Let $C = V(x^2 - y^3)$ be the cuspidate cubic which sits inside $\mathbb{C}^2$. Let $\pi: \text{Bl}_0(\mathbb{C}) \to \mathbb{C}^2$ be the blow up of the origin. The reduction of the total transform of $C$ has local equation $y^2(t^2-1)=0$ and the total transform is $\widehat{C} = 2 E_1 + C_1$. Blow up again, the local equation is $t^3 u^2(t-u)=0$ with $\widehat{C} = 2E_1 + 3E_2 + C_2$. Finally, blow up again, the local equation is $t^6(v-1)^3v^2(1-2v)=0, $ with $\widehat{C} = 2E_1 + 3E_2 + 6E_3 + C_3.$
With $$K_{X/Y} := K_{\text{Bl}_0(\mathbb{C}^2)/\mathbb{C}^2} = E_1 + 2E_2 +4E_3,$$ and recalling that the log canonical threshold is given by $$\min_j \frac{k_j+1}{a_j},$$ where $$K_{X/Y} = \sum_j k_j E_j$$ and $$\widehat{C} = \sum_j a_j E_j,$$ we see that $$\text{lct} = \min \left \{ \frac{1+1}{2}, \frac{2+1}{3}, \frac{4+1}{6} \right \} = \frac{5}{6}.$$
Question: Are the numbers $\frac{k_j+1}{a_j}$ birational invariants?
These numbers are known as the discrepancies, and they are not even invariants of $C$, much less birational invariants ($C$ is birational to $\Bbb A^1$, for instance). To be more precise, the numbers you get can depend on the resolution of singularities you take, but they always behave somewhat nicely - if for one resolution of singularities, all $a_i$ are $>0$, $\geq 0$, $>-1$, or $\geq -1$, then the same is true for any other resolution of singularities. One says that if all $a_i$ are $>0$, the singularity is terminal; $\geq 0$, canonical; $>-1$, log terminal; $\geq -1$, log canonical.