<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[AcToy]]></title><description><![CDATA[Where there is a will there is a way ]]></description><link>https://blog.youkuaiyun.com/AcToy</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; AcToy]]></copyright><item><title><![CDATA[Uva10106 - Product]]></title><link>https://blog.youkuaiyun.com/AcToy/article/details/13291009</link><guid>https://blog.youkuaiyun.com/AcToy/article/details/13291009</guid><author>AcToy</author><pubDate>Mon, 28 Oct 2013 12:30:03 +0800</pubDate><description><![CDATA[求两个大数的乘积

#include 
#include 
#include 
#include 
#include 
const int maxn = 505;
using namespace std;

struct bign{
  int len, s[maxn];

  bign() {
    memset(s, 0, sizeof(s));
    len = 1;
  }

  ]]></description><category></category></item><item><title><![CDATA[Uva424 - Integer Inquiry]]></title><link>https://blog.youkuaiyun.com/AcToy/article/details/13290889</link><guid>https://blog.youkuaiyun.com/AcToy/article/details/13290889</guid><author>AcToy</author><pubDate>Mon, 28 Oct 2013 12:16:19 +0800</pubDate><description><![CDATA[#include 
#include 
#include 
#include 
using namespace std;
const int maxn = 10005;

struct bign {
    int len, s[maxn];
    bign() { len = 0; memset(s, 0, sizeof(s)); s[0] = 0; }

    bign operator ]]></description><category></category></item><item><title><![CDATA[UVa 11524 - In-Circle]]></title><link>https://blog.youkuaiyun.com/AcToy/article/details/9388615</link><guid>https://blog.youkuaiyun.com/AcToy/article/details/9388615</guid><author>AcToy</author><pubDate>Sat, 20 Jul 2013 09:56:11 +0800</pubDate><description><![CDATA[不需要算法，几何计算题。方法如同Ac_coral http://www.cnblogs.com/Accoral/p/3142038.html
/*************************************************************************
 > File Name: UVa11524.cpp
 > Author: AcToy
 > Mail: ]]></description><category></category></item><item><title><![CDATA[UVa 11817 - Tunnelling the Earth]]></title><link>https://blog.youkuaiyun.com/AcToy/article/details/9385299</link><guid>https://blog.youkuaiyun.com/AcToy/article/details/9385299</guid><author>AcToy</author><pubDate>Fri, 19 Jul 2013 20:08:34 +0800</pubDate><description><![CDATA[将平面上的亮点坐标转化成空间坐标，为球面上的两点。求所对应的弧长与弦长的差值，转化为空间坐标的函数为
double torad(double deg) {
  return deg / 180 * M_PI;
}
void get_coord(double lat, double lng, double& x, double& y, double& z) {
  lat = torad(lat)]]></description><category></category></item><item><title><![CDATA[UVa 11646 - Athletics Track]]></title><link>https://blog.youkuaiyun.com/AcToy/article/details/9384777</link><guid>https://blog.youkuaiyun.com/AcToy/article/details/9384777</guid><author>AcToy</author><pubDate>Fri, 19 Jul 2013 18:39:27 +0800</pubDate><description><![CDATA[在圆中有个矩形，知道长：宽 = a ： b。设一值x，这样我们可以求出圆半径R，2*R*x = sqrt（a*x*a*x +b*x*b*x),x可化简没，sin（ang1）= b / 2R，矩形对角划分的一个角，画出图来就会知道是那个角了，这个我可以求出圆弧所对应的圆心角为 ang = 2 * ang1 。所以200 = a * x + R * x *ang，可以求出x，这样长为a * x， 宽为]]></description><category></category></item><item><title><![CDATA[UVa 11800 - Determine the Shape]]></title><link>https://blog.youkuaiyun.com/AcToy/article/details/9378623</link><guid>https://blog.youkuaiyun.com/AcToy/article/details/9378623</guid><author>AcToy</author><pubDate>Fri, 19 Jul 2013 12:02:58 +0800</pubDate><description><![CDATA[计算几何入门题，直接模拟即可，这个题关键是判断四个点的位置，我的方法如下：
if(SegmentProperIntersection(z[1],z[2], z[3], z[4])) A = z[1], D = z[2], C = z[3], B = z[4], flag = 1;
    else if(SegmentProperIntersection(z[1], z[3], z[2], z[]]></description><category></category></item><item><title><![CDATA[UVa 11437 - Triangle Fun]]></title><link>https://blog.youkuaiyun.com/AcToy/article/details/9372327</link><guid>https://blog.youkuaiyun.com/AcToy/article/details/9372327</guid><author>AcToy</author><pubDate>Thu, 18 Jul 2013 19:42:18 +0800</pubDate><description><![CDATA[入门题，给了A，B，C三点的坐标，通过参数方程可以求得，在BC上点的坐标 p = B + ( C - B ) t , 因为D是三等分点，所以t = 1 / 3. 可以求得D = 2 / 3 * B + 1 / 3 * C，同理可以求得E，F。
这样就可以求得点P， Q， R三点的坐标。例如求P点坐标方法如下
Point GetLineIntersection(Point p, Vector v]]></description><category></category></item><item><title><![CDATA[UVa 11178 - Morley's Theorem]]></title><link>https://blog.youkuaiyun.com/AcToy/article/details/9370595</link><guid>https://blog.youkuaiyun.com/AcToy/article/details/9370595</guid><author>AcToy</author><pubDate>Thu, 18 Jul 2013 16:56:39 +0800</pubDate><description><![CDATA[例题，全是模板，知道了怎么计算，就可以做出来，只是Rotate（）函数要注意的是，弧度rad为正时逆时针旋转，为负时顺时针旋转
/*************************************************************************
 > File Name: UVa11178.cpp
 > Author: AcToy
 > Mail: ycsgldy]]></description><category></category></item><item><title><![CDATA[poj2096 - Collecting Bugs]]></title><link>https://blog.youkuaiyun.com/AcToy/article/details/9366485</link><guid>https://blog.youkuaiyun.com/AcToy/article/details/9366485</guid><author>AcToy</author><pubDate>Thu, 18 Jul 2013 13:59:03 +0800</pubDate><description><![CDATA[http://poj.org/problem?id=2096

深深的被这个题给坑了，题意叙述的就比较麻烦。最纠结的是在输出的时候用%.4lf就是WA，用%.4f就AC了，不明白什么原因。
题意：给n个bug，s个子系统，某人一天会发现一个bug，这个bug属于某种类型的bug，发生在某个子系统上，这样dp[i][j]表示的是j个子系统中找到了i个bug，这样概率应该为i / n * j /]]></description><category></category></item><item><title><![CDATA[UVa11584-Partitioning by Palindromes]]></title><link>https://blog.youkuaiyun.com/AcToy/article/details/9364049</link><guid>https://blog.youkuaiyun.com/AcToy/article/details/9364049</guid><author>AcToy</author><pubDate>Thu, 18 Jul 2013 10:23:34 +0800</pubDate><description><![CDATA[http://uva.onlinejudge.org/external/115/11584.html
将所给的字符串，最少可以分成多少个回文串，dp[i]表示0～i之间最少可以分成的值，然后判断j～i（j 
/*************************************************************************
 > File Name: UVa11]]></description><category></category></item><item><title><![CDATA[hdoj4405-Aeroplane chess]]></title><link>https://blog.youkuaiyun.com/AcToy/article/details/9356375</link><guid>https://blog.youkuaiyun.com/AcToy/article/details/9356375</guid><author>AcToy</author><pubDate>Wed, 17 Jul 2013 15:35:12 +0800</pubDate><description><![CDATA[求期望值，体重说明，如果可以flight的话就直接到达规定那个点，用f数组将两个点连在一起就可以，反之，就要求打骰子后的期望啦，e[i] =
∑(e[i + k] * 1 / 6) + 1（k为点数1~6,1/6为致筛子的概率，因为不可以flight就要置一次筛子，所以+1）。
/*********************************************************]]></description><category></category></item><item><title><![CDATA[hdu3853-LOOPS]]></title><link>https://blog.youkuaiyun.com/AcToy/article/details/9350581</link><guid>https://blog.youkuaiyun.com/AcToy/article/details/9350581</guid><author>AcToy</author><pubDate>Wed, 17 Jul 2013 09:38:27 +0800</pubDate><description><![CDATA[接触的第一道概率dp，这个题和以前做的唯一区别就是加了一个概率，但是也是参照了别人的思路做出来的。
dp[i][j]表示在第i，j格子所期望的。p[i][j][0], p[i][j][1], p[i][j][2]分别表示三种传送的概率。
dp[i][j] = p[i][j][0] * dp[i][j] + p[i][j][1] * dp[i][j + 1]  + p[i][j][2] * dp]]></description><category></category></item><item><title><![CDATA[NOI2004 - 郁闷的出纳员]]></title><link>https://blog.youkuaiyun.com/AcToy/article/details/9343251</link><guid>https://blog.youkuaiyun.com/AcToy/article/details/9343251</guid><author>AcToy</author><pubDate>Tue, 16 Jul 2013 15:21:41 +0800</pubDate><description><![CDATA[就是维护一个有序的序列，对这个序列进行插入，删除，加和，查找第k大的数值，但值得注意的是，在插入的时候，如果他的工资起始就比工资下届底就不需要插入了，这表示他每到公司，也不算离开。删除的时候也要判断下，ss是否是起始的节点，如果是的话同样也不需要删除的操作


/*****************************************************************]]></description><category></category></item><item><title><![CDATA[HNOI2002 - 营业额统计]]></title><link>https://blog.youkuaiyun.com/AcToy/article/details/9342395</link><guid>https://blog.youkuaiyun.com/AcToy/article/details/9342395</guid><author>AcToy</author><pubDate>Tue, 16 Jul 2013 14:19:56 +0800</pubDate><description><![CDATA[找出比num小的最大数和比num大的最小数，然后计算差值和最小。只是两个查找的过程，和插入的过程，结果在控制a， b的位置wa了N次，
/*************************************************************************
    > File Name: HNOI2002.cpp
    > Author: Ac_Toy
    >]]></description><category></category></item><item><title><![CDATA[poj3481-Double Queue]]></title><link>https://blog.youkuaiyun.com/AcToy/article/details/9341155</link><guid>https://blog.youkuaiyun.com/AcToy/article/details/9341155</guid><author>AcToy</author><pubDate>Tue, 16 Jul 2013 11:45:36 +0800</pubDate><description><![CDATA[根据优先值的大小，输出它所对应的节点，在建树的过程中，再加一个kk变量，找到优先值p的位置，输出kk即可
/*************************************************************************
    > File Name: 3481.cpp
    > Author: Ac_Toy
    > Mail: ycsgldy@163.]]></description><category></category></item><item><title><![CDATA[HNOI2004-宠物收养所]]></title><link>https://blog.youkuaiyun.com/AcToy/article/details/9341045</link><guid>https://blog.youkuaiyun.com/AcToy/article/details/9341045</guid><author>AcToy</author><pubDate>Tue, 16 Jul 2013 11:37:43 +0800</pubDate><description><![CDATA[/*************************************************************************
    > File Name: hysbz1208.cpp
    > Author: Ac_Toy
    > Mail: ycsgldy@163.com 
    > Created Time: 2013年06月21日 星期五 06时48分17]]></description><category></category></item><item><title><![CDATA[UVa 10539 - Almost Prime Numbers]]></title><link>https://blog.youkuaiyun.com/AcToy/article/details/9069537</link><guid>https://blog.youkuaiyun.com/AcToy/article/details/9069537</guid><author>AcToy</author><pubDate>Mon, 10 Jun 2013 08:55:53 +0800</pubDate><description><![CDATA[求low~high之间有多少个质数分解后由一个素数组成，将10^6内的素数全部打出来，然后枚举每个素数的n（n > 2 ）次幂在low和high之间个数的和。
/*************************************************************************
    > File Name: 10539.cpp
    > Author: Toy]]></description><category></category></item><item><title><![CDATA[UVa 382	-Perfection]]></title><link>https://blog.youkuaiyun.com/AcToy/article/details/9068027</link><guid>https://blog.youkuaiyun.com/AcToy/article/details/9068027</guid><author>AcToy</author><pubDate>Sun, 09 Jun 2013 21:31:47 +0800</pubDate><description><![CDATA[求num的所有因子的和tmp（不包括num），tmp和num的大小关系输出不同的结果。
tmp==num  输出 PERFECT

tmp>num     输出ABUNDANT

tmp
/*************************************************************************
    > File Name: 382.cpp]]></description><category></category></item><item><title><![CDATA[UVa 160	-Factors and Factorials]]></title><link>https://blog.youkuaiyun.com/AcToy/article/details/9067769</link><guid>https://blog.youkuaiyun.com/AcToy/article/details/9067769</guid><author>AcToy</author><pubDate>Sun, 09 Jun 2013 21:01:38 +0800</pubDate><description><![CDATA[就是求num！中对从2开始的每个素数的指数幂是多少，然后输出指数。注意输出格式
/*************************************************************************
    > File Name: 160.cpp
    > Author: Toy
    > Mail: ycsgldy@163.com 
    > Create]]></description><category></category></item><item><title><![CDATA[UVa 10139 - Factovisors]]></title><link>https://blog.youkuaiyun.com/AcToy/article/details/9052907</link><guid>https://blog.youkuaiyun.com/AcToy/article/details/9052907</guid><author>AcToy</author><pubDate>Sat, 08 Jun 2013 07:59:41 +0800</pubDate><description><![CDATA[输入Case和num，求Case！是否能整除num。
思路：将num质数分解，然后用Case！对每个num质数分解后的素数求指数，和num这个素数比较，如果小于，说明不能整除。
/*************************************************************************
    > File Name: 10139.cpp
    > A]]></description><category></category></item></channel></rss>