<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[Fatal Fake Field]]></title><description><![CDATA[来吧，少年~！Unlimited Code Works!!]]></description><link>https://blog.youkuaiyun.com/aroslhy</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; aroslhy]]></copyright><item><title><![CDATA[发现一种JavaScript“类的继承”的方法]]></title><link>https://blog.youkuaiyun.com/aroslhy/article/details/39676737</link><guid>https://blog.youkuaiyun.com/aroslhy/article/details/39676737</guid><author>aroslhy</author><pubDate>Mon, 29 Sep 2014 21:32:03 +0800</pubDate><description><![CDATA[function First(a, b) {
	this.a = a;
	this.b = b;
}
var first = new First(1, 2);
console.log(first);

function Second(c, d) {
	First.apply(this, [1, 2]);
	this.c = c;
	this.d = d;
}
var second = new Se]]></description><category></category></item><item><title><![CDATA[NJUST 1743 Boring Game（2013南京邀请赛G题）（thx to _LT_zyc）]]></title><link>https://blog.youkuaiyun.com/aroslhy/article/details/8964744</link><guid>https://blog.youkuaiyun.com/aroslhy/article/details/8964744</guid><author>aroslhy</author><pubDate>Thu, 23 May 2013 13:01:19 +0800</pubDate><description><![CDATA[本做法是基于暴力枚举第一行走法的O((n^2)*(2^n))的方法的预处理优化版。
　　首先做一个预处理，对于空白的棋盘，枚举第一行的走法，然后以把前n-1行复原成白的为目标，逐行向下推出下面n-1行的走法，记录下最后第n行的状态和对应走法的映射（可以用邻接表）。
　　对于每组case，第一行不动，同样地，以把前n-1行复原成白的为目标，逐行向下推出下面n-1行的走法x，这时第n行残留下一个状]]></description><category></category></item><item><title><![CDATA[HDU 4445 Crazy Tank]]></title><link>https://blog.youkuaiyun.com/aroslhy/article/details/8842170</link><guid>https://blog.youkuaiyun.com/aroslhy/article/details/8842170</guid><author>aroslhy</author><pubDate>Tue, 23 Apr 2013 22:20:02 +0800</pubDate><description><![CDATA[先反求可行的和不可行的出射角度区间，再扫描线求最大覆盖区间数。

// 2013-04-23 21:46:47  Accepted    4445    15MS    256K    2249 B  G++ Aros
#include
#include
#include
#include
using namespace std;
const int MAXN = 200+5;
const d]]></description><category></category></item><item><title><![CDATA[UVALive 4004 Space Beacon]]></title><link>https://blog.youkuaiyun.com/aroslhy/article/details/8727966</link><guid>https://blog.youkuaiyun.com/aroslhy/article/details/8727966</guid><author>aroslhy</author><pubDate>Wed, 27 Mar 2013 19:01:58 +0800</pubDate><description><![CDATA[先DP预处理出以i开头、长度为j、先升/降（0/1）的序列的个数：d[i][j][0/1]。再数给出的串是第几个串。
// 4004	Space Beacon	Accepted	C++	0.009	2013-03-27 10:52:06
#include
#include
#include
#include
using namespace std;
const int MAXN = 18+5;]]></description><category></category></item><item><title><![CDATA[Codeforces 235C Cyclical Quest]]></title><link>https://blog.youkuaiyun.com/aroslhy/article/details/8597798</link><guid>https://blog.youkuaiyun.com/aroslhy/article/details/8597798</guid><author>aroslhy</author><pubDate>Thu, 21 Feb 2013 11:46:08 +0800</pubDate><description><![CDATA[对第一个串s建立后缀自动机，用父亲指针求出每个节点代表的子串在s中出现的次数。对于串x，复制一个循环节加在后面。在后缀自动机上跑一遍，如果某个节点匹配长度大于x原来的长度，检查父节点是否也大于，是的话用父节点代替当前节点，把预处理出来的该节点的出现次数累加就是答案。

//Feb 21, 2013 7:32:16 AM 	bigoceanlhy 	235C - Cyclical Quest]]></description><category></category></item><item><title><![CDATA[HYSBZ 2806 [Ctsc2012]Cheat]]></title><link>https://blog.youkuaiyun.com/aroslhy/article/details/8596128</link><guid>https://blog.youkuaiyun.com/aroslhy/article/details/8596128</guid><author>aroslhy</author><pubDate>Wed, 20 Feb 2013 20:50:43 +0800</pubDate><description><![CDATA[后缀自动机+单调队列优化的DP。参考了http://hi.baidu.com/wyl8899/item/568baeed5034eec2baf37d69，其实后缀处理和DP转移自己都想到了，只差单调队列的优化。做的第一道SAM+DP的题，纪念一下～
//bigoceanlhy	2806	Accepted	54516 kb	2336 ms	C++/Edit	2346 B	2013-02-20 20]]></description><category></category></item><item><title><![CDATA[HDU 4436 str2int]]></title><link>https://blog.youkuaiyun.com/aroslhy/article/details/8596118</link><guid>https://blog.youkuaiyun.com/aroslhy/article/details/8596118</guid><author>aroslhy</author><pubDate>Wed, 20 Feb 2013 20:46:34 +0800</pubDate><description><![CDATA[把所有串用没出现过的符号相连，建立后缀自动机。记录每个节点所表示的子串对于此问题所求的sum值，再记录这个节点能收到多少个有效子串，于是就能从根结点往后递推sum值了。最后把所有有效节点的sum加起来就是答案。注意：（1）带有前缀零的子串无效；（2）带有分隔符的子串无效。

//2013-02-18 13:07:47	Accepted	4436	250MS	12624K	1780 B	G++]]></description><category></category></item><item><title><![CDATA[UVA 709 Formatting Text]]></title><link>https://blog.youkuaiyun.com/aroslhy/article/details/8330913</link><guid>https://blog.youkuaiyun.com/aroslhy/article/details/8330913</guid><author>aroslhy</author><pubDate>Wed, 19 Dec 2012 11:10:17 +0800</pubDate><description><![CDATA[d[i][j]表示第i个词于第j列开始，且后面的词按规则摆放的最小代价。因为怕搞错多解时的答案，所以用了逆推。
//709 	Formatting Text 	Accepted 	C++ 	0.020 	2012-12-19 03:06:24
#include
#include
#include
#include
using namespace std;
const int MAXN = 80+]]></description><category></category></item><item><title><![CDATA[UVA 10163 Storage Keepers]]></title><link>https://blog.youkuaiyun.com/aroslhy/article/details/8296857</link><guid>https://blog.youkuaiyun.com/aroslhy/article/details/8296857</guid><author>aroslhy</author><pubDate>Fri, 14 Dec 2012 21:06:14 +0800</pubDate><description><![CDATA[方法一、d[i][j][k]表示前i个人看j个货总保险值为k时的最小花费。时间复杂度较高。
//10163 	Storage Keepers 	Accepted 	C++ 	0.236 	2012-12-14 03:23:11
#include
#include
#include
using namespace std;
const int MAXN = 100+5, MAXM = 30+5,]]></description><category></category></item><item><title><![CDATA[UVA 10817 Headmaster's Headache]]></title><link>https://blog.youkuaiyun.com/aroslhy/article/details/8292277</link><guid>https://blog.youkuaiyun.com/aroslhy/article/details/8292277</guid><author>aroslhy</author><pubDate>Thu, 13 Dec 2012 19:21:41 +0800</pubDate><description><![CDATA[状态压缩DP，d[i][j]表示在前i个人中聘用某些人达到状态j所需的最小花费。
//10817 	Headmaster's Headache 	Accepted 	C++ 	0.312 	2012-12-13 11:14:13
#include
#include
#include
#include
#include
using namespace std;
const int MAX = 16]]></description><category></category></item><item><title><![CDATA[UVA 11008 Antimatter Ray Clearcutting]]></title><link>https://blog.youkuaiyun.com/aroslhy/article/details/8285474</link><guid>https://blog.youkuaiyun.com/aroslhy/article/details/8285474</guid><author>aroslhy</author><pubDate>Wed, 12 Dec 2012 10:34:16 +0800</pubDate><description><![CDATA[明显的状态压缩DP，状态转移时要分打掉一个点或者打掉一条线，用最朴素的转移，总时间复杂度是O(2^N*N^3)，姿势优越的话可以水过这题。

//11008 	Antimatter Ray Clearcutting 	Accepted 	C++ 	2.656 	2012-12-12 02:15:41
#include
#include
#include
using namespace std;]]></description><category></category></item><item><title><![CDATA[UVA 10913 Walking on a Grid]]></title><link>https://blog.youkuaiyun.com/aroslhy/article/details/8280052</link><guid>https://blog.youkuaiyun.com/aroslhy/article/details/8280052</guid><author>aroslhy</author><pubDate>Mon, 10 Dec 2012 21:47:57 +0800</pubDate><description><![CDATA[设d[i][j][k]为走到(i,j)经过k个负数的收益，负无穷表示无法达到这个状态。由于只能向下、左、右且不能走回头路，于是按层DP就好了。注意用long long。

//10913 	Walking on a Grid 	Accepted 	C++ 	0.032 	2012-12-10 13:29:24
#include
#include
#include
using namespace]]></description><category></category></item><item><title><![CDATA[UVA 10313 Pay the Price]]></title><link>https://blog.youkuaiyun.com/aroslhy/article/details/8205159</link><guid>https://blog.youkuaiyun.com/aroslhy/article/details/8205159</guid><author>aroslhy</author><pubDate>Tue, 20 Nov 2012 19:40:47 +0800</pubDate><description><![CDATA[算法：用前缀和把O(n^3)的dp优化成O(n^2)。注意：（1）long long。（2）L1、L2不一定在[1,300]内。（3）没有空行（不用处理空行能过）。
//10313 	Pay the Price 	Accepted 	C++ 	0.056 	2012-11-20 11:23:14
#include
#include
#include
using namespace std;
co]]></description><category></category></item><item><title><![CDATA[UVA 10534 Wavio Sequence]]></title><link>https://blog.youkuaiyun.com/aroslhy/article/details/8183945</link><guid>https://blog.youkuaiyun.com/aroslhy/article/details/8183945</guid><author>aroslhy</author><pubDate>Wed, 14 Nov 2012 18:48:43 +0800</pubDate><description><![CDATA[正反两个方向各求一遍LIS（需要O(nlogn)算法），枚举中心。
//10534 	Wavio Sequence 	Accepted 	C++ 	0.096 	2012-11-14 10:34:30
#include
#include
#include
using namespace std;
const int MAXN = 10000+5;
int N, a[MAXN];
int d[2]]]></description><category></category></item><item><title><![CDATA[UVA 10564 Paths through the Hourglass]]></title><link>https://blog.youkuaiyun.com/aroslhy/article/details/8183938</link><guid>https://blog.youkuaiyun.com/aroslhy/article/details/8183938</guid><author>aroslhy</author><pubDate>Wed, 14 Nov 2012 18:46:54 +0800</pubDate><description><![CDATA[设d[i][j][k]表示从i行j列往下走到底路径上的和为k的方案数。
//10564 	Paths through the Hourglass 	Accepted 	C++ 	0.072 	2012-11-14 01:49:52
#include
#include
#include
using namespace std;
const int MAXN = 40+5, MAXM = 400+5]]></description><category></category></item><item><title><![CDATA[HDU 4433 locker]]></title><link>https://blog.youkuaiyun.com/aroslhy/article/details/8126019</link><guid>https://blog.youkuaiyun.com/aroslhy/article/details/8126019</guid><author>aroslhy</author><pubDate>Mon, 29 Oct 2012 21:12:20 +0800</pubDate><description><![CDATA[设状态d[i][j][k]表示前i位匹配，之后两位为j、k，达到此状态的最小代价。
        （1）现场写的状态转移：
//2012-10-29 21:07:26	Accepted	4433	687MS	4552K	3686 B	G++	Aros
#include
#include
#include
#include
#define MAX 10
#define up(x) (x+1)%]]></description><category></category></item><item><title><![CDATA[Tianjin Regional——区域赛初体验]]></title><link>https://blog.youkuaiyun.com/aroslhy/article/details/8105971</link><guid>https://blog.youkuaiyun.com/aroslhy/article/details/8105971</guid><author>aroslhy</author><pubDate>Wed, 24 Oct 2012 11:11:27 +0800</pubDate><description><![CDATA[终于迎来了向往已久的区域赛。20日早上出发，上午抵达天津，领了坑爹的贴纸+秋衣的参赛服、没有印名字的胸卡、新生手册……囧。中午吃饭的时候，教练说了一些让人压力很大的话（强校强队很多什么的），这时我才感觉到此行之凶险。于是我们果断改变目标，改保铜争银为保铁争铜（太没骨气了！）。
        下午睡过开幕式，便是热身赛了。这桌子真是小得一逼，施展开来很不方便呐。前两题很简单，数列求和，于是我们各]]></description><category></category></item><item><title><![CDATA[Codeforces Round #142 (Div. 2)]]></title><link>https://blog.youkuaiyun.com/aroslhy/article/details/8036976</link><guid>https://blog.youkuaiyun.com/aroslhy/article/details/8036976</guid><author>aroslhy</author><pubDate>Tue, 02 Oct 2012 18:18:18 +0800</pubDate><description><![CDATA[A.贪心，每次都选剩下的里面x最小的。

//Oct 1, 2012 7:42:03 PM 	bigoceanlhy 	230A - Dragons 	GNU C++ 	Accepted 	46 ms 	0 KB
#include
#include
#include
using namespace std;
const int MAXN = 1000+5;
int N, s, x[MAXN],]]></description><category></category></item><item><title><![CDATA[HDU 4416 Good Article Good sentence]]></title><link>https://blog.youkuaiyun.com/aroslhy/article/details/8020554</link><guid>https://blog.youkuaiyun.com/aroslhy/article/details/8020554</guid><author>aroslhy</author><pubDate>Wed, 26 Sep 2012 13:36:32 +0800</pubDate><description><![CDATA[求多串的子串并集元素的个数，先用没出现过的不同的字符把多个串拼接，用后缀数组求这个串的不同子串的个数，再减去含有拼接字符的子串的个数。用上述方法求『A、B1、……、BN』中不同子串的个数sumAB和『B1、……、BN』中不同子串的个数sumB，答案就是sumAB-sumB。
//2012-09-26 13:22:35	Accepted	4416	671MS	7752K	2071 B	G++	Ar]]></description><category></category></item><item><title><![CDATA[HDU 4419 Colourful Rectangle]]></title><link>https://blog.youkuaiyun.com/aroslhy/article/details/8014308</link><guid>https://blog.youkuaiyun.com/aroslhy/article/details/8014308</guid><author>aroslhy</author><pubDate>Mon, 24 Sep 2012 20:16:50 +0800</pubDate><description><![CDATA[分别求出7种矩形并的面积（离散化+扫描线+线段树），再根据容斥关系计算所求颜色面积。
//2012-09-24 20:30:42	Accepted	4419	359MS	4740K	2707 B	G++	Aros
#include
#include
#include
#include
using namespace std;
const int MAXN = 10000+5, MAXM = 20]]></description><category></category></item></channel></rss>