1.百分号
95\%
2.下标
b_{ij}
3.表格(excel2latex)
- 符号说明
egin{table}[H] centering egin{tabular}{m{100pt}<{centering} m{100pt}<{centering} m{200pt}<{centering}} oprule 序列 & 符号& 符号说明\ midrule 1 & $t_{ij}$ & 第i个会员第j种DVD的态度\ 2 & $s_j$ & 愿意观看第j张的DVD的会员数 \ 3 & $p_j$ & 会员愿意观看第j种DVD的概率 \ 4 & $x_{ij}$ & 第i个会员是否被分配到第j张DVD \ 5 & $a_{ij}$ & 第i个会员对第j张DVD的满意度 \ 6 & $k_{i}$ & 第i个会员有没有被分配到DVD \ 7 & $c_j$ & 第j种DVD购买的数量 \ 8 & $b_{ij}$ & 第i个会员是否返还第j张DVD \ 9 & $w_j$ & 两阶段共需要购买的j种DVD数量 \ 10 & $z_{ij}$ & 第二阶段针第i个会员需求的第j种DVD \ ottomrule end{tabular} end{table}
- 一般表格【有表头】
egin{table}[H] caption{会员对5种DVD的愿意观看概率} centering egin{tabular}{|c|c|c|c|c|c|} hline $mbox{DVD名称}$ & $mbox{DVD1}$ & $mbox{DVD2}$ & $mbox{DVD3}$ & $mbox{DVD4}$ & $mbox{DVD5}$\ hline $mbox{第j张愿意观看概率}$ & $0.2$ & 0.1 & $0.05$ & $0.025$ & $0.01$ \ hline end{tabular} end{table}
- 改变字体大小
-
iny
- 双栏
egin{table}[H] centering caption{前30个会员分配情况} egin{tabular}{r|rrr||r|rrr} oprule 会员编号 &DVD1&DVD2&DVD3& 会员编号&DVD1&DVD2&DVD3 \ midrule C001 & D008 & D041 & D098 & C016 & D010 & D084 & D097 \ C002 & D006 & D044 & D062 & C017 & D047 & D051 & D067 \ C003 & D032 & D050 & D080 & C018 & D041 & D060 & D078 \ C004 & D007 & D018 & D041 & C019 & D066 & D084 & D086 \ C005 & D011 & D066 & D068 & C020 & D045 & D061 & D089 \ C006 & D019 & D053 & D066 & C021 & D045 & D050 & D053 \ C007 & D026 & D066 & D081 & C022 & D038 & D055 & D057 \ C008 & D031 & D035 & D071 & C023 & D029 & D081 & D095 \ C009 & D053 & D078 & D100 & C024 & D037 & D041 & D076 \ C010 & D041 & D055 & D085 & C025 & D009 & D069 & D094 \ C011 & D059 & D063 & D066 & C026 & D022 & D068 & D095 \ C012 & D002 & D031 & D041 & C027 & D050 & D058 & D078 \ C013 & D021 & D078 & D096 & C028 & D008 & D034 & D082 \ C014 & D023 & D052 & D089 & C029 & D026 & D030 & D055 \ C015 & D013 & D052 & D085 & C030 & D037 & D062 & D098 \ ottomrule end{tabular}% label{tab:addlabel}% end{table}%
4.等式
egin{equation} E=0.6*2+0.44*1=1.6 end{equation}
egin{equation*} t_{ij}= egin{cases} 1,& ext{第i个会员愿意观看第j种DVD}\ 0,& ext{第i个会员不愿意观看第j种DVD}\ end{cases} end{equation*}
- 求和
$$s_j=sum_{i=1}^{100000}t_{ij},j=1,2,ldots,5$$
- 极限
egin{equation} lim_{n o infty}P(frac{eta_{x}-np}{sqrt{np(1-p)}}leq x)=int _{-infty}^{x}frac{1}{sqrt{2pi}}e^{-frac{t^2}{2}} dt =Phi (x) end{equation}
5.另起一段,另起一页
par
ewpage
6.字体加粗
extbf{(2)三个月内95\%}
7.矩阵
$$X=egin{pmatrix} x_{1,1}&x_{1,2}&ldots&x_{1,100}\ x_{2,1}&x_{2,2}&ldots&x_{2,100}\ vdots &vdots & &vdots\ x_{1000,1}&x_{1000,2}&ldots&x_{1000,100}\ end{pmatrix}$$
- 相乘
$$ left(egin{array}{c} X_1 \ X_2 \ dots \ X_p end{array} ight) = left(egin{array}{cccc} a_{11} & a_{12} & dots &a_{1m} \ a_{21} & a_{22} & dots &a_{2m} \ dots & dots & dots &dots \ a_{p1} & a_{p2} & dots &a_{pm} \ end{array} ight) left(egin{array}{c} F_1 \ F_2 \ dots \ F_m end{array} ight) left(egin{array}{c} varepsilon_1 \ varepsilon_2 \ dots \ varepsilon_p end{array} ight) $$
8.空格
quad
9.线性规划
egin{center} $$maxsumlimits_{i=1}^{1000}sumlimits_{j=1}^{100} = x_{ij}a_{ij} $$ egin{equation*} s.t. egin{cases} sumlimits_{j=1}^{100}x_{ij} = 3k_{i}\ sumlimits_{i=1}^{1000}x_{ij} leq num_j,j = 1,2,ldots,100 \ x_{ij},k_{ij} = 0 ext{或} 1; \ i = 1,2,ldots,1000; \ j = 1,2,ldots,100 \ end{cases} end{equation*} end{center}
10.加帽子
widehat{D_t}
11.参考文献
ibitem{1}
12.附录
zihao{5}songti hesection {附录1 quad 第二题lingo实现代码} egin{verbatim} end{verbatim}