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.