Хроніки Архіваріуса R2-A2 (Розбір)

S_{0} S_{1} S_{2} S_{3} S_{4} S_{5} : A, : C, : B, : B C A = S_{2} S_{1} S_{0}, : B C A B C = S_{3} S_{2} S_{1}, : B C A B C B C A B = S_{4} S_{3} S_{2}, тощо .

\begin{aligned} S_{0}&: A,\\ S_{1}&: C,\\ S_{2}&: B,\\ S_{3}&: BCA=S_{2}S_{1}S_{0},\\ S_{4}&: BCABC=S_{3}S_{2}S_{1},\\ S_{5}&: BCABCBCAB=S_{4}S_{3}S_{2},\;\text{тощо.} \end{aligned}

S_{t} = S_{t - 1} S_{t - 2} S_{t - 3} .

\boxed{\, S_{t}=S_{t-1}\,S_{t-2}\,S_{t-3}\,}.

T_{0} = T_{1} = T_{2} = 1, T_{t} = T_{t - 1} + T_{t - 2} + T_{t - 3} (t \geq 3),

T_{0}=T_{1}=T_{2}=1,\qquad T_{t}=T_{t-1}+T_{t-2}+T_{t-3}\;\;(t\ge 3),

T_{t} \equiv T_{t - 1} + T_{t - 2} + T_{t - 3} (mod M) .

T_{t}\equiv T_{t-1}+T_{t-2}+T_{t-3}\pmod{M}.

f (0, 1) = A, f (1, 1) = C, f (2, 1) = B .

f(0,1)=A,\quad f(1,1)=C,\quad f(2,1)=B.

f (t, k) = ⎩ ⎨ ⎧ f (t - 1, k), f (t - 2, k - T_{t - 1}), f (t - 3, k - T_{t - 1} - T_{t - 2}), k \leq T_{t - 1}, T_{t - 1} < k \leq T_{t - 1} + T_{t - 2}, інакше .

f(t,k)= \begin{cases} f(t-1,k), & k\le T_{t-1},\\ f(t-2,k-T_{t-1}), & T_{t-1}<k\le T_{t-1}+T_{t-2},\\ f(t-3,k-T_{t-1}-T_{t-2}), & \text{інакше.} \end{cases}