1. Given sets $M=\left\{a_{1}, a_{2}, \cdots, a_{2 n+1}\right\}, N$ $=\left\{-2^{2 n},-2^{2 n-1}, \cdots,-2,0,2, \cdots, 2^{2 n}\right\}$. If the injection $f: M \rightarrow N$ satisfies
$$
f\left(a_{1}\right)+f\left(a_{2}\right)+\cdots+f\left(a_{2 n+1}\right)=0,
$$

then the number of such injections $f$ is $(\quad)$.
(A) $(2 n+1)!\mathrm{C}_{2 n}^{n}$
(B) $(2 n+1)!\mathrm{C}_{2 n+1}^{n}$
(C) $(2 n+1)!\mathrm{C}_{4 n+1}^{2 n+1}$
(D) $(2 n+1)!\mathrm{C}_{4 n}^{2 n}$
-.1.A.
For any positive integer $m$, we have
$$
2^{m}>2^{m-1}+\cdots+2=2^{m}-2 \text {. }
$$

Thus, the highest power of 2 in $\left|f\left(a_{j}\right)\right| j=1,2, \cdots, 2 n+1$ is the same as the highest power of -2. Otherwise, the sum would not be 0. By canceling these two terms, similarly, $2^{i}$ and $-2^{i}$ either both appear or do not appear in $\left|f\left(a_{j}\right)\right| j=1,2, \cdots, 2 n+1 \mid$ $(i=1,2, \cdots, 2 n)$.
Since there are an odd number of terms, there exists an $i$ such that $f\left(a_{i}\right)=0$.
Given that $\left\{2,2^{2}, \cdots, 2^{2 n}\left\{, \mid-2,-2^{2}, \cdots,-2^{2 n}\right\}\right.$ each appear $n$ times in $\left\{f\left(a_{j}\right) \mid j=1,2, \cdots, 2 n+1\right\}$, the number of such injections is $(2 n+1)!\mathrm{C}_{n, n}^{n}$.