$5 \cdot 37$ If the graph of the quadratic function $y=a x^{2}+b x+c(a \neq 0)$ has its vertex in the first quadrant, and passes through the points $(0,1)$ and $(-1,0)$, then the range of values for $S=a+b+c$ is
(A) $0<S<1$.
(B) $0<S<2$.
(C) $1<S<2$.
(D) $-1<S<1$.
(Shaanxi Province, China Junior High School Mathematics Competition, 1997)
[Solution 1] Let $x=0, y=1$ and $x=-1, y=0$,
we obtain
$$
\begin{aligned}
c & =1, a=b-1 . \\
\therefore \quad S=a+b+c & =2 b .
\end{aligned}
$$

From the problem, we know $-\frac{b}{2 a}>0$, and $a<0$.
Also, from $b=a+1$ and $a<0$, it leads to a contradiction. Therefore, option $(\mathrm{D})$ can be eliminated.
Thus, the answer is $(B)$.
[Solution 3] Eliminate $(A)$ and $(C)$ as above.
If $a+b+c=0$, then when $x=-1$ and $x=1$, the $y$ values are both 0, indicating that the y-axis is the axis of symmetry of the parabola, which contradicts the fact that the vertex of the parabola is in the first quadrant. Therefore, option (D) can be eliminated.
Thus, the answer is $(B)$.