<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[plsxysam的专栏]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/plsxysam</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; plsxysam]]></copyright><item><title><![CDATA[LAMP初次搭建]]></title><link>https://blog.youkuaiyun.com/plsxysam/article/details/48087051</link><guid>https://blog.youkuaiyun.com/plsxysam/article/details/48087051</guid><author>plsxysam</author><pubDate>Sat, 29 Aug 2015 15:29:00 +0800</pubDate><description><![CDATA[查看liunx版本信息
在linux下面察看版本信息，通常使用的命令是uname：
$ uname -a
-a选项表示察看所有的信息，但是uname看到的版本信息，只是内核版本的信息，而不是发行版的版本信息



哪有没有一个更加通用的方法呢？那就是/etc/issue文件
issue文件是登录前的信息和标识文件,里面存放了发行版的版本信息
$cat /etc/issue
这样]]></description><category></category></item><item><title><![CDATA[组合c(m,n)的计算方法]]></title><link>https://blog.youkuaiyun.com/plsxysam/article/details/46982647</link><guid>https://blog.youkuaiyun.com/plsxysam/article/details/46982647</guid><author>plsxysam</author><pubDate>Tue, 21 Jul 2015 10:52:38 +0800</pubDate><description><![CDATA[问题：求解组合数C(n,m)，即从n个相同物品中取出m个的方案数，由于结果可能非常大，对结果模10007即可。

方案1:

暴力求解，C(n,m)=n*(n-1)*...*(n-m+1)/m!，n

 int Combination(int n, int m) 
{ 
    const int M = 10007; 
    int ans = 1; 
    for(int i=]]></description><category></category></item><item><title><![CDATA[zznuoj 1909: yifan的数组]]></title><link>https://blog.youkuaiyun.com/plsxysam/article/details/46967321</link><guid>https://blog.youkuaiyun.com/plsxysam/article/details/46967321</guid><author>plsxysam</author><pubDate>Mon, 20 Jul 2015 13:50:48 +0800</pubDate><description><![CDATA[1909: yifan的数组


题目描述

给你一个数组，初始值都是0，然后有N个操作，每次在一段区间L，R上加W，操作完后，求出这个数组中最大值是多少?


输入

多组样例，一个数字N表示N个操作，接下来N个数字L，R, W，表示给区间L，R加W，(1 


输出

一个数字代表答案


样例输入
3
1 2 1
1 3 2
1 3 4
4
1 1 2
]]></description><category></category></item><item><title><![CDATA[二分查找——zznuoj1906yifan and Zeroes 解题题解]]></title><link>https://blog.youkuaiyun.com/plsxysam/article/details/46965797</link><guid>https://blog.youkuaiyun.com/plsxysam/article/details/46965797</guid><author>plsxysam</author><pubDate>Mon, 20 Jul 2015 11:12:33 +0800</pubDate><description><![CDATA[最简单二分查找，分析题意，某数阶乘的末尾有几个零，然而其实就是求阶乘展开式有5的几次方，一个5代表一个零。而且规律可得n个零时，5*n的阶乘中5的x次方大于等于n，顾二分初始边界取1和n，然后不断通过mid来判断，最后缩小范围然后确定。
题目：


问题 F: yifan and Zeroes


题目描述

 One of My Teammates asd encountere]]></description><category></category></item><item><title><![CDATA[NBUT 2015 ACM Trainning Contest-11038 Problem  A How many Points  解题报告]]></title><link>https://blog.youkuaiyun.com/plsxysam/article/details/46862319</link><guid>https://blog.youkuaiyun.com/plsxysam/article/details/46862319</guid><author>plsxysam</author><pubDate>Mon, 13 Jul 2015 14:44:16 +0800</pubDate><description><![CDATA[由于刚刚练习了线段树，直接第一反应用线段树敲了，WA，把数据改成long ，WA，最后才发现错在没有考虑0不在线段树的节点上，所以又在输出时增加判断包含0的情况。最后超大空间超长时间AC。
       而且读题发现All the integers are distinct 。。。忽略了这句话


Problem A: How many Points

Description

W]]></description><category></category></item><item><title><![CDATA[线段树——HDOJ 1166 敌兵布阵 解题报告]]></title><link>https://blog.youkuaiyun.com/plsxysam/article/details/46852913</link><guid>https://blog.youkuaiyun.com/plsxysam/article/details/46852913</guid><author>plsxysam</author><pubDate>Sun, 12 Jul 2015 20:21:40 +0800</pubDate><description><![CDATA[线段树入门题目]]></description><category></category></item><item><title><![CDATA[C++使用localtime函数需要注意的地方]]></title><link>https://blog.youkuaiyun.com/plsxysam/article/details/46537147</link><guid>https://blog.youkuaiyun.com/plsxysam/article/details/46537147</guid><author>plsxysam</author><pubDate>Wed, 17 Jun 2015 19:06:58 +0800</pubDate><description><![CDATA[VS2013中出现：

This function or variable may be unsafe. Consider using localtime_s instead. To disable deprecation, use _CRT_SECURE_NO_WARNINGS. See online help for details.


查阅后使用了errno_t localtim]]></description><category></category></item><item><title><![CDATA[Debug和Release区别]]></title><link>https://blog.youkuaiyun.com/plsxysam/article/details/46536995</link><guid>https://blog.youkuaiyun.com/plsxysam/article/details/46536995</guid><author>plsxysam</author><pubDate>Wed, 17 Jun 2015 18:51:52 +0800</pubDate><description><![CDATA[VC下Debug和Release区别

最近写代码过程中，发现 Debug 下运行正常，Release 下就会出现问题，百思不得其解，而Release 下又无法进行调试，于是只能采用printf方式逐步定位到问题所在处，才发现原来是给定的一个数组未初始化，导致后面处理异常。网上查找了些资料，在这 罗列汇总下，做为备忘～ 
一、Debug 和 Release 的区别 
        Deb]]></description><category></category></item><item><title><![CDATA[CIPAddressCtrl类的使用(IP地址与CString的互相转化)]]></title><link>https://blog.youkuaiyun.com/plsxysam/article/details/46536721</link><guid>https://blog.youkuaiyun.com/plsxysam/article/details/46536721</guid><author>plsxysam</author><pubDate>Wed, 17 Jun 2015 18:18:40 +0800</pubDate><description><![CDATA[//将CString   型IP地址在IPAddressCtrl中显示  
CString   strIP="192.168.0.10";  
DWORD   dwIP;  
dwIP   =   inet_addr(strIP);  
unsigned   char   *pIP   =   (unsigned   char*)&dwIP;  
m_ipAddr.SetAddress(]]></description><category></category></item><item><title><![CDATA[带负权的最短路bellman_ford——POJ 3259 Wormholes题解]]></title><link>https://blog.youkuaiyun.com/plsxysam/article/details/45228659</link><guid>https://blog.youkuaiyun.com/plsxysam/article/details/45228659</guid><author>plsxysam</author><pubDate>Thu, 23 Apr 2015 22:23:37 +0800</pubDate><description><![CDATA[Description

While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way path that delivers you to its destination a]]></description><category></category></item><item><title><![CDATA[最短路dij——POJ 2387 Til the Cows Come Home题解]]></title><link>https://blog.youkuaiyun.com/plsxysam/article/details/45182817</link><guid>https://blog.youkuaiyun.com/plsxysam/article/details/45182817</guid><author>plsxysam</author><pubDate>Wed, 22 Apr 2015 00:31:20 +0800</pubDate><description><![CDATA[Description

Bessie is out in the field and wants to get back to the barn to get as much sleep as possible before Farmer John wakes her for the morning milking. Bessie needs her beauty sleep, so she]]></description><category></category></item><item><title><![CDATA[BFS——POJ 3984 迷宫问题 解题题解]]></title><link>https://blog.youkuaiyun.com/plsxysam/article/details/45131811</link><guid>https://blog.youkuaiyun.com/plsxysam/article/details/45131811</guid><author>plsxysam</author><pubDate>Sun, 19 Apr 2015 19:43:01 +0800</pubDate><description><![CDATA[迷宫问题，求最短路径，BFS解题]]></description><category></category></item><item><title><![CDATA[最小生成树——HDOJ 2988 Dark roads解题报告]]></title><link>https://blog.youkuaiyun.com/plsxysam/article/details/45110741</link><guid>https://blog.youkuaiyun.com/plsxysam/article/details/45110741</guid><author>plsxysam</author><pubDate>Sat, 18 Apr 2015 09:55:18 +0800</pubDate><description><![CDATA[Problem Description
Economic times these days are tough, even in Byteland. To reduce the operating costs, the government of Byteland has decided to optimize the road lighting. Till now every road was illuminated all night long, which costs 1 Bytelandian Do]]></description><category></category></item><item><title><![CDATA[并查集——HDOJ 1213How Many Tables解题报告]]></title><link>https://blog.youkuaiyun.com/plsxysam/article/details/45093457</link><guid>https://blog.youkuaiyun.com/plsxysam/article/details/45093457</guid><author>plsxysam</author><pubDate>Fri, 17 Apr 2015 10:19:04 +0800</pubDate><description><![CDATA[Problem Description

Today is Ignatius' birthday. He invites a lot of friends. Now it's dinner time. Ignatius wants to know how many tables he needs at least. You have to notice that not all the friends know each other, and all the friends do not want to s]]></description><category></category></item><item><title><![CDATA[最小生成树——HDOJ 1233 还是畅通工程 解题报告]]></title><link>https://blog.youkuaiyun.com/plsxysam/article/details/45080517</link><guid>https://blog.youkuaiyun.com/plsxysam/article/details/45080517</guid><author>plsxysam</author><pubDate>Thu, 16 Apr 2015 20:31:09 +0800</pubDate><description><![CDATA[Problem Description
某省调查乡村交通状况，得到的统计表中列出了任意两村庄间的距离。省政府“畅通工程”的目标是使全省任何两个村庄间都可以实现公路交通（但不一定有直接的公路相连，只要能间接通过公路可达即可），并要求铺设的公路总长度为最小。请计算最小的公路总长度。]]></description><category></category></item><item><title><![CDATA[最小生成树——HDOJ 1863 畅通工程 解题报告]]></title><link>https://blog.youkuaiyun.com/plsxysam/article/details/45080363</link><guid>https://blog.youkuaiyun.com/plsxysam/article/details/45080363</guid><author>plsxysam</author><pubDate>Thu, 16 Apr 2015 20:19:15 +0800</pubDate><description><![CDATA[Problem Description 
省政府“畅通工程”的目标是使全省任何两个村庄间都可以实现公路交通（但不一定有直接的公路相连，只要能间接通过公路可达即可）。经过调查评估，得到的统计表中列出了有可能建设公路的若干条道路的成本。现请你编写程序，计算出全省畅通需要的最低成本。]]></description><category></category></item></channel></rss>