1. Let $S_{n}=1-2+3-4+\cdots+(-1)^{n-1} n, n=1,2, \cdots$, then $S_{17}+S_{33}+S_{50}$ equals
A. 0
B. 1
C. 6
D. -1
E. 2
1. $\mathrm{B}$
(1) When $n$ is even
$$
\begin{aligned}
S_{n} & =1-2+3-4+\cdots+(-1)^{n-1} n=(1-2)+(3-4)+\cdots+[(n-1)-n] \\
& =-1-1-\cdots-1=-\frac{n}{2}
\end{aligned}
$$
(2) When $n$ is odd
$$
\begin{array}{l}
\begin{aligned}
S_{n} & =1+\{(-2+3)+(-4+5)+\cdots+[-(n-1)+n]\} \\
& =1+\{1+1+\cdots+1\}=1+\frac{n-1}{2}=\frac{n+1}{2} .
\end{aligned} \\
\text { From the above, } S_{17}+S_{33}+S_{50}=\frac{18}{2}+\frac{34}{2}-\frac{50}{2}=1 .
\end{array}
$$

Therefore, the answer is B.