Lost Array (Codeforces 1713F)
Tags: Math, bitmasks

My orzlers, we can optimize this problem from $$$O(S^3)$$$ to $$$O\left(T^\frac{5}{9}\right)$$$!— Spyofgame, founder of Orzlim religionA long time ago, Spyofgame invented the famous array $$$a$$$ ($$$1$$$-indexed) of length $$$n$$$ that contains information about the world and life. After that, he decided to convert it into the matrix $$$b$$$ ($$$0$$$-indexed) of size $$$(n + 1) \times (n + 1)$$$ which contains information about the world, life and beyond.Spyofgame converted $$$a$$$ into $$$b$$$ with the following rules.  $$$b_{i,0} = 0$$$ if $$$0 \leq i \leq n$$$;  $$$b_{0,i} = a_{i}$$$ if $$$1 \leq i \leq n$$$;  $$$b_{i,j} = b_{i,j-1} \oplus b_{i-1,j}$$$ if $$$1 \leq i, j \leq n$$$. Here $$$\oplus$$$ denotes the bitwise XOR operation.Today, archaeologists have discovered the famous matrix $$$b$$$. However, many elements of the matrix has been lost. They only know the values of $$$b_{i,n}$$$ for $$$1 \leq i \leq n$$$ (note that these are some elements of the last column, not the last row).The archaeologists want to know what a possible array of $$$a$$$ is. Can you help them reconstruct any array that could be $$$a$$$?

Input: The first line contains a single integer $$$n$$$ ($$$1 \leq n \leq 5 \cdot 10^5$$$).The second line contains $$$n$$$ integers $$$b_{1,n}, b_{2,n}, \ldots, b_{n,n}$$$ ($$$0 \leq b_{i,n} < 2^{30}$$$).

Output: If some array $$$a$$$ is consistent with the information, print a line containing $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$. If there are multiple solutions, output any.If such an array does not exist, output $$$-1$$$ instead.

Note: If we let $$$a = [1,2,3]$$$, then $$$b$$$ will be: $$$\bf{0}$$$$$$\bf{1}$$$$$$\bf{2}$$$$$$\bf{3}$$$$$$\bf{0}$$$$$$1$$$$$$3$$$$$$0$$$$$$\bf{0}$$$$$$1$$$$$$2$$$$$$2$$$$$$\bf{0}$$$$$$1$$$$$$3$$$$$$1$$$ The values of $$$b_{1,n}, b_{2,n}, \ldots, b_{n,n}$$$ generated are $$$[0,2,1]$$$ which is consistent with what the archaeologists have discovered.

Sample Inputs:
3
0 2 1

1
199633

10
346484077 532933626 858787727 369947090 299437981 416813461 865836801 141384800 157794568 691345607

Sample Outputs:
1 2 3

199633

725081944 922153789 481174947 427448285 516570428 509717938 855104873 280317429 281091129 1050390365

Solution (Python):
a = [*map(int, [*open(0)][1].split())]
for k in (0, 1):
    for i in range(19):
        z = 1 << i
        for j in range(len(a)):
            if j & z:
                a[j - k * z] ^= a[j + k * z - z]
print(*reversed(a))

Solution (Python):
a = [*map(int, [*open(0)][1].split())]
for k in (0, 1):
    for i in range(19):
        z = 1 << i
        for j in range(len(a)):
            if j & z:
                a[j - k * z] ^= a[j + k * z - z]
print(*reversed(a))

Solution (Python):
a = [*map(int, [*open(0)][1].split())]
n = len(a)
for k in (0, 1):
    for i in range(19):
        z = 1 << i
        for j in range(n):
            if j & z:
                a[j - k * z] ^= a[j + k * z - z]
print(*reversed(a))