你好，这是我的表格，但我无法弄清楚如何排列（x，y，z）坐标以便好看（我的意思是它们不在一条直线上：））并且每个框，它们开始出现在中心北，为了好看，我想要居中：）抱歉我的语言不好。提前谢谢。
这是我的代码，抱歉没有整理好，我还没有习惯 TeX.SE :)
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline
Step(k) & I.strategy & II.strategy & $\,$ Return $\,$ & Total Return &$\quad$ $g_{i}^r$ $\quad$ &$\quad$ $g_{k}^c$ $\quad$ & $\,$ $U_k$ $\,$ & $\,$ $V_k$ $\,$ & $\displaystyle \frac{minU_k}{k}$ & $ \displaystyle \frac{maxV_k}{k}$ \\[2ex] \hline
1 & $I_1$ & $II_1$ & 9 & 9 & (9,2,-1) & (9,-9,6) & (9,2,{\color{blue}-1}) & ({\color{red}9},-9,6) & -1 & 9 \\[1.5ex] \hline
2 & $I_1$ & $II_3$ &-1 & 8 & (9,2,-1) & (-1,0,2) & (18,4,{\color{blue}-2}) & ({\color{red}8},-9,8) & -1 & 4 \\[1.5ex] \hline
3 & $I_1$ & $II_3$ &-1 & 7 & (9,2,-1) & (-1,0,2) & (27,6,{\color{blue}-3}) & (7,-9,{\color{red}10}) & -1 & $\frac{10}{3}$ \\[1.5ex] \hline
4 & $I_3$ & $II_3$ & 2 & 9 & (6,-6,2) & (-1,0,2) & (33,0,{\color{blue}-1}) & (6,-9,{\color{red}12}) & $-\frac{1}{4}$ & $3$ \\[1.5ex] \hline
5 & $I_3$ & $II_3$ & 2 &11 & (6,-6,2) & (-1,0,2) & (39,{\color{blue}-6},1) & (5,-9,{\color{red}14}) & $-\frac{6}{5}$ & $\frac{14}{5}$ \\[1.5ex] \hline
6 & $I_3$ & $II_2$ &-6 & 5 & (6,-6,2) & (2,9,-6) & (45,{\color{blue}-12},3) & (7,0,{\color{red}8}) & $-2$ & $\frac{4}{3}$ \\[1.5ex] \hline
7 & $I_3$ & $II_2$ &-6 &-1 & (6,-6,2) & (2,9,-6) & (51,{\color{blue}-18},5) & ({\color{red}9},9,2) & $-\frac{18}{7}$ & $\frac{9}{7}$ \\[1.5ex] \hline
8 & $I_1$ & $II_2$ & 2 & 1 & (9,2,-1) & (2,9,-6) & (60,{\color{blue}-16},4) & (11,{\color{red}18},-4) & $-2$ & $\frac{9}{4}$ \\[1.5ex] \hline
9 & $I_2$ & $II_2$ & 9 &10 & (-9,9,0) & (2,9,-6) & (51,{\color{blue}-7},4) & (13,{\color{red}17},-10) & $-\frac{7}{9}$ & $\frac{17}{9}$ \\[1.5ex] \hline
10 & $I_2$ & $II_2$ & 9 &19 & (-9,9,0) & (2,9,-6) & (42,{\color{blue}2},4) & (15,{\color{red}26},-16) & $\frac{1}{5}$ & $\frac{13}{5}$ \\[1.5ex] \hline
11 & $I_2$ & $II_2$ & 9 &28 & (-9,9,0) & (2,9,-6) & (33,11,{\color{blue}4}) & (17,{\color{red}35},-22) & $\frac{4}{11}$ & $\frac{35}{11}$ \\[1.5ex] \hline
12 & $I_2$ & $II_3$ & 0 &28 & (-9,9,0) & (-1,0,2) & (24,20,{\color{blue}4}) & (16,{\color{red}35},-20) & $\frac{1}{3}$ & $\frac{35}{12}$ \\[1.5ex] \hline
13 & $I_2$ & $II_3$ & 0 &28 & (-9,9,0) & (-1,0,2) & (15,29,{\color{blue}4}) & (15,{\color{red}35},-18) & $\frac{4}{13}$ & $\frac{35}{13}$ \\[1.5ex] \hline
14 & $I_2$ & $II_3$ & 0 &28 & (-9,9,0) & (-1,0,2) & (6,38,{\color{blue}4}) & (14,{\color{red}35},-16) & $\frac{4}{14}$ & $\frac{35}{14}$ \\[1.5ex] \hline
15 & $I_2$ & $II_3$ & 0 &28 & (-9,9,0) & (-1,0,2) & ({\color{blue}-3},47,4) & (13,{\color{red}35},-14) & $-\frac{3}{5}$ & $\frac{7}{3}$ \\[1.5ex] \hline
\end{tabular}
答案1
我建议以更开放的眼光看待表格，并采用一些简化输入的技巧。
不过，我相信斜线形式的10/3渲染效果会比 更好\frac{10}{3}。
\documentclass{article}
\usepackage[landscape,a4paper]{geometry}
\usepackage{array,booktabs}
\usepackage{amsmath}
\usepackage{xcolor}
\newcommand{\bn}[1]{% blue number
\textcolor{blue}{#1}%
}
\newcommand{\rn}[1]{% red number
\textcolor{red}{#1}%
}
\begin{document}
\begin{table}
\centering
\renewcommand{\arraystretch}{1.2}
\addtolength{\tabcolsep}{-.2pt}
\begin{tabular}{
*{3}{c}
*{8}{>{$}c<{$}}
}
\toprule
Step($k$) &
Strategy I &
Strategy II &
\multicolumn{1}{c}{Return} &
\multicolumn{1}{c}{Total Return} &
g_{i}^r &
g_{k}^c &
U_k &
V_k &
\dfrac{\min U_k}{k} &
\dfrac{\max V_k}{k} \\
\midrule
1 & I$_1$ & II$_1$ & 9 & 9 & (9,2,-1) & (9,-9,6) & (9,2,\bn{-1}) & (\rn{9},-9,6) & -1 & 9 \\
2 & I$_1$ & II$_3$ &-1 & 8 & (9,2,-1) & (-1,0,2) & (18,4,\bn{-2}) & (\rn{8},-9,8) & -1 & 4 \\
3 & I$_1$ & II$_3$ &-1 & 7 & (9,2,-1) & (-1,0,2) & (27,6,\bn{-3}) & (7,-9,\rn{10}) & -1 & \frac{10}{3} \\
4 & I$_3$ & II$_3$ & 2 & 9 & (6,-6,2) & (-1,0,2) & (33,0,\bn{-1}) & (6,-9,\rn{12}) & -\frac{1}{4} & 3 \\
5 & I$_3$ & II$_3$ & 2 &11 & (6,-6,2) & (-1,0,2) & (39,\bn{-6},1) & (5,-9,\rn{14}) & -\frac{6}{5} & \frac{14}{5} \\
6 & I$_3$ & II$_2$ &-6 & 5 & (6,-6,2) & (2,9,-6) & (45,\bn{-12},3) & (7,0,\rn{8}) & -2 & \frac{4}{3} \\
7 & I$_3$ & II$_2$ &-6 &-1 & (6,-6,2) & (2,9,-6) & (51,\bn{-18},5) & (\rn{9},9,2) & -\frac{18}{7} & \frac{9}{7} \\
8 & I$_1$ & II$_2$ & 2 & 1 & (9,2,-1) & (2,9,-6) & (60,\bn{-16},4) & (11,\rn{18},-4) & -2 & \frac{9}{4} \\
9 & I$_2$ & II$_2$ & 9 &10 & (-9,9,0) & (2,9,-6) & (51,\bn{-7},4) & (13,\rn{17},-10) & -\frac{7}{9} & \frac{17}{9} \\
10 & I$_2$ & II$_2$ & 9 &19 & (-9,9,0) & (2,9,-6) & (42,\bn{2},4) & (15,\rn{26},-16) & \frac{1}{5} & \frac{13}{5} \\
11 & I$_2$ & II$_2$ & 9 &28 & (-9,9,0) & (2,9,-6) & (33,11,\bn{4}) & (17,\rn{35},-22) & \frac{4}{11} & \frac{35}{11} \\
12 & I$_2$ & II$_3$ & 0 &28 & (-9,9,0) & (-1,0,2) & (24,20,\bn{4}) & (16,\rn{35},-20) & \frac{1}{3} & \frac{35}{12} \\
13 & I$_2$ & II$_3$ & 0 &28 & (-9,9,0) & (-1,0,2) & (15,29,\bn{4}) & (15,\rn{35},-18) & \frac{4}{13} & \frac{35}{13} \\
14 & I$_2$ & II$_3$ & 0 &28 & (-9,9,0) & (-1,0,2) & (6,38,\bn{4}) & (14,\rn{35},-16) & \frac{4}{14} & \frac{35}{14} \\
15 & I$_2$ & II$_3$ & 0 &28 & (-9,9,0) & (-1,0,2) & (\bn{-3},47,4) & (13,\rn{35},-14) & -\frac{3}{5} & \frac{7}{3} \\
\bottomrule
\end{tabular}
\end{table}
\end{document}
