1.111 $111222222-333333=x^{2}$ and $x>$ 0. Then $x=(\quad)$.
(A) 332233
(B) 223377
(C)333 333
(D) 331177
$\begin{array}{l}\text {-1.C. } \\ 111111222222-333333 \\ =111111 \times(10^{6}+2)-111111 \times 3 \\ =111111 \times(10^{6}-1)=\frac{999999}{9}(10^{6}-1) \\ =\frac{10^{6}-1}{9}(10^{6}-1)=\left(\frac{10^{6}-1}{3}\right)^{2}=333333^{2} .\end{array}$