<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[二十一画生的专栏]]></title><description><![CDATA[小Q]]></description><link>https://blog.youkuaiyun.com/u010679062</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; u010679062]]></copyright><item><title><![CDATA[pyqt5 将控件截图]]></title><link>https://blog.youkuaiyun.com/u010679062/article/details/121834346</link><guid>https://blog.youkuaiyun.com/u010679062/article/details/121834346</guid><author>u010679062</author><pubDate>Thu, 09 Dec 2021 17:24:20 +0800</pubDate><description><![CDATA[pyqt5 将控件截图]]></description><category></category></item><item><title><![CDATA[2014 ACM/ICPC Asia Regional Shanghai Online 1006 Sawtooth]]></title><link>https://blog.youkuaiyun.com/u010679062/article/details/39612539</link><guid>https://blog.youkuaiyun.com/u010679062/article/details/39612539</guid><author>u010679062</author><pubDate>Sat, 27 Sep 2014 16:10:30 +0800</pubDate><description><![CDATA[#include   
#include   
#include   
#include   
#include
#include
using namespace std;   
  
#define MAXN 9999  
#define MAXSIZE 10  
#define DLEN 4  
  
class BigNum  
{   
private:   
    int a[500]]]></description><category></category></item><item><title><![CDATA[uva 11235 - Frequent values（RMQ问题）]]></title><link>https://blog.youkuaiyun.com/u010679062/article/details/14468569</link><guid>https://blog.youkuaiyun.com/u010679062/article/details/14468569</guid><author>u010679062</author><pubDate>Thu, 07 Nov 2013 20:17:50 +0800</pubDate><description><![CDATA[题意：给出一个非降序排列的整数数组a1，a2，。。。an，你的任务是对于一系列询问（i，j），回答ai，ai+1，。。。，aj中出现次数最多的值所出现的次数。

#include
#include
#include
#define mx 100000+10
#define max(x,y) (x)>(y) ? (x) : (y)

using namespace std;

int n,m,d]]></description><category></category></item><item><title><![CDATA[uva 1428 - Ping pong (二叉索引树)]]></title><link>https://blog.youkuaiyun.com/u010679062/article/details/14454347</link><guid>https://blog.youkuaiyun.com/u010679062/article/details/14454347</guid><author>u010679062</author><pubDate>Thu, 07 Nov 2013 17:06:37 +0800</pubDate><description><![CDATA[题意：一条大街上住着n个乒乓球爱好者，经常组织比赛切磋技术。每个人都有一个不同的技能值ai。每场比赛需要3个人：两名选手，一名裁判。他们有一个奇怪的规定，即裁判必须住在两名选手的中间，并且技能值也在两名选手之间。问一共能组织多少种比赛。

#include
#include
#include
#define mx 100000+10
#define mm 20000+10

using nam]]></description><category></category></item><item><title><![CDATA[uva 1329 - Corporative Network]]></title><link>https://blog.youkuaiyun.com/u010679062/article/details/14381753</link><guid>https://blog.youkuaiyun.com/u010679062/article/details/14381753</guid><author>u010679062</author><pubDate>Wed, 06 Nov 2013 20:48:22 +0800</pubDate><description><![CDATA[题意：有n个结点，初始时每个结点的父节点都不存在。你的任务是执行如下操作：
           I u v：把结点u的父节点设为v，距离为|u -v|除以1000的余数。输入保证执行指令前u没有父节点。
          E u ：询问u到根节点的距离。

#include
#include
#define mx 20010

using namespace std;

int fa[]]></description><category></category></item><item><title><![CDATA[uva 1160 - X-Plosives（并查集）]]></title><link>https://blog.youkuaiyun.com/u010679062/article/details/14310283</link><guid>https://blog.youkuaiyun.com/u010679062/article/details/14310283</guid><author>u010679062</author><pubDate>Wed, 06 Nov 2013 19:50:55 +0800</pubDate><description><![CDATA[简单的并查集。

#include
#define mx 100010

int fa[mx];

int find(int x)
{
	while(fa[x]!=x) x=fa[x];
	return x;
}

int main()
{
	int i,x,y,cnt;
	while(scanf("%d",&x)!=EOF)
	{
		for(i=0;i<mx;i++) fa[i]=i;]]></description><category></category></item><item><title><![CDATA[uva 11997 - K Smallest Sums]]></title><link>https://blog.youkuaiyun.com/u010679062/article/details/14253471</link><guid>https://blog.youkuaiyun.com/u010679062/article/details/14253471</guid><author>u010679062</author><pubDate>Wed, 06 Nov 2013 19:05:42 +0800</pubDate><description><![CDATA[题意：有k个整数数组，各包含k个元素。在每个数组中取一个元素加起来，可以得到 kk个和。求这些和中最小的k个值（从小到大输出）。

#include
#include
#include

using namespace std;

struct node
{
	int s,b;
	node(int s,int b):s(s),b(b){}
};
bool operator < (node x,]]></description><category></category></item><item><title><![CDATA[uva 1203 - Argus]]></title><link>https://blog.youkuaiyun.com/u010679062/article/details/14230857</link><guid>https://blog.youkuaiyun.com/u010679062/article/details/14230857</guid><author>u010679062</author><pubDate>Wed, 06 Nov 2013 17:39:03 +0800</pubDate><description><![CDATA[题意：Argus系统支持一个Register命令：Register Q_num Period
该命令注册了一个触发器，它每Period秒就会产生一次编号为Q_num的事件。你的任务是模拟出前k个事件。如果事件同时发生，先处理Q_num小的事件。

#include
#include

using namespace std;

struct node
{
	int num,period,ti]]></description><category></category></item><item><title><![CDATA[uva 11991 - Easy Problem from Rujia Liu?]]></title><link>https://blog.youkuaiyun.com/u010679062/article/details/14229509</link><guid>https://blog.youkuaiyun.com/u010679062/article/details/14229509</guid><author>u010679062</author><pubDate>Wed, 06 Nov 2013 16:29:07 +0800</pubDate><description><![CDATA[题意：给出一个包含n个整数的数组，你需要回答若干查询。每次询问两个整数k和v，shuc]]></description><category></category></item><item><title><![CDATA[uva 11995 - I Can Guess the Data Structure!]]></title><link>https://blog.youkuaiyun.com/u010679062/article/details/14184967</link><guid>https://blog.youkuaiyun.com/u010679062/article/details/14184967</guid><author>u010679062</author><pubDate>Tue, 05 Nov 2013 20:25:02 +0800</pubDate><description><![CDATA[大致题意：输入1表示push，2表示pop，输出：
stack：一定是个栈；
queue：一定是个队列；
priority：一定是个优先队列；
impossible：都不是；
not sure：至少有两种可能。
注意：它有可能在pop前已为空，所以要先判断empty()。

#include
#include
#include
#include

using namespace s]]></description><category></category></item><item><title><![CDATA[青年文摘-往世书]]></title><link>https://blog.youkuaiyun.com/u010679062/article/details/14133689</link><guid>https://blog.youkuaiyun.com/u010679062/article/details/14133689</guid><author>u010679062</author><pubDate>Mon, 04 Nov 2013 19:47:54 +0800</pubDate><description><![CDATA[往事书

傀儡师能做出和人几无差别的傀儡，而付出的代价，是远离凡尘。他们在死的时候，刻一只与自己一模一样的傀儡，将生命延续下去。



一 重逢

没有谁的箭比谢洛白更快，没有谁的枪比姬郁非更强，也没有谁，敢和曲南陵比眼睛的锐利。

对于风行于晋北军中的这一说法，通常谢洛白只低头看自己的双手，姬郁非会露出看似忧郁实则骄傲的笑容，而那个倒霉的曲南陵，多半是苦着脸做一个噤声的手势：]]></description><category></category></item><item><title><![CDATA[hdu 1007 Quoit Design(分治法求最近点对)]]></title><link>https://blog.youkuaiyun.com/u010679062/article/details/13773167</link><guid>https://blog.youkuaiyun.com/u010679062/article/details/13773167</guid><author>u010679062</author><pubDate>Thu, 31 Oct 2013 15:22:19 +0800</pubDate><description><![CDATA[大致题意：给N个点，求最近点对的距离 d ；输出：r = d/2。

// Time 2093 ms; Memory 1812 K
#include
#include
#include
#include
#define eps 1e-8
#define maxn 100010
#define sqr(a) ((a)*(a))

using namespace std;

int sig(do]]></description><category></category></item><item><title><![CDATA[hdu 4717 The Moving Points(三分法)]]></title><link>https://blog.youkuaiyun.com/u010679062/article/details/11865177</link><guid>https://blog.youkuaiyun.com/u010679062/article/details/11865177</guid><author>u010679062</author><pubDate>Sat, 21 Sep 2013 11:42:05 +0800</pubDate><description><![CDATA[大致题意：给定 n 个起点的二维坐标和速度的大小和方向；问在哪一时刻所有两点间的最大距离最小。

// Time 78 ms; Memory 1316K
#include
#include
#include
#define maxn 50000
#define maxm 305
#define sqr(x) ((x)*(x))
#define eps 1e-5

using namespac]]></description><category></category></item><item><title><![CDATA[hdu 4741 Save Labman No.004 [2013年杭州ACM网络赛]]]></title><link>https://blog.youkuaiyun.com/u010679062/article/details/11712771</link><guid>https://blog.youkuaiyun.com/u010679062/article/details/11712771</guid><author>u010679062</author><pubDate>Sun, 15 Sep 2013 19:14:30 +0800</pubDate><description><![CDATA[// Time 234 ms; Memory 244 K
#include
#include
#include

using namespace std;

typedef struct point
{
	double x,y,z;
	point(double xx=0,double yy=0,double zz=0):x(xx),y(yy),z(zz){}
}vector;
vector op]]></description><category></category></item><item><title><![CDATA[hdu 4736  This Is The Job The Bear Finds(2013年成都ACM网络赛)]]></title><link>https://blog.youkuaiyun.com/u010679062/article/details/11703359</link><guid>https://blog.youkuaiyun.com/u010679062/article/details/11703359</guid><author>u010679062</author><pubDate>Sun, 15 Sep 2013 10:31:36 +0800</pubDate><description><![CDATA[#include
#include
#include
#include
#define eps 1e-10
#define sqr(a) ((a)*(a))
#define pi (2.0*asin(1.0))

using namespace std;

double ma[100010];

int sig(double a)
{
	return (a>eps)-(a<-eps);
}

ty]]></description><category></category></item><item><title><![CDATA[hdu 4007 Dave (2011年大连ACM网络赛)]]></title><link>https://blog.youkuaiyun.com/u010679062/article/details/10932551</link><guid>https://blog.youkuaiyun.com/u010679062/article/details/10932551</guid><author>u010679062</author><pubDate>Mon, 02 Sep 2013 17:02:50 +0800</pubDate><description><![CDATA[题意：给定正方形的边长 r ，在平面内寻找正方形可以圈住的点的最大的个数。
分析：先对点排序，然后固定一条边，再平移另一条垂直边，得到点的个数，最后比较大小即可。
注意：不包含正方形倾斜的情况！
// Time 125ms; Memory 240k#include
#include
#include

using namespace std;

struct point
{
    int]]></description><category></category></item><item><title><![CDATA[uva 11796 - Dog Distance]]></title><link>https://blog.youkuaiyun.com/u010679062/article/details/10364965</link><guid>https://blog.youkuaiyun.com/u010679062/article/details/10364965</guid><author>u010679062</author><pubDate>Mon, 26 Aug 2013 22:13:58 +0800</pubDate><description><![CDATA[#include
#include
#include
#define eps 1e-8
#define max(a,b) a>b?a:b

using namespace std;

int sig(double a)
{
    return (a>eps)-(a<-eps);
}

typedef struct point
{
    double x,y;
    point(double ]]></description><category></category></item><item><title><![CDATA[uva 1342 - That Nice Euler Circuit]]></title><link>https://blog.youkuaiyun.com/u010679062/article/details/10362283</link><guid>https://blog.youkuaiyun.com/u010679062/article/details/10362283</guid><author>u010679062</author><pubDate>Mon, 26 Aug 2013 19:52:13 +0800</pubDate><description><![CDATA[题意：平面上有一个包含n个点的一笔画，图案是封闭的。线段可以相交，但不会部分重叠，如图，求平面被分成多少部分。



#include
#include
#include
#define eps 1e-8

using namespace std;

int sig(double a)
{
	return (a>eps)-(aeps);
}

typedef struct point
{]]></description><category></category></item><item><title><![CDATA[uva 12165 - Triangle Hazard]]></title><link>https://blog.youkuaiyun.com/u010679062/article/details/10241817</link><guid>https://blog.youkuaiyun.com/u010679062/article/details/10241817</guid><author>u010679062</author><pubDate>Fri, 23 Aug 2013 21:44:20 +0800</pubDate><description><![CDATA[#include
#include
#include

using namespace std;

typedef struct point
{
    double x,y;
    point(double xx=0,double yy=0):x(xx),y(yy){}
}vector;
vector operator - (point a,point b)
{
    return vect]]></description><category></category></item><item><title><![CDATA[uva 1447 - Malfatti Circles]]></title><link>https://blog.youkuaiyun.com/u010679062/article/details/10216637</link><guid>https://blog.youkuaiyun.com/u010679062/article/details/10216637</guid><author>u010679062</author><pubDate>Fri, 23 Aug 2013 12:03:10 +0800</pubDate><description><![CDATA[题意：给出一个三角形的3个顶点的坐标，求3个圆，使得每个圆和三角形的两条边以及另两个圆均相切，如图，输出这3个圆的半径。


 #include
#include
#include
#define sqr(a) ((a)*(a))
#define eps 1e-8
#define min(a,b) (a)<(b)?(a):(b)

using namespace std;

int sig(]]></description><category></category></item></channel></rss>