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DBwater的博客2016 Multi-University Training Contest4题解. 2016 Multi-University Training Contest 1题解.
http://www.dbwater.net/
2016 Multi-University Training Contest4题解. 2016 Multi-University Training Contest 1题解.
http://www.dbwater.net/
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DBwater的博客 | dbwater.net Reviews
https://dbwater.net
2016 Multi-University Training Contest4题解. 2016 Multi-University Training Contest 1题解.
用vim打造py开发环境 | DBwater的博客
http://www.dbwater.net/2016/10/30/python-vim
垃圾电脑打开pycharm实在是太卡了,于是就想通过配置一些vim插件来使vim更加好使用一些,现在记录下我的经历以免以后忘记,c,c ,html,其实也可以用vim来打造相应的环境。 Git clone https:/ github.com/VundleVim/Vundle.vim.git. Set the runtime path to include Vundle and initialize. Set rtp = /.vim/bundle/Vundle.vim. Alternatively, pass a path where Vundle should install plugins. Call vundle#begin(' /some/path/here'). Let Vundle manage Vundle, required. Plugin 'gmarik/Vundle.vim'. Plugin 'tmhedberg/SimpylFold' 防止折叠超过限制. Plugin 'vim-scripts/indentpython.vim' 解决自动缩进的一些小问题. Let g:ycm se...
lunix命令行学习第一篇 | DBwater的博客
http://www.dbwater.net/2016/10/18/bash
大家知道命令行能够连用比如:cd /urs; ls; cd -首先更改目录到/usr,然后列出目录 内容,最后回到原始目录,那么现在我们可以用alias命令把这一串命令变为一个命令使用之前可以先用type命令查看你取的别名是已经存在,如果存在你就只能换一个名字了。 Eg: alias foo=’cd/usr; ls; cd-‘把刚才我们的操作取名为foo,那么输入foo,则表示连用了三个命令。 2016 Multi-University Training Contest4题解. 2016 Multi-University Training Contest 1题解.
Archives | DBwater的博客
http://www.dbwater.net/archives
2016 Multi-University Training Contest4题解. 2016 Multi-University Training Contest 1题解.
ubutn上常用软件的配置 | DBwater的博客
http://www.dbwater.net/2016/10/31/ubuntu
Sudo apt-get install git-core. Sudo apt-get install zsh. Wget https:/ github.com/robbyrussell/oh-my-zsh/raw/master/tools/install.sh. Sudo apt-get install python-pip. Trusted-host = pypi.douban.com. Index-url = http:/ pypi.douban.com/simple. Sudo apt-get install update. Sudo apt-get install shadowsocks-qt5. Sudo dpkg -i wps-office 8.1.0.3724 b1p2 i386.deb. Sudo apt-get install -f. Sudo dpkg -i wps-office 8.1.0.3724 b1p2 i386.deb. Sudo cp * /usr/share/fonts. Cnpm install hexo-cli -g.
正则表达式引擎的构建(1) | DBwater的博客
http://www.dbwater.net/2016/10/19/regex1
第一步我们可以先把所给的正则表达式 中缀表达式 转换为后缀表达式(这一部分就不做描述,可以直接观看完整源代码),你可以用一个特殊符号来表示连接预算符号,这里离我们用.来表示,如 a(bb) a 转换一个等效的后缀表达式 abb. .a。 例如,编译完abb的abb. .a,栈中包含了a,b,b的NFA局部。 State *start; / 表示局部的开始状态节点. Ptrlist *out; / 表示局部的尾节点因为尾节点可能分叉,所以需要用一个list来表示。 Stackp, e1, e2, e;. Fprintf(stderr, "postfix: %s n", postfix);. Push(s) *stackp = s. P=postfix; *p; p ){. E = pop();. Patch(e.out, &matchstate);. S = state(*p, NULL. Push(frag(s, list1(&s- out) ;. E2 = pop();. E1 = pop();. Patch(e1.out, e2.start);. E2 = pop();. E1 = pop();.
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hexo使用注意事项 | Kingofprank
http://kingofprank.com/2016/08/07/hexo_config
用hexo s后无法进入127.0.0.1:4000. 刚开始装完系统后完全没什么问题,突然有一天就不能进入127.0.0.1:4000了。 Hexo s -p 5000. 这样在浏览器上输入127.0.0.1:5000就可以访问了。 Codeforces Round 365 (Div. 2) 题解. 用hexo s后无法进入127.0.0.1:4000. 主题 - NexT.Mist.
Kingofprank
http://kingofprank.com/page/2
博弈论又被称为对策论 Game Theory 既是现代数学的一个新分支,也是运筹学的一个重要学科。 主题 - NexT.Mist.
2016多校联合训练第6场 | Kingofprank
http://kingofprank.com/2016/08/02/muti-uni-contest6
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others). Total Submission(s): 2522 Accepted Submission(s): 326. Intput contains multiple test cases. The first line is an integer 1 T 100, the number of test cases. Each case begins with an integer n, indicating the number of the heaps, the next line contains N integers s[0],s[1],.,s[n 1], representing heaps with s[0],s[1], s[n 1] objects respectively.(1 n 106,1 s[i] 109). Second player wins. First player wins. I n;i ){. 因为这个组合数非常庞大...
博弈论(下)——SG函数 | Kingofprank
http://kingofprank.com/2016/08/03/Game(2)
Codeforces Round 365 (Div. 2) 题解. 主题 - NexT.Mist.
华硕Y581LD4200安装SSD教程 | Kingofprank
http://kingofprank.com/2016/07/31/SSD-install
Cn windows 7 ultimate with sp1 x64 dvd u 677408.iso. 主题 - NexT.Mist.
归档 | Kingofprank
http://kingofprank.com/archives
Codeforces Round 365 (Div. 2) 题解. Educational Codeforces Round 15 题解. 主题 - NexT.Mist.
2016多校联合训练第4场 | Kingofprank
http://kingofprank.com/2016/07/28/multi-uni-contest4
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others). Total Submission(s): 0 Accepted Submission(s): 0. P is a permutation of the integers from 1 to N(index starting from 1). Here is the code of Bubble Sort in C . For(int i=1;i =N; i). For(int j=N,t;j i; j). T=P[j],P[j]=P[j-1],P[j-1]=t;. After the sort, the array is in increasing order? Wants to know the absolute values of difference of rightmost place and leftmost place for every number it reached. 1 = N = 100000. I=n;i = 1.
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DBwater的博客
2016 Multi-University Training Contest4题解. 2016 Multi-University Training Contest 1题解.
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