2010年6月12日
#
sudo cp /etc/apt/sources.list /etc/apt/sources.list_backup 備份現有更新源
sudo gedit /etc/apt/sources.list 編輯更新源列表
從下面各服務器列表內容中選擇一段替換文件中的所有內容
保存編輯好的文件，執行以下命令更新。
sudo apt-get update (修改sources.list后必須運行！)
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
2010年6月4日
#
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?
2010年5月29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;
}
2009年6月20日
#
2008年12月23日
#
在這個網絡時代，幾乎任何知識性的問題都可以迅速搜索或請教到答案。不過，如何在已知知識之外發掘出未知知識，如何解決未知問題，那就還是要看個人的能力了，需要我們發揮自己的思維。
用優秀的思維去解未知問題，這樣可以成就大的學問。我們需要培養自己的思維。開始時，是有意識的去用這些思維方式；然后，發展到潛意識地去用這些方式，這就到一個境界了。 http://blog.csdn.net/pongba/archive/2008/12/18/3549560.aspx
當然，我們還要
先養成自己解知識性問題的基本功，并且要花大力氣，下苦功夫。這是基礎。
2008年10月22日
#
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