(⊻):{2/~=/+2\x,y}                  / bitwise xor
(≫):{_x%*/y#2}                     / right shift

(a;b;c;z;p):(.*|" "\)'0:"i/17.in"  / initial A, B, C, discard blank line, program
c:{((!4),A,B,C)@x}                 / combo operand
I :{A::A≫c x},{B::B⊻x}             / adv, bxl
I,:{B::8!c x},{N::(x;N)@~A}        / bst, jnz
I,:{B::B⊻C}  ,{O,:8!c x}           / bxc, out
I,:{B::A≫c x},{C::A≫c x}           / bdv, cdv
r:{(A;B;C;O)::x,b,c,,!0;           / run[A]. uses globals A, B, C, Output
 >[#p;]{N::x+2;I.2#x_p;N}/0;O}     / N = index of next instruction

`0:","/$r a                        / part 1: program's output
&/0{(y=*'r')#,/(!8)+/:8*x}/|p      / part 2: minimum A that outputs the program

Very imperative, but there's no way around it for this problem (you might even say that the imperative parts are... imperative).

The part 2 solution assumes that the program loops until A is 0, and that every time it does, it shifts A 3 bits to the right.

~/aoc24k/17.html