设函数 $f(x)$ 连续, 满足 $\int_0^1 f(x) \mathrm{d} x=0$. 若 $\int_0^1 \mathrm{e}^{1-x} f\left(x \mathrm{e}^{1-x}\right) \mathrm{d} x=1$, 则 $\int_0^1 x \mathrm{e}^{1-x} f\left(x \mathrm{e}^{1-x}\right) \mathrm{d} x$ $= $
$\text{A.}$ -1
$\text{B.}$ 0
$\text{C.}$ 1
$\text{D.}$ e