Clearly, from the units part of equation
$R= E-E =0$
and from the $10$s
$C \equiv - E (\text{mod } 10)$.
Also, as before, we can take advantage of the fact that $D=S-1$ so that we must have a pair of consecutive digits at the end of calculation (provided we proceed from right to left as below).
As with the previous alphametic, if we select $E$ the rest is determined to a large extent. This time, I'll go through each of the cases explicitly.
$E=1 \rightarrow C=9$ and since $A-N-1 \equiv 9 (\text{mod } 10)$ we get $A \equiv N (\text{mod } 10)$ which is not allowed.
$E=2 \rightarrow C=8$ which means $A-N-1 \equiv 8(\text{mod } 10)$ and $A = N-1$. Then, $U-A-1 \equiv N (\text{mod } 10) \rightarrow U=2N(\text{mod }10)$.
Since $0$, $2$ and $8$ are taken, this means that $U=4$ or $6$ but $U=6$ isn't feasible since then $N=3$ (since it can't be $8$) and $A=2$ (already taken). Hence, $U=4 \rightarrow N=7 \rightarrow A=6$. But this solution fails to preserve a pair of consecutive digits for $D$ and $S$.
$E=3 \rightarrow C=7 \rightarrow A-N-1 \equiv 7 (\text{mod } 10)$. Now, we could have $(A,N) = (9,1)$ or $A=N-2$. If the former, then $U=2$ and $Q \equiv D (\text{mod }10)$ which is not allowed.
If the latter then $U-A-1 \equiv N (\text{mod } 10) \rightarrow U \equiv 2N-1(\text{mod }10)$ which mean that $U$ must be $1, 5$ or $9$. We divide this into subcases
(i) $U=1 \rightarrow N=6 \rightarrow A=4 \rightarrow S=9$ and $D=8$ (since they are the only remaining consecutive digits). But then $Q-9 \equiv 4 (\text{mod }10) \rightarrow Q=E=3$ which is not allowed.
(ii) $U=5 \rightarrow N=8 \rightarrow A=6 \rightarrow S=2, D=1$ but then $Q-2 = 6 \rightarrow Q=N=8$ also not allowed
(iii) $U=9 \rightarrow N=5 \rightarrow A=E=3$ which is not allowed.
$E=4 \rightarrow C=6 \rightarrow A-N-1 \equiv 6 (\text{mod }10)$. So $(A,N) = (9,2), (8,1), (2,5)$ or $(5,8)$
(i) $A=9 \rightarrow U=1 \rightarrow Q=D$ which is not allowed.
(ii) $A=8 \rightarrow U=9 \rightarrow S=3, D=2 \rightarrow Q-2 \equiv 8 (\text{mod } 10) \rightarrow Q=R=0$ not allowed.
(iii) $A=2 \rightarrow U=8$ which means there are no consecutive digits left for $S$ and $D$.
(iv) $A=5 \rightarrow U=4=E$, not allowed.
$E=5 \rightarrow C=5$, not allowed.
$E=6 \rightarrow C=4 \rightarrow (A,N) = (2,7), (3,8), (7,2)$ or $(8,3)$ then
(i) $A=2 \rightarrow U=0=R$, not allowed
(ii) $A=3 \rightarrow U=2$ but there are no remaining consecutive digits for $S$ and $D$.
(iii) $A=7 \rightarrow U=9$ but again no remaining consecutive digits for $S$ and $D$.
(iv) $A=8 \rightarrow U=1 \rightarrow S=3, D=2 \rightarrow Q=1=U$, not allowed.
$E=7 \rightarrow C=3 \rightarrow (A,N) = (2,8), (5,1), (6,2), (8,4) or (9,5)$ then
(i) $A=2 \rightarrow U=1 \rightarrow D=4,5$ and $Q-D=3$ which means $Q=7$ or $8$ so $Q=E$ or $Q=N$, both of which are not allowed.
(ii) $A=5 \rightarrow U=6 \rightarrow S=9, D=8$ but then $Q=3=C$, not allowed.
(iii) $A=6 \rightarrow U=8 \rightarrow S=5, D=4$ but then $Q=0=R$, not allowed
(iv) $A=8 \rightarrow U=2 \rightarrow S=6, D=5$ but then $Q=4=N$, not allowed.
(v) $A=9 \rightarrow U=4 \rightarrow S=2, D=1 \rightarrow Q=1=D$, not allowed.
$E=8 \rightarrow C=2 \rightarrow (A,N) = (2,9), (4,1), (6,3), (7,4)$ or $(9,6)$ then
(i)$A=2 \rightarrow U=2=A$, not allowed
(ii) $A=4 \rightarrow U=5 \rightarrow S=7, D=6 \rightarrow Q=0=R$, not allowed
(iii) $A=6 \rightarrow U=9 \rightarrow S=5, D=4 \rightarrow Q=0=R$, not allowed
(iv) $A=7 \rightarrow U=1 \rightarrow S=6, D=5 \rightarrow Q=3$ which is a valid solution
(v) $A=9 \rightarrow U=5 \rightarrow S=4, D=3 \rightarrow Q=3=D$, not allowed.
Finally $E=9 \rightarrow C=1 \rightarrow (A,N) = (4,2), (5,3), (6,4), (7,5)$ or $(8,6)$ then
(i) $A=4 \rightarrow U=6 \rightarrow S=8, D=7$ but then $Q=1=C$, not allowed.
(ii) $A=5 \rightarrow U=8 \rightarrow S=7, D=6 \rightarrow Q=1=C$, not allowed
(iii) $A=6 \rightarrow U=0=R$, not allowed.
(iv) $A=7 \rightarrow U=2 \rightarrow S=4, D=3 \rightarrow Q=0=R$, not allowed.
(v) $A=8 \rightarrow U=4 \rightarrow S=3, D=2 \rightarrow Q=1=C$, not allowed.