<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[Cheky_的专栏]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/Cheky_</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; Cheky_]]></copyright><item><title><![CDATA[sgu499]]></title><link>https://blog.youkuaiyun.com/Cheky_/article/details/50726098</link><guid>https://blog.youkuaiyun.com/Cheky_/article/details/50726098</guid><author>Cheky_</author><pubDate>Tue, 23 Feb 2016 21:07:00 +0800</pubDate><description><![CDATA[//暴力求解复杂度为O（n(n-1)/2）

//假设所有输入中最大值为maxp,枚举i=maxp to 1（从大到小枚举）,
//如果i是某两个数的约数，则i是所有数的最大最大公约数。
//复杂度 ？？

#include  
#include  

#define maxn 1000005   

int num[maxn];  
int main()  
{  
	int n,i,j,k; ]]></description><category></category></item><item><title><![CDATA[sgu 318]]></title><link>https://blog.youkuaiyun.com/Cheky_/article/details/49046709</link><guid>https://blog.youkuaiyun.com/Cheky_/article/details/49046709</guid><author>Cheky_</author><pubDate>Sun, 11 Oct 2015 11:36:38 +0800</pubDate><description><![CDATA[/*** * 以资源为列，用户为行，统计资源列有多少个不同值，去掉空值，即为答案。 */import java.util.*;import java.io.*;public class Solution {public static void main(String[] argv) throws IOException {BufferedReader in = new BufferedReader]]></description><category></category></item><item><title><![CDATA[some useful softwares in linux]]></title><link>https://blog.youkuaiyun.com/Cheky_/article/details/17795255</link><guid>https://blog.youkuaiyun.com/Cheky_/article/details/17795255</guid><author>Cheky_</author><pubDate>Fri, 03 Jan 2014 17:39:20 +0800</pubDate><description><![CDATA[1. graphic user interfase
xorg / xinit - xserver system
openbox - a window manager


2. tools
gcc g++ gdb make 
git - version control tool
mit-scheme  - scheme interpretor
emacs - a editor]]></description><category></category></item><item><title><![CDATA[debian wheezy sources.list]]></title><link>https://blog.youkuaiyun.com/Cheky_/article/details/17794525</link><guid>https://blog.youkuaiyun.com/Cheky_/article/details/17794525</guid><author>Cheky_</author><pubDate>Fri, 03 Jan 2014 17:14:39 +0800</pubDate><description><![CDATA[open the setting file: /etc/apt/sources.list
add the follow code:
deb http://http.debian.net/debian wheezy main contrib non-free
deb http://mirrors.163.com/debian wheezy main contrib non-free 
deb h]]></description><category></category></item><item><title><![CDATA[openbox shortcut keybind and menu settings]]></title><link>https://blog.youkuaiyun.com/Cheky_/article/details/17794361</link><guid>https://blog.youkuaiyun.com/Cheky_/article/details/17794361</guid><author>Cheky_</author><pubDate>Fri, 03 Jan 2014 17:08:20 +0800</pubDate><description><![CDATA[1. keybind
find the file ~/.config/openbox/rc.xml
in the node , add below code:
    
      
        emacs
      
    
    
      
        x-www-browser
      
    
    
      
        x-terminal-em]]></description><category></category></item><item><title><![CDATA[sgu 178]]></title><link>https://blog.youkuaiyun.com/Cheky_/article/details/9423975</link><guid>https://blog.youkuaiyun.com/Cheky_/article/details/9423975</guid><author>Cheky_</author><pubDate>Tue, 23 Jul 2013 17:17:30 +0800</pubDate><description><![CDATA[f（n）表示问题的解
 
第一步：将n0 = n分成三部分a1 ，b1= 1， n1；其中a1 = n0 - n1 - 1，a1 〉=  n1；
n1 是之后要分解的 ，可行解为 f（n1）
我们得到  a1，b1， f（n1）

前n1天，用n1（会拆开）支付，解为f（n1）第n1+1天用 b1支付第n1+2天用a1（完整的）支付，必须找零a1 - 1，店主有n1+1个，所以必须有]]></description><category></category></item><item><title><![CDATA[setup  emacs for windows]]></title><link>https://blog.youkuaiyun.com/Cheky_/article/details/9370857</link><guid>https://blog.youkuaiyun.com/Cheky_/article/details/9370857</guid><author>Cheky_</author><pubDate>Thu, 18 Jul 2013 17:22:13 +0800</pubDate><description><![CDATA[1. download 
emacs-24.3-bin-i386.zip, form  http://mirrors.ustc.edu.cn/gnu/emacs/windows/
 
2. unzip it to a  directory,  such as d:\,  you will get d:\emacs-24.3\
 
3. add shortcut to start menu]]></description><category></category></item><item><title><![CDATA[stl support for gdb: gdb-stl-views]]></title><link>https://blog.youkuaiyun.com/Cheky_/article/details/9351833</link><guid>https://blog.youkuaiyun.com/Cheky_/article/details/9351833</guid><author>Cheky_</author><pubDate>Wed, 17 Jul 2013 10:46:11 +0800</pubDate><description><![CDATA[#                                                                                                        
#   STL GDB evaluators/views/utilities - 1.03
#
#   The new GDB commands:]]></description><category></category></item><item><title><![CDATA[windows emacs R ess]]></title><link>https://blog.youkuaiyun.com/Cheky_/article/details/9193883</link><guid>https://blog.youkuaiyun.com/Cheky_/article/details/9193883</guid><author>Cheky_</author><pubDate>Fri, 28 Jun 2013 09:44:36 +0800</pubDate><description><![CDATA[- setup emacs
 - setup R
 - setup ess
 - add Rterm.exe to path
 - finish.]]></description><category></category></item><item><title><![CDATA[setup virtual box extention for lubuntu]]></title><link>https://blog.youkuaiyun.com/Cheky_/article/details/9191741</link><guid>https://blog.youkuaiyun.com/Cheky_/article/details/9191741</guid><author>Cheky_</author><pubDate>Thu, 27 Jun 2013 21:18:35 +0800</pubDate><description><![CDATA[# before setup virtual box extention for lubuntu
sudp apt-get install dkms gcc make]]></description><category></category></item><item><title><![CDATA[WARNING: gnome-keyring:: couldn't connect to]]></title><link>https://blog.youkuaiyun.com/Cheky_/article/details/9191679</link><guid>https://blog.youkuaiyun.com/Cheky_/article/details/9191679</guid><author>Cheky_</author><pubDate>Thu, 27 Jun 2013 21:15:13 +0800</pubDate><description><![CDATA[- problem:
WARNING: gnome-keyring:: couldn't connect to: /tmp/keyring-rof1VB/pkcs11:
 No such file or directory

 - solution
/usr/bin/gnome-keyring-daemon
 --start --components=pkcs11


 - re]]></description><category></category></item><item><title><![CDATA[sgu 125]]></title><link>https://blog.youkuaiyun.com/Cheky_/article/details/9140997</link><guid>https://blog.youkuaiyun.com/Cheky_/article/details/9140997</guid><author>Cheky_</author><pubDate>Thu, 20 Jun 2013 21:49:42 +0800</pubDate><description><![CDATA[AC之后看大网上别人的基本是dfs、暴力搜索
因为我的算法不是这样，贴出来看看
sgu结果：  Accepted  31 ms   835 kb
 
英文很烂，注释看不懂可email
 
/**
 * @file
 * @brief my c solution for problem 125 on SGU OJ
 * @author ck
 * @date 2013-06-19
 *]]></description><category></category></item><item><title><![CDATA[sgu 120]]></title><link>https://blog.youkuaiyun.com/Cheky_/article/details/8810821</link><guid>https://blog.youkuaiyun.com/Cheky_/article/details/8810821</guid><author>Cheky_</author><pubDate>Tue, 16 Apr 2013 20:30:23 +0800</pubDate><description><![CDATA[#include 
#include 

const double PI = 3.1415926535897932384626;
int n, n1, n2;
double n1x, n1y, n2x, n2y;


int main() {
  scanf("%d%d%d", &n, &n1, &n2);
  scanf("%lf%lf%lf%lf", &n1x, &n1y, &n2x, &n2]]></description><category></category></item><item><title><![CDATA[sgu 193]]></title><link>https://blog.youkuaiyun.com/Cheky_/article/details/8804096</link><guid>https://blog.youkuaiyun.com/Cheky_/article/details/8804096</guid><author>Cheky_</author><pubDate>Mon, 15 Apr 2013 15:16:19 +0800</pubDate><description><![CDATA[/*
 if n = a*b + r and gcd(n, r) = 1 then
 	ans = max{a, b} (a, b <= n/2)
 
 
 if n = 2p + 1 then 
 	ans = p;
 else  n = 2p 
 	let p = 2q+r
 	if r=0 then 
 		n = 4q =2(2q - 1) + 2, gcd(n, 2q-1) = gcd(]]></description><category></category></item><item><title><![CDATA[sgu 226]]></title><link>https://blog.youkuaiyun.com/Cheky_/article/details/8781835</link><guid>https://blog.youkuaiyun.com/Cheky_/article/details/8781835</guid><author>Cheky_</author><pubDate>Wed, 10 Apr 2013 11:16:52 +0800</pubDate><description><![CDATA[/*
notice:
  the graph may contain loops
  between two point,there may be many ways that have different color
 */

#include 
#include 

using namespace std;

const int INF = 0x7ffffffe;
int n, m;
int]]></description><category></category></item><item><title><![CDATA[sgu 403]]></title><link>https://blog.youkuaiyun.com/Cheky_/article/details/8768559</link><guid>https://blog.youkuaiyun.com/Cheky_/article/details/8768559</guid><author>Cheky_</author><pubDate>Sun, 07 Apr 2013 17:01:26 +0800</pubDate><description><![CDATA[/*
  是某某的1/x 居然被翻译成：
  which is x times less than ...
  
 */

#include 

int main() {
  int x;
  scanf("%d", &x);
  printf("%d\n", (x<<1) + 1);
  return 0;
}]]></description><category></category></item><item><title><![CDATA[sgu 224]]></title><link>https://blog.youkuaiyun.com/Cheky_/article/details/8767596</link><guid>https://blog.youkuaiyun.com/Cheky_/article/details/8767596</guid><author>Cheky_</author><pubDate>Sun, 07 Apr 2013 14:46:37 +0800</pubDate><description><![CDATA[//dfs

//   k = 0， ans = 1；
//   k = 1， ans = n*n ；
//   k > n,  ans = 0;
//还可以进一步优化  
//   if (n - ps < k - count) rollback;  当剩余列数小于剩余棋子数时，回滚
//   对称性优化, 第一个棋子放在某一列m行，和放在该列n-m+1
//   直接用数组保存所有结果 ---]]></description><category></category></item><item><title><![CDATA[sgu 259]]></title><link>https://blog.youkuaiyun.com/Cheky_/article/details/8761960</link><guid>https://blog.youkuaiyun.com/Cheky_/article/details/8761960</guid><author>Cheky_</author><pubDate>Fri, 05 Apr 2013 17:38:12 +0800</pubDate><description><![CDATA[/*
  greedy method
  sort the leaflets by Li by descending order
  let sumTk = Ti + T2 + ..... Tk   
  answer = max ( sumTk + Lk),  k = 1...n


proofproof by contradiction :
  假设最优序列为 a = { 1, 2, 3 ..]]></description><category></category></item><item><title><![CDATA[sgu 223]]></title><link>https://blog.youkuaiyun.com/Cheky_/article/details/8754568</link><guid>https://blog.youkuaiyun.com/Cheky_/article/details/8754568</guid><author>Cheky_</author><pubDate>Wed, 03 Apr 2013 10:10:28 +0800</pubDate><description><![CDATA[/*
状态压缩dp


每一行的可行状态进行压缩 ， n=3时 s = {000， 001， 010， 100， 101}
d[row][si][p] 表示 第row行状态为si，前row-1行已放置且不冲突，总共放了p个棋子的方案数
初始状态 d[0][0][0] = 1， others = 0；
递推公式 d[row][s[i]][p + c[i]] = sigma（d[row-1][s[j]]]></description><category></category></item><item><title><![CDATA[sgu 296]]></title><link>https://blog.youkuaiyun.com/Cheky_/article/details/8659750</link><guid>https://blog.youkuaiyun.com/Cheky_/article/details/8659750</guid><author>Cheky_</author><pubDate>Mon, 11 Mar 2013 15:57:18 +0800</pubDate><description><![CDATA[#include 

#define NONE 'n'

/*
升序数列，删除前面的高位
降序数列，删除后面的低位

先处理升序
*/



char n[1009];
int k;

// find the next digit after digit a, return -1 when can not find.
int next(int a) {]]></description><category></category></item></channel></rss>