Let $C$-type language be a regular language iff exists a deterministic finite automaton (DFA) $A$ which has a single accepting state such that $L=L(A)$. An example of a regular non-empty language which is not of $C$-type is $L=\{1,10\}$.
This is the proof:
Suppose that $L$ is a $C$-type language, that is a DFA exists: $A=\big(\sum, Q, q_0, \{q_f\}, \delta\big)$ which accepts the language $L$.
Then $\delta(q_0,1)=q_f$ and $\delta(q_0,10)=q_f$.
Therefore $q_f=\delta(q_0,10)=\delta(\delta(q_0,1),0)=\delta(q_f,0)$.
Since $\delta(q_f,0)=q_f$ then $\delta(q_0,100)=\delta(\delta(q_0,10),0)=\delta(q_f,0)=q_f$ so $100\in L$ in contradiction to the given details.
According to my intuition the DFA looks like along these lines:
I do understand that $100$ is also part of $L$ but I as far as I see the DFA has a single accepting state $q_f$ as is the definition of $C$-type language. It doesn't have more than one accepting state. So why is this a contradiction?
By definition, $L = \{1, 10\}$.
If $L$ is a $C$-type language, $100 \in L$.
Therefore $L$ is not a $C$-type language.