<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[MakingMaker的专栏]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/MakingMaker</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; MakingMaker]]></copyright><item><title><![CDATA[关于置换的一点收获]]></title><link>https://blog.youkuaiyun.com/MakingMaker/article/details/41307671</link><guid>https://blog.youkuaiyun.com/MakingMaker/article/details/41307671</guid><author>MakingMaker</author><pubDate>Thu, 20 Nov 2014 08:53:53 +0800</pubDate><description><![CDATA[对于一个排列{1， 2，3，....n}，可以通过某一种方式打乱其顺序，]]></description><category></category></item><item><title><![CDATA[Mac Apache Django WSGI 部署]]></title><link>https://blog.youkuaiyun.com/MakingMaker/article/details/28600325</link><guid>https://blog.youkuaiyun.com/MakingMaker/article/details/28600325</guid><author>MakingMaker</author><pubDate>Thu, 05 Jun 2014 11:15:47 +0800</pubDate><description><![CDATA[话说Django＋Apache也太难部署了，先试了mod_python后来试了几个小时都不行，]]></description><category></category></item><item><title><![CDATA[cocos2d-x随机数的问题]]></title><link>https://blog.youkuaiyun.com/MakingMaker/article/details/26288335</link><guid>https://blog.youkuaiyun.com/MakingMaker/article/details/26288335</guid><author>MakingMaker</author><pubDate>Mon, 19 May 2014 22:50:23 +0800</pubDate><description><![CDATA[大家都知道游戏里经常会涉及到随机数的问题，
cocos2d-x]]></description><category></category></item><item><title><![CDATA[cocos2d-x 3.0 常见问题及解决]]></title><link>https://blog.youkuaiyun.com/MakingMaker/article/details/25572375</link><guid>https://blog.youkuaiyun.com/MakingMaker/article/details/25572375</guid><author>MakingMaker</author><pubDate>Mon, 12 May 2014 00:46:55 +0800</pubDate><description><![CDATA[1.打包图片生成plist时、、、，如果图片挨在一起会出现黑线，打包时将Extrude设为1即可]]></description><category></category></item><item><title><![CDATA[cocos2dx3.0添加模块方面的一些问题]]></title><link>https://blog.youkuaiyun.com/MakingMaker/article/details/24688765</link><guid>https://blog.youkuaiyun.com/MakingMaker/article/details/24688765</guid><author>MakingMaker</author><pubDate>Tue, 29 Apr 2014 08:42:35 +0800</pubDate><description><![CDATA[昨天使用cocos2dx3.0版本添加了cocostudio模块，]]></description><category></category></item><item><title><![CDATA[Unity3D学习笔记]]></title><link>https://blog.youkuaiyun.com/MakingMaker/article/details/18107221</link><guid>https://blog.youkuaiyun.com/MakingMaker/article/details/18107221</guid><author>MakingMaker</author><pubDate>Sat, 11 Jan 2014 01:15:31 +0800</pubDate><description><![CDATA[1.Debug.Log很耗时，尽量不要循环使用]]></description><category></category></item><item><title><![CDATA[书上说我应该学习写博客]]></title><link>https://blog.youkuaiyun.com/MakingMaker/article/details/17558215</link><guid>https://blog.youkuaiyun.com/MakingMaker/article/details/17558215</guid><author>MakingMaker</author><pubDate>Wed, 25 Dec 2013 16:45:54 +0800</pubDate><description><![CDATA[我将来可能是一个程序员（废话，来优快云的还能是啥，至少我是这么认为的），作为准程序员，我确实在沟通上存在很大问题。
感觉说话的时候有的事能不说绝不废话，能少说绝不多说。
俗称宅（穷人好意思说宅？），每次都有这个感觉，但就是不知道怎么改进啊。
今天看了一本书（《高效能程序员的修炼》），上面说要经常写作，我就上来写一篇文章（如果能叫文章的话）。
平时呢经常逛QQ空间，但很少发说说上面的，唉！]]></description><category></category></item><item><title><![CDATA[神奇代码段]]></title><link>https://blog.youkuaiyun.com/MakingMaker/article/details/12567815</link><guid>https://blog.youkuaiyun.com/MakingMaker/article/details/12567815</guid><author>MakingMaker</author><pubDate>Thu, 10 Oct 2013 13:45:46 +0800</pubDate><description><![CDATA[1.已经匹配了i个后再加一个字符j变成匹配e[i][j]个

memset(e, 0, sizeof(e));
for(int i = 1; i <= len; i++) {
    int fail = e[i-1][s[i]-'A'];
    e[i-1][s[i]-'A'] = i;
    memcpy(e[i], e[fail], sizeof(e[i]));
}]]></description><category></category></item><item><title><![CDATA[动态规划优化整理]]></title><link>https://blog.youkuaiyun.com/MakingMaker/article/details/12555205</link><guid>https://blog.youkuaiyun.com/MakingMaker/article/details/12555205</guid><author>MakingMaker</author><pubDate>Thu, 10 Oct 2013 10:07:18 +0800</pubDate><description><![CDATA[1.四边形优化
 m(i, j)状态方程满足：
i 
i = j: 0
i > j: INF
对于i 
令s(i, j)为对应决策变量的最大值,即m(i, j)通过s(i, j)达到最优, 并且s(i, j)最大
s(i, j) = MAX(i
如果m(i, j)满足四边形定理, 则s(i, j) 
于是m(i, j) = MIN(s(i, j-1) 
2.斜率优化
此处以d]]></description><category></category></item><item><title><![CDATA[MakingMaker ACM]]></title><link>https://blog.youkuaiyun.com/MakingMaker/article/details/10079779</link><guid>https://blog.youkuaiyun.com/MakingMaker/article/details/10079779</guid><author>MakingMaker</author><pubDate>Mon, 19 Aug 2013 17:50:25 +0800</pubDate><description><![CDATA[#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;

#define Fi first
#define Se second
#define Mpair make_pair
#d]]></description><category></category></item><item><title><![CDATA[HDU4676 Sum Of Gcd]]></title><link>https://blog.youkuaiyun.com/MakingMaker/article/details/9990929</link><guid>https://blog.youkuaiyun.com/MakingMaker/article/details/9990929</guid><author>MakingMaker</author><pubDate>Thu, 15 Aug 2013 22:07:06 +0800</pubDate><description><![CDATA[题意：给定一个n的排列，询问L到R之间任意两个数的gcd之和
以下是解题报告的思路，解题报告看了n久才看懂， 记一下，以免以后忘记了~~
对于L到R之间的一个数A，如果数A的一个因子B在L到R之间出现了K次，则结果加上K*F[B]。例如4有因子2，出现了K次，加上K*F[2]。
当然A有因子B，则A有B的因子C，又要加上K*F[C]。 假如A有因子2，除去A之外，L到R还有有1个因子2，1个]]></description><category></category></item><item><title><![CDATA[TwoSat]]></title><link>https://blog.youkuaiyun.com/MakingMaker/article/details/9830393</link><guid>https://blog.youkuaiyun.com/MakingMaker/article/details/9830393</guid><author>MakingMaker</author><pubDate>Thu, 08 Aug 2013 09:16:07 +0800</pubDate><description><![CDATA[有N个布尔变量Xi，另外有M个需要满足的条件，每个条件形式为“Xi为真/假 则 Xj必为真/假”。是否能给所有的Xi赋值并且满足所有的条件。struct TwoSat {
    vector nxt[MAXN<<1];
    int mark[MAXN<<1];
    int stk[MAXN<<1], top;
    int n;
    
    void init(int tn) {]]></description><category></category></item><item><title><![CDATA[Codeforces Round #182 (Div. 1)  D. Yaroslav and Divisors]]></title><link>https://blog.youkuaiyun.com/MakingMaker/article/details/9034221</link><guid>https://blog.youkuaiyun.com/MakingMaker/article/details/9034221</guid><author>MakingMaker</author><pubDate>Wed, 05 Jun 2013 21:35:32 +0800</pubDate><description><![CDATA[数学分析很重要啊~~ 


求l，r之间有多少个序列对：
道题给出的是1-N的一个排列，因此是一个非常特殊的问题，倍数关系的对只有NLOGN(N/1 + N/2 + ..+N/N)
预处理出所有的序列对，可以统计出前缀和sum[l]表示1~l之间的对数
对于每个查询(X,Y)，我们将所有关系对XI>=X的，YI 
可以用总数sum[y]-sum[x] - (xI=x&&yI
需要对]]></description><category></category></item><item><title><![CDATA[Uva 11248 Frequency Hopping]]></title><link>https://blog.youkuaiyun.com/MakingMaker/article/details/8349543</link><guid>https://blog.youkuaiyun.com/MakingMaker/article/details/8349543</guid><author>MakingMaker</author><pubDate>Thu, 20 Dec 2012 13:55:15 +0800</pubDate><description><![CDATA[坑爹啊 c可能为0......又是两小时
先求出最大流 然后可能修改的边只能是最小割的边， 枚举没条边
保存修改之前的最大流 每次枚举时不重新找最大流直接在初始最大流上跑
只要最大流超过c就可以停止了
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
ty]]></description><category></category></item><item><title><![CDATA[POJ 3565 Ants]]></title><link>https://blog.youkuaiyun.com/MakingMaker/article/details/8332347</link><guid>https://blog.youkuaiyun.com/MakingMaker/article/details/8332347</guid><author>MakingMaker</author><pubDate>Wed, 19 Dec 2012 16:04:42 +0800</pubDate><description><![CDATA[KM 模板题
#include
#include
#include
#include
#include
#include
using namespace std;

const int MAXN = 107;
const int INF = 1<<27;
const double eps = 1e-6;
struct KM {
	double dt[MAXN][MAXN];
	double lx]]></description><category></category></item><item><title><![CDATA[Uva 11090 Going in Cycle!!]]></title><link>https://blog.youkuaiyun.com/MakingMaker/article/details/8330696</link><guid>https://blog.youkuaiyun.com/MakingMaker/article/details/8330696</guid><author>MakingMaker</author><pubDate>Wed, 19 Dec 2012 10:41:08 +0800</pubDate><description><![CDATA[又WA无数。。。。。坑爹啊，就不能让我一次A吗？？
#include
#include
#include
#include
#include
using namespace std;

const int MAXN = 107;
const int INF = 1 << 27;
struct BellmanFord {
	vector G[MAXN];
	vector W[MAXN];
	dou]]></description><category></category></item><item><title><![CDATA[Uva 10917 Walk Through the Forest]]></title><link>https://blog.youkuaiyun.com/MakingMaker/article/details/8316041</link><guid>https://blog.youkuaiyun.com/MakingMaker/article/details/8316041</guid><author>MakingMaker</author><pubDate>Tue, 18 Dec 2012 20:00:00 +0800</pubDate><description><![CDATA[完全没考虑原理 瞬间敲出来然后WA了。。。。。坑爹啊
#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long ll;

const int MAXN = 1007;
const ll INF = 1LL<<56;
struct Dijkstra {
	struct H]]></description><category></category></item><item><title><![CDATA[Uva 11374 Airpot Express]]></title><link>https://blog.youkuaiyun.com/MakingMaker/article/details/8315543</link><guid>https://blog.youkuaiyun.com/MakingMaker/article/details/8315543</guid><author>MakingMaker</author><pubDate>Tue, 18 Dec 2012 17:47:43 +0800</pubDate><description><![CDATA[坑爹啊 INF开大了要错，还没有PE
#include
#include
#include
#include
#include
#include
#include
using namespace std;

const int MAXN = 1007;
const int INF = 1 << 27;
struct Edge {
	int from, to, dist;
	Edge() {}]]></description><category></category></item><item><title><![CDATA[UVA 11324 The Largest Clique]]></title><link>https://blog.youkuaiyun.com/MakingMaker/article/details/8309846</link><guid>https://blog.youkuaiyun.com/MakingMaker/article/details/8309846</guid><author>MakingMaker</author><pubDate>Mon, 17 Dec 2012 23:06:30 +0800</pubDate><description><![CDATA[。。。 错了很多次啊 细节问题啊。。
坑爹啊 两小时就这样没了
#include
#include
#include
#include
#include
using namespace std;

const int MAXN = 1007;
vector G[MAXN], scc[MAXN], D[MAXN];
int pt[MAXN], low[MAXN], dt;
int sccno[M]]></description><category></category></item><item><title><![CDATA[LA 4287 Proving Equivalences]]></title><link>https://blog.youkuaiyun.com/MakingMaker/article/details/8307709</link><guid>https://blog.youkuaiyun.com/MakingMaker/article/details/8307709</guid><author>MakingMaker</author><pubDate>Mon, 17 Dec 2012 15:33:10 +0800</pubDate><description><![CDATA[强连通分量的题，tarjan算法
然后缩点，记a为入度为0的点的个数，b为出度为0的点的个数，ans = max(a, b);
特殊情况缩点后只有一个点 ans = 0;
#include
#include
#include
#include
#include
using namespace std;

const int MAXN = 20007;
int pt[MAXN], low[MA]]></description><category></category></item></channel></rss>