XOR = Average (Codeforces 1758B)
Tags: Others, constructive algorithms

You are given an integer $$$n$$$. Find a sequence of $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ such that $$$1 \leq a_i \leq 10^9$$$ for all $$$i$$$ and $$$$$$a_1 \oplus a_2 \oplus \dots \oplus a_n = \frac{a_1 + a_2 + \dots + a_n}{n},$$$$$$ where $$$\oplus$$$ represents the bitwise XOR.It can be proven that there exists a sequence of integers that satisfies all the conditions above.

Input: The first line of input contains $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases.The first and only line of each test case contains one integer $$$n$$$ ($$$1 \leq n \leq 10^5$$$) — the length of the sequence you have to find.The sum of $$$n$$$ over all test cases does not exceed $$$10^5$$$.

Output: For each test case, output $$$n$$$ space-separated integers $$$a_1, a_2, \dots, a_n$$$ satisfying the conditions in the statement. If there are several possible answers, you can output any of them.

Note: In the first test case, $$$69 = \frac{69}{1} = 69$$$.In the second test case, $$$13 \oplus 2 \oplus 8 \oplus 1 = \frac{13 + 2 + 8 + 1}{4} = 6$$$.

Sample Inputs:
3
1
4
3

Sample Outputs:
69
13 2 8 1
7 7 7

Solution (Python):
t = int(input())
output = []
for _ in range(t):
    n = int(input())
    if n % 2 == 1:
        res = ['1'] * n
        output.append(' '.join(res))
    else:
        res = ['2' for _ in range(n)]
        res[0] = '1'
        res[1] = '3'
        output.append(' '.join(res))
for o in output:
    print(o)

Solution (Python):
for n in [int(input()) for _ in range(int(input()))]:
    print('1 ' * n if n % 2 else '1 3 ' + '2 ' * (n - 2))

Solution (Python):
print('\n'.join([' '.join((['1'] * int(i), ['2'] * (int(i) - 2) + ['1', '3'])[int(i) % 2 == 0]) for i in [*open(0)][1:]]))