yes
No

Source

HDU3177

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[HDU_ACM_Steps]Chapter One Section One

蓝飞 蓝飞 | 时间:2012-07-23, Mon | 10,845 views
编程算法 

A+B for Input-Output Practice (I)

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13098 Accepted Submission(s): 6187

Description

Your task is to Calculate a + b.
Too easy?! Of course! I specially designed the problem for acm beginners.
You must have found that some problems have the same titles with this one, yes, all these problems were designed for the same aim.

Input

The input will consist of a series of pairs of integers a and b, separated by a space, one pair of integers per line.

Output

For each pair of input integers a and b you should output the sum of a and b in one line, and with one line of output for each line in input.

Sample Input


1 5
10 20

Sample Output

6
30

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初识PhoneGap

蓝飞 蓝飞 | 时间:2012-06-07, Thu | 3,881 views
前端开发, 编程算法 

初识phonegap

一、PhoneGap是什么?

PhoneGap是一个标准的开源框架,用PhoneGap开发移动应用是免费的,无论是商业或是开源;一个用基于HTMLCSSJavaScript的,创建跨平台移动应用的快速开发平台。它使开发者能够利用iPhone、Android、Palm、Symbian、WP7、Bada和Blackberry智能手机的核心功能——包括地理定位,加速器,联系人,声音和振动等,此外PhoneGap拥有丰富的插件,可以以此扩展无限的功能。

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[ACM实验八]ACM程序设计基础(6)

蓝飞 蓝飞 | 时间:2012-06-03, Sun | 10,956 views
编程算法 

实验项目:ACM程序设计基础(6)
实验目的:掌握C++程序设计基础。
实验要求:使用VC++6.0实现实验要求。
实验内容:

1.猜牌游戏问题,请看简单回溯——猜牌游戏(提示:可以参考实验六的最后一题Crashing Balloon)。
2. 给定n个作业的集合Jn,每一个作业Ji都有两项任务分别在2台机器上完成。每个作业必须先由机器1处理,然后再由机器2处理。求所有作业在机器2上完成处理的时间和最少,并输出最佳调度方案。如:

机器1 机器2
作业1 2 1
作业2 3 1
作业3 2 3

最佳调度方案为:1 3 2,其完成时间为18。
3. 电路布线问题,请看简单动态规划——电路布线

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