已知等差数列$\{a_{n}\}$的前$n$项和为$S_{n}.$若数列$\{b_{n}\}$满足：对任意的$n\in {\bf N}^{*}$，都有$a_{n}+b_{n}=n-1$，且${S}_{n}={b}_{n}^{2}$，则$a_{10}=(\qquad)$．

$10$

$19$

$20$

$39$

$\because $ 等差数列$\{a_{n}\}$的前$n$项和为$S_{n}$，对任意的$n\in {\bf N}^{*}$，都有$a_{n}+b_{n}=n-1$，且${S}_{n}={b}_{n}^{2}$，

$\therefore \begin{cases}{a}_{1}+{b}_{1}=0\\ {S}_{1}={a}_{1}={b}_{1}^{2}\end{cases}\Rightarrow \begin{cases}{a}_{1}=1\\ {b}_{1}=-1\end{cases}$或$\begin{cases}{a}_{1}=0\\ {b}_{1}=0\end{cases}$，

设$\{a_{n}\}$公差为$d$，则$b_{n}=-a_{n}+n-1=(1-d)n+d-a_{1}-1$，

又${S}_{n}={b}_{n}^{2}$，

由${S}_{n}=\dfrac{d}{2}{n}^{2}+\left({a}_{1}-\dfrac{d}{2}\right)n$

$={(1-d)}^{2}{n}^{2}+2(1-d)(d-{a}_{1}-1)n+{(d-{a}_{1}-1)}^{2}$，

对照系数得：$\begin{cases}\dfrac{d}{2}={(1-d)}^{2}\\ d-{a}_{1}-1=0\end{cases}\Rightarrow \begin{cases}\dfrac{d}{2}={(1-d)}^{2}\\ {a}_{1}=\dfrac{d}{2}\\ d-{a}_{1}-1=0\end{cases}$，解得$\begin{cases}d=2\\ {a}_{1}=1\end{cases}$，

$\therefore a_{10}=a_{1}+9d=19$．

故选：$\rm B$