1. A triangle with all sides as integers, and the longest side being 11, has
$\qquad$ possibilities.
Ni, 1.36.
Let the three sides be $x, y, z$, and $x \leqslant y \leqslant z, z=11, x+$ $y>z$, then
$$
6 \leqslant y \leqslant 11 .
$$

When $y=6, x=6$, it forms 1 triangle;
When $y=7, x$ is $5, 6, 7$, it forms 3 triangles; .....
When $y=11, x$ is $1, 2, \cdots, 11$, it forms 11 triangles. Therefore, there are a total of $1+3+5+\cdots+11=36$ triangles.