This contest was over so here we go:- precomputaion:- store smallest prime factor for each number from one to 1000000 let call this sm(x)

1) first solve a(x) for all x 1 to 1000000.
how?--> using above sm(x) and using totient function logic
2) Using the sieve logic calculate b(x) for each x hint if(i divide x then add a[i] to b[x])
3) again using the logic of sieve calculate c[x] using b[x] hint let t= b[x] divide t by sm[t] until t=1. and calculate number of iteration.
4) for d(x) just use the prefix sum

If u notice B(x) = x,so using A and B is useless, then C(x) is sum of exponent of each prime in factorisation of x, and D(x) is prefix sum of C(x).