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            雪之精靈

            從吐絲結(jié)繭到破繭而出

            2010年6月12日 #

            Ubuntu 8.04 source

            sudo cp /etc/apt/sources.list /etc/apt/sources.list_backup 備份現(xiàn)有更新源
            sudo gedit /etc/apt/sources.list    編輯更新源列表
            從下面各服務(wù)器列表內(nèi)容中選擇一段替換文件中的所有內(nèi)容
            保存編輯好的文件,執(zhí)行以下命令更新。
            sudo apt-get update (修改sources.list后必須運(yùn)行!)
            sudo apt-get dist-upgrade (更新所有軟件)
            # Shanghai Jiaotong Unverisity. APT source for Ubuntu 8.04 (hardy heron)
            deb http://ftp.sjtu.edu.cn/ubuntu/ hardy main multiverse restricted universe
            deb http://ftp.sjtu.edu.cn/ubuntu/ hardy-backports main multiverse restricted universe
            deb http://ftp.sjtu.edu.cn/ubuntu/ hardy-proposed main multiverse restricted universe
            deb http://ftp.sjtu.edu.cn/ubuntu/ hardy-security main multiverse restricted universe
            deb http://ftp.sjtu.edu.cn/ubuntu/ hardy-updates main multiverse restricted universe
            deb-src http://ftp.sjtu.edu.cn/ubuntu/ hardy main multiverse restricted universe
            deb-src http://ftp.sjtu.edu.cn/ubuntu/ hardy-backports main multiverse restricted universe
            deb-src http://ftp.sjtu.edu.cn/ubuntu/ hardy-proposed main multiverse restricted universe
            deb-src http://ftp.sjtu.edu.cn/ubuntu/ hardy-security main multiverse restricted universe
            deb-src http://ftp.sjtu.edu.cn/ubuntu/ hardy-updates main multiverse restricted universe

            # University of Science and Technology of China. APT source for Ubuntu 8.04 (hardy heron)
            deb http://debian.ustc.edu.cn/ubuntu/ hardy main multiverse restricted universe
            deb http://debian.ustc.edu.cn/ubuntu/ hardy-backports main multiverse restricted universe
            deb http://debian.ustc.edu.cn/ubuntu/ hardy-proposed main multiverse restricted universe
            deb http://debian.ustc.edu.cn/ubuntu/ hardy-security main multiverse restricted universe
            deb http://debian.ustc.edu.cn/ubuntu/ hardy-updates main multiverse restricted universe
            deb-src http://debian.ustc.edu.cn/ubuntu/ hardy main multiverse restricted universe
            deb-src http://debian.ustc.edu.cn/ubuntu/ hardy-backports main multiverse restricted universe
            deb-src http://debian.ustc.edu.cn/ubuntu/ hardy-proposed main multiverse restricted universe
            deb-src http://debian.ustc.edu.cn/ubuntu/ hardy-security main multiverse restricted universe
            deb-src http://debian.ustc.edu.cn/ubuntu/ hardy-updates main multiverse restricted universe

            posted @ 2010-06-12 15:43 雪之精靈 閱讀(476) | 評(píng)論 (0)編輯 收藏

            2010年6月4日 #

            Suggestions for presentations about reading academic paper

            1. Plan to spend 30-40 minutes on the talk (10 minutes of motivation, 10 minutes explaining the main idea and results, 10 minutes summarizing the paper and providing your perspective.)
            2. This will leave 10-15 minutes for a discussion on the topic. At the end of the discussion, we should have answered some of the following questions:

              Why is this a significant problem?
              Are there alternate approaches to solve this problem?
              Can we improve the techniques proposed in the paper?

            posted @ 2010-06-04 13:16 雪之精靈 閱讀(230) | 評(píng)論 (0)編輯 收藏

            2010年5月29日 #

            把自1970年1月1日以來(lái)的秒數(shù)轉(zhuǎn)化成年月日

            #include <iostream>
            using namespace std;

            bool IsRound(int year){
             if((year%100)&&(year%4==0)) return 1;
             if((year%100==0)&&(year%400==0)) return 1;
             return 0;
            }
            int main() {
             int n;
             int year_s[2]= {365*24*60*60, 366*24*60*60};
             int month_s[2][12]={{31,28,31,30,31,30,31,31,30,31,30,31},
               {31,29,31,30,31,30,31,31,30,31,30,31}};
             int day_s = 24*60*60;
             int hour_s = 60*60;
             int minute_s = 60;

             

             while(cin>>n){
                int temp = n;
                int year=1970;
                     int month=1;
                     int day=1;
                     int hour=0;
                     int minute=0;
                     int second=0;

                while(temp>=60){

                 int flag= IsRound(year);
                 if(temp>=year_s[flag]) { year++; temp-=year_s[flag]; }
                 else if(temp>=day_s){
                  int days = temp/day_s;
                  temp=temp%day_s;
                  int i=0;
                  int flag = IsRound(year);
                  int hh=31;
                  while(days>=hh){
                   days-=month_s[flag][i++];
                   hh=month_s[flag][i];
                   }
                  month+= i;
                  day+=days;

                 }else if(temp>=hour_s){
                      hour=temp/hour_s;
                      temp%=hour_s;
                 }else if(temp>=minute_s){
                       minute = temp/minute_s;
                       temp%=minute_s;
                 }
                 }
                 second = temp;

                    cout<<year<<"-";
                  if(month/10==0) cout<<"0";
                  cout<<month<<"-";
                  if(day/10==0) cout<<"0";
                  cout<<day<<" ";
                  if(hour/10==0) cout<<"0";
                  cout<<hour<<":";
                  if(minute/10==0) cout<<"0";
                  cout<<minute<<":";
                  if(second/10==0) cout<<"0";
                  cout<<second<<endl;
             }


             return 0;
            }

            posted @ 2010-05-29 13:54 雪之精靈 閱讀(2211) | 評(píng)論 (0)編輯 收藏

            2009年6月20日 #

            Don Knuth's Home Page

            http://www-cs-faculty.stanford.edu/~knuth/

            posted @ 2009-06-20 11:33 雪之精靈 閱讀(256) | 評(píng)論 (0)編輯 收藏

            2008年12月23日 #

            思維的方法

                     在這個(gè)網(wǎng)絡(luò)時(shí)代,幾乎任何知識(shí)性的問(wèn)題都可以迅速搜索或請(qǐng)教到答案。不過(guò),如何在已知知識(shí)之外發(fā)掘出未知知識(shí),如何解決未知問(wèn)題,那就還是要看個(gè)人的能力了,需要我們發(fā)揮自己的思維。
                     用優(yōu)秀的思維去解未知問(wèn)題,這樣可以成就大的學(xué)問(wèn)。我們需要培養(yǎng)自己的思維。開(kāi)始時(shí),是有意識(shí)的去用這些思維方式;然后,發(fā)展到潛意識(shí)地去用這些方式,這就到一個(gè)境界了。 http://blog.csdn.net/pongba/archive/2008/12/18/3549560.aspx

                      當(dāng)然,我們還要先養(yǎng)成自己解知識(shí)性問(wèn)題的基本功,并且要花大力氣,下苦功夫。這是基礎(chǔ)。

            posted @ 2008-12-23 11:09 雪之精靈 閱讀(223) | 評(píng)論 (0)編輯 收藏

            2008年10月22日 #

            DFS,BFS,DFS+ID

            In a Nutshell Search Time Space When to use
                DFS O(c k) O(k) Must search tree anyway, know the level the answers are on, or you aren't looking for the shallowest number.
                BFS O(c d ) O(c d ) Know answers are very near top of tree, or want shallowest answer.
                DFS+ID O(c d) O(d) Want to do BFS, don't have enough space, and can spare the time.
            d is the depth of the answer k is the depth searched d <= k Remember the ordering properties of each search. If the program needs to produce a list sorted shortest solution first (in terms of distance from the root node), use breadth first search or iterative deepening. For other orders, depth first search is the right strategy. If there isn't enough time to search the entire tree, use the algorithm that is more likely to find the answer. If the answer is expected to be in one of the rows of nodes closest to the root, use breadth first search or iterative deepening. Conversely, if the answer is expected to be in the leaves, use the simpler depth first search. Be sure to keep space constraints in mind. If memory is insufficient to maintain the queue for breadth first search but time is available, use iterative deepening.
                quote from http://ace.delos.com/usacotext2?a=y9SZdbB6WeB&S=rec

            posted @ 2008-10-22 19:02 雪之精靈 閱讀(564) | 評(píng)論 (0)編輯 收藏

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