问题
Given a base array, I need to compute the value of a function given below:
A[] = { a0, a1, a2, a3, .. an }
F(0,i) = ai [Base case]
F(1,i) = F(0,i) xor F(0,i+1) xor F(0,i+2) ... xor F(0,n)
F(2,i) = F(1,i) xor F(1,i+1) xor F(1,i+2) ... xor F(1,n)
.
.
F(x,i) = F(x-1,i) xor F(x-1,i+1) xor F(x-1,i+2) ... xor F(x-1,n)
0 < x < 10^18
0 < n < 10^5
I need to find F(x,0).
I am trying to solve this equation for the past 3 days. I have failed to optimise it and come up with a feasible solution. Any help to find F(x,0) in less than linear time is appreciated.
My observation (if it is of any importance) :
F(0,0) = a0
F(1,0) = a0^a1^a2^a3 ....
F(2,0) = a0^a2^a4 ....
F(3,0) = (a0^a4^a8...) ^ (a1^a5^a9...)
F(4,0) = a0^a4^a8 ....
F(5,0) = (a0^a1^a2^a3) ^ (a8^a9^a10^a11) ^ (a16^..a19) ^ ...
F(6,0) = (a0^a8^a16...) ^ (a2^a10^a18...)
F(7,0) = (a0^a8^a16...) ^ (a1^a9^a17...)
F(8,0) = a0^a8^a16^ ....
回答1:
Maybe it becomes easier if you reverse the array, with the recurrence relation
F(0, i) = a[n-i]
F(x, i) = XOR[0 <= j < i]( F(x-1, j) )
that is, going from 0 to i instead of i to n. We are then looking for F(x, n). If we make a table:
| 0 1 2 3
--+-------------------------------------------------------------------------------------------
0 | A=a[n] B=a[n-1] C=a[n-2] D=a[n-3]
1 | A A^B A^B^C A^B^C^D
2 | A A^(A^B) A^(A^B)^(A^B^C) A^(A^B)^(A^B^C)^(A^B^C^D)^B^C^D
3 | A A^(A^(A^B)) A^(A^(A^B))^(A^(A^B)^(A^B^C)) A^(A^(A^B))^(A^(A^B)^(A^B^C))^(A^(A^B)^(A^B^C)^(A^B^C^D))
or let's stop using the XOR sign and just count how often each term appears:
| 0 1 2 3 4
--+-----------------------------------------------
0 | 1A 1B 1C 1D 1E
1 | 1A 1A1B 1A1B1C 1A 1B1C1D 1A 1B 1C1D1E
2 | 1A 2A1B 3A2B1C 4A 3B2C1D 5A 4B 3C2D1E
3 | 1A 3A1B 6A3B1C 10A 6B3C1D 15A10B 6C3D1E
4 | 1A 4A1B 10A4B1C 20A10B4C1D 35A20B10C4D1E
We can also see that the recurrence relation is the same as
F(x, i) = F(x-1, i) + F(x, i-1)
This is Pascal's Triangle. Specifically, the values we are after are the nth diagonal, which can of course be calculated on its own. Then check whether the number is even or odd, so that you know whether the XOR operations on that element cancel themselves out or not.
来源：https://stackoverflow.com/questions/46090710/how-to-find-fx-0-when-fx-i-fx-1-i-xor-fx-1-i1-xor-fx-1-n-in-le