如何将Average BER右边的数据与average ber两行中间对齐，谢谢各位大佬！ 代码如下： \begin{table}[!ht] \centering \caption{\label{tab:2}Average BER and average channel capacity of an FSO communication system under weak and moderate turbulence conditions with various types of noise ($I_o$= $\rm{ - }20$ dBm).} \begin{tabular}{ccccccc} \\\hline \noalign{\smallskip} \tabincell{c}{Turbulence \\ conditions} & Performance & \tabincell{c}{Thermal \\ noise} & \tabincell{c}{Background \\ noise} & \tabincell{c}{Quantum \\ noise} & \tabincell{c}{Total \\ noise} \\ \noalign{\smallskip}\hline\noalign{\smallskip} Weak & \tabincell{c}{Average \\ BER \\ Average \\ capacity \\ (bps)} & \tabincell{c}{$2 \times {10^{{\rm{ - }}7}}$ \\ \\ 10.50} & \tabincell{c}{$9.5 \times {10^{{\rm{ - }}14}}$ \\ \\ 14.72} & \tabincell{c}{$8.5 \times {10^{{\rm{ - }}19}}$ \\ \\ 17.19} & \tabincell{c}{$2.5 \times {10^{{\rm{ - }}7}}$ \\ \\ 10.41} \\ Moderate & \tabincell{c}{Average \\ BER \\ Average \\ capacity \\ (bps)} & \tabincell{c}{$1.7 \times {10^{{\rm{ - }}4}}$ \\ \\ 5.11} & \tabincell{c}{$1.4 \times {10^{{\rm{ - }}6}}$ \\ \\ 7.22} & \tabincell{c}{$7.2 \times {10^{{\rm{ - }}8}}$ \\ \\ 8.46} & \tabincell{c}{$1.8 \times {10^{{\rm{ - }}4}}$ \\ \\5.07} \\\hline \noalign{\smallskip} \end{tabular} \end{table}