8. The sequence $\left\{x_{n}\right\}: 1,3,3,3,5,5,5,5,5, \cdots$ is formed by arranging all positive odd numbers in ascending order, and each odd number $k(k=1,3,5, \cdots)$ appears consecutively $k$ times. If the general term formula of this sequence is $x_{n}=a[\sqrt{b n+c}]+d$ (where $[x]$ denotes the greatest integer not greater than $x$), then $a+b+c+d=$ . $\qquad$
8.3.

Given $x_{k^{2}+1}=x_{k^{2}+2}=\cdots=x_{(k+1)^{2}}=2 k+1$, that is, when $k^{2}+1 \leqslant n \leqslant(k+1)^{2}$, we have $x_{n}=2 k+1(k=[\sqrt{n-1}])$.
Therefore, $x_{n}=2[\sqrt{n-1}]+1$.
Hence, $(a, b, c, d)=(2,1,-1,1), a+b+c+d=3$.