Seven, (10 points) A, B, and C each have several pieces of candy. They are required to give each other candies, starting with A giving B and C, the number of candies given being equal to the number of candies B and C originally had; then, in the same manner, B gives A and C the number of candies they currently have, and finally, C gives A and B the number of candies they currently have. After the mutual giving, each person ends up with exactly

64 pieces of candy. How many pieces of candy did each person originally have?
Seven, let the original number of sugar blocks that A, B, and C have be $x, y, z$. According to the problem, after the first distribution, they each have sugar blocks: $x-y-z, 2 y, 2 z$.
After the second distribution, they each have sugar blocks: $2(x-y-z), 2 y-(x-y-z)$ $-2 z, 4 z$.
After the third distribution, they each have sugar blocks: $4(x-y-z), 2[2 y(x-y-z)$ $-2 z], 4 z-2(x-y-z)-[2 y-(x-y-z)-2 z]$.
Solving the system of equations, we get
$$
x=104, y=56, z=32 \text {. }
$$