Kodaira type vanishing for an ample divisor + effective divisor with SNC support Norimatsu's lemma says that on a smooth projective complex variety $X$ of dimension $n$, then we have $H^i(X,\mathcal O_X(-A-E))=0$  for $i < n$ when $A$ is an ample divisor and $E$ is a simple normal crossings (SNC) divisor.  Does this statement remain true if $E$ is an effective divisor with SNC support?  In other words, if $\sum E_i$ is SNC, do we still have these vanishings for $\mathcal O_X(-A-\sum a_i E_i)$ where $a_i >0$?  Or is there a counterexample?
 A: Let $X$ be an arbitrary smooth projective surface, $A$ an arbitrary ample divisor and $E\subset X$ a smooth proper curve. Consider the short exact sequence
$$
0\to \mathscr O_X(-A-(a+1)E)\to \mathscr O_X(-A-aE)\to \mathscr O_X(-A-aE)|_E\to 0 
$$
If $H^0(X, \mathscr O_X(-A-aE))=0$, then 
$H^1(X, \mathscr O_X(-A-(a+1)E))= 0$ only if
$H^0(E, \mathscr O_X(-A-aE)|_E)= 0$. Therefore, if $(A+aE)\cdot E\ll 0$, then 
the desired vanishing will fail.
So, for example if $E$ is such that $E^2<0$, then this will happen for $a\gg 0$. 
If, as in Francesco's example, $X$ is a Del Pezzo surface, $A=-K_X$, and $E$ is an exceptional curve, then $(-A-aE)\cdot E=K_X\cdot E -aE^2=-1+a$ and hence $\mathscr O_X(-A-aE)|_E\simeq \mathscr O_E(a-1)$, so if $a>0$, then $H^0(E, \mathscr O_X(-A-aE)|_E)\neq 0$ and hence this gives another proof that in Francesco's example the desired vanishing fails as soon as the coefficient of $E$ is at least $2$. 

A similar example works in higher dimensions as long as $-E|_E$ is ample, for instance if it is the exceptional divisor of the blow up of a point (probably even under more general conditions). In other words, you cannot expect the vanishing you would like unless $E$ has some semi-positivity property (say $E$ is nef), but in that case you probably get it directly (since for example if $E$ is nef, then $A+aE$ is ample).
A: Here is a counterexample.
Let $X$ be a smooth cubic surface in $\mathbf{P}^3$ and $E$ a line on it; moreover take $A=-K_X$.
Then by Serre duality $$H^1(X, -A-aE)=H^1(X, K_X + A + aE)=H^1(X, aE).$$
On the other hand $h^0(X, aE)=1$ since $E^2=-1$ and $h^2(X, aE)=h^0(X, K_X-aE)=0$ since $K_X$ is not effective. 
Now Riemann-Roch yields $$\chi(X, aE)=\frac{aE(aE-K_X)}{2}+\chi(\mathcal{O}_X)=\frac{-a^2+a}{2}+1,$$
hence $$h^1(X, aE)=\frac{a(a-1)}{2}$$ which is not zero for $a \geq 2$.
