Short exact sequence and cohomology group Given a short exact sequence for some finite groups $A,B,C$,
$$1\to A \to B \to C \to 1,$$
how could we construct an exact sequence of their cohomology group out of it?
One version of the story I found is here,
$$1\to H^0(G,A) \to H^0(G,B) \to H^0(G,C)\to H^1(G,A) \to H^1(G,B) \to H^1(G,C)$$
but I am looking for a relation of an exact sequence relating the following cohomology groups by group homomorphism:
$H^d(A,\mathbb{R}/\mathbb{Z})$, $H^d(B,\mathbb{R}/\mathbb{Z})$, $H^d(C,\mathbb{R}/\mathbb{Z})$, $H^{d-1}(A,\mathbb{R}/\mathbb{Z})$, $H^{d-1}(B,\mathbb{R}/\mathbb{Z})$, $H^{d-1}(C,\mathbb{R}/\mathbb{Z})$.

Is there a relation like 
$$\to H^d(A,\mathbb{R}/\mathbb{Z}) \to H^d(B,\mathbb{R}/\mathbb{Z}) \to H^d(C,\mathbb{R}/\mathbb{Z})\to ?$$
or the other way around
$$\to H^d(C,\mathbb{R}/\mathbb{Z}) \to H^d(B,\mathbb{R}/\mathbb{Z}) \to H^d(A,\mathbb{R}/\mathbb{Z})\to ?$$
and how do we proceed the sequence?

 A: As requested, I am turning my comments into an answer.  
First, given a short exact sequence $$1 \to A \to B \to C \to 1$$ of finite discrete groups, one can think of them as a sequence of topological spaces $$K(A,1) \to K(B,1) \to K(C,1).$$  If this were a cofiber sequence, then we would get a long exact sequence in cohomology as usual, but unfortunately this is not the case.  Instead, this is a fibration, so we have the (Lyndon-Hochschild-)Serre spectral sequence.  More precisely, suppose we are given compatible systems of local coefficients, which in this case amounts to a $B$-module $M$.  Then we have a spectral sequence $$E_2^{p,q} = H^p(C, H^q(A, M)) \Rightarrow H^{p+q}(B,M).$$
This is a first-quadrant spectral sequence, and some of the terms at the lower left corner of he $E_2$ page are unaffected by diffentials and hence survives intact to the $E_\infty$ page.  Then, when reconstructing $H^*(B, M)$ from the associated graded on the $E_\infty$ page, we find that we get a short "long" exact sequence $$0 \to H^1(C, M^A) \to H^1(B, M) \to H^1(A, M)^C \xrightarrow{\delta} H^2(C, M^A) \to H^2(B, M).$$  The map $\delta$ is given by the $d_2$ differential $E_2^{0,1} \to E_2^{2,0}$.  In group cohomology, this sequence is called the inflation-restriction sequence.  However, this sequence can be extended only for a couple more terms, but even then one of the terms would be less explicit.  For example, the next term in the sequence would be $H^1(C, H^1(A,M))$.  
On the other hand, if you can show that $E_2^{p,q} = 0$ for $0 < p \leq d$ and $0 < q < d$, then you would have a five-term exact sequence starting in a higher degree $$0 \to H^d(C, M^A) \to H^d(B, M) \to H^d(A, M)^C \to H^{d+1}(C, M^A) \to H^{d+1}(B,M).$$
Finally, you asked when you have maps $H^d(C, M) \to H^d(B, M)$.  The answer is always, and this map is called the restriction.  (There is also sometimes a wrong-way map $H^d(B,M) \to H^d(C,M)$ called the transfer.)  The problem is that they don't knit together to form an exact sequence, and there certainly isn't going to be a "connecting homomorphism" in general, as we see from the spectral sequence.  
