116 Let $f(x)=a x^{2}+b x+c, a, b, c \in \mathbf{R}, a>100$, such that $|f(x)| \leqslant 50$ holds for different integer $x$ at most
A. 0
B. 1
C. 2
D. At least 3
116 C. When $x=0$ and $x=2$, $|f(x)|=0 \leqslant 50$ holds.
Assume there are 3 different integers $x_{1}, x_{2}, x_{3}$ such that
$$
\left|f\left(x_{i}\right)\right| \leqslant 50, i=1,2,3,
$$

Then at least 2 of $x_{1}, x_{2}, x_{3}$ are on the same side of $-\frac{b}{2 a}$. Without loss of generality, assume $x_{1}>x_{2} \geqslant-\frac{b}{2 a}$. Since $a > 0$, $f(x)$ is an increasing function on $\left(-\frac{b}{2 a},+\infty\right)$, so $50 \geqslant f\left(x_{1}\right) \geqslant f\left(x_{2}+1\right)$. That is,
$$
a\left(x_{2}+1\right)^{2}+b\left(x_{2}+1\right)+c \leqslant 50 .
$$

Also, $a x_{2}^{2}+b x_{2}+c \geqslant-50$,
Subtracting the two equations gives $a\left(2 x_{2}+1\right)+b \leqslant 100$. Therefore,
$$
100 \geqslant a\left(2 x_{2}+1\right)+b \geqslant a\left(-\frac{b}{a}+1\right)+b=-b+a+b=a .
$$

This contradicts $a>100$.