Laura and Operations
Contest: Codeforces Round 911 (Div. 2)
Difficulty rating: 900
Tags: dp, math

Laura is a girl who does not like combinatorics. Nemanja will try to convince her otherwise.

Nemanja wrote some digits on the board. All of them are either $$$1$$$, $$$2$$$, or $$$3$$$. The number of digits $$$1$$$ is $$$a$$$. The number of digits $$$2$$$ is $$$b$$$ and the number of digits $$$3$$$ is $$$c$$$. He told Laura that in one operation she can do the following:

- Select two different digits and erase them from the board. After that, write the digit ($$$1$$$, $$$2$$$, or $$$3$$$) different from both erased digits.

For example, let the digits be $$$1$$$, $$$1$$$, $$$1$$$, $$$2$$$, $$$3$$$, $$$3$$$. She can choose digits $$$1$$$ and $$$3$$$ and erase them. Then the board will look like this $$$1$$$, $$$1$$$, $$$2$$$, $$$3$$$. After that, she has to write another digit $$$2$$$, so at the end of the operation, the board will look like $$$1$$$, $$$1$$$, $$$2$$$, $$$3$$$, $$$2$$$.

Nemanja asked her whether it was possible for only digits of one type to remain written on the board after some operations. If so, which digits can they be?

Laura was unable to solve this problem and asked you for help. As an award for helping her, she will convince Nemanja to give you some points.

Input format:
Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^5$$$). The description of the test cases follows.

The first and only line of each test case contains three integers $$$a$$$, $$$b$$$, $$$c$$$ ($$$1 \le a, b, c \le 100$$$) — the number of ones, number of twos, and number of threes, respectively.

Output format:
For each test case, output one line containing $$$3$$$ integers.

The first one should be $$$1$$$ if it is possible that after some operations only digits $$$1$$$ remain on the board, and $$$0$$$ otherwise.

Similarly, the second one should be $$$1$$$ if it is possible that after some operations only digits $$$2$$$ remain on the board, and $$$0$$$ otherwise.

Similarly, the third one should be $$$1$$$ if it is possible that after some operations only digits $$$3$$$ remain on the board, and $$$0$$$ otherwise.

Constraints:
Time limit: 2.0 s
Memory limit: 256.0 MB

Note:
In the first test case, Laura can remove digits $$$2$$$ and $$$3$$$ and write digit $$$1$$$. After that, the board will have $$$2$$$ digits $$$1$$$. She can make it have only digits $$$2$$$ or $$$3$$$ left by performing a similar operation.

In the second test case, she can remove digits $$$1$$$ and $$$3$$$ and write a digit $$$2$$$. After performing that operation $$$2$$$ times, the board will have only digits $$$2$$$ left. It can be proven that there is no way to have only digits $$$1$$$ or only digits $$$3$$$ left.

In the third test case, there is a sequence of operations that leaves only digits $$$1$$$ on the board. It can be proven that there is no way to have only digits $$$2$$$ or only digits $$$3$$$ left.

Example input:
3
1 1 1
2 3 2
82 47 59

Example output:
1 1 1
0 1 0
1 0 0

Editorial:
Let us assume that we want only the digits $$$1$$$ to be left on the board. Solution works analogically for the digits $$$2$$$ and $$$3$$$.
Define $$$x$$$, $$$y$$$ and $$$z$$$ as current number of digits $$$1$$$, $$$2$$$ and $$$3$$$ on the board respectively.
Performing the operation changes the parity of $$$x$$$, $$$y$$$ and $$$z$$$. In every moment $$$x$$$, $$$y$$$ and $$$z$$$ will all have the same parity or all will have different parity as numbers $$$a$$$, $$$b$$$ and $$$c$$$ respectively. We conclude that in order for only $$$1$$$ to be left on the board, $$$b$$$ and $$$c$$$ must have same parity, if they do not $$$y$$$ and $$$z$$$ can't have value $$$0$$$ in the same moment.
We conclude that $$$y$$$ and $$$z$$$ must have the same parity.
Let us assume that $$$z$$$ is greater than $$$y$$$. Solution works analogically if $$$y$$$ is greater than $$$z$$$. We can perform the following process until $$$y$$$ and $$$z$$$ become same: We perform the operation on digits $$$2$$$ and $$$3$$$, then we perform the operation on digits $$$1$$$ and $$$3$$$. This way $$$x$$$ and $$$y$$$ will remain the same and $$$z$$$ will be decreased by $$$2$$$. Now, when $$$y = z$$$, we can perform the operation on digits $$$2$$$ and $$$3$$$, $$$y$$$ times, until only the digits $$$1$$$ are left on the board. We conclude that it is possible to have only digits $$$1$$$ left on the board if and only if $$$b$$$ and $$$c$$$ have the same parity.
There also exists dynamic programming solution using precalculation.