63. A, B, and C originally had several candies each. The first time, A gave B and C candies equal to the number of candies B and C originally had, respectively; the second time, B gave A and C candies equal to the number of candies A and C currently had, respectively; the third time, C gave A and B candies equal to the number of candies A and B currently had, respectively. In this way, each person ended up with exactly 64 candies. Originally, A had $\qquad$ candies, B had $\qquad$ candies, and C had $\qquad$ candies.
Answer: $104,56,32$.
Solution: List and reverse the process:
\begin{tabular}{|c|c|c|c|}
\hline & A & B & C \\
\hline After the third operation & 64 & 64 & 64 \\
\hline After the second operation & $64 \div 2$ & $64 \div 2$ & $64+32+32$ \\
\hline After the first operation & $32 \div 2$ & $32+16+64$ & 128 \\
\hline Before the first operation & $16+56+32$ & 112 & $64 \div 2$ \\
\hline Original number of candies & 104 & 56 & $64 \div 2$ \\
\hline
\end{tabular}

That is, A, B, and C originally had $104,56,32$ candies each.