First let me state a result of Kazdan and Warner

Let $M$ be a compact orientable two dimensional manifold. Let $f:M \rightarrow \mathbb{R}$ be a function that has the same sign as the Euler characteristic of $M$ at some point $p$. Then $M$ admits a Riemannian metric such that the Gaussian Curvature is equal to $f$.

I want to know if there is a combinatorial analogue of this statement. Namely, is the following statement true:

Let $M$ be a compact orientable two dimensional manifold. Let $f:M \rightarrow \mathbb{R}$ be a function that is zero except at a finite set of points $p_1 \ldots p_n$. At one of those points, $f$ has the same sign as the Euler characteristic of $M$. Then does $M$ admit a triangulation with $p_1 \ldots p_n$ as vertices such that the ``Curvature'' at each of these points is same as $f$?

Here by curvature we mean the angle deficit from $2 \pi$. And the curvature of $M$ at any point that is not a vertex is defined to be zero.