Codeforces Round 658 (Div. 1) |
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Finished |
This is the easy version of the problem. The difference between the versions is the constraint on $$$n$$$ and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings $$$a$$$ and $$$b$$$ of length $$$n$$$ (a binary string is a string consisting of symbols $$$0$$$ and $$$1$$$). In an operation, you select a prefix of $$$a$$$, and simultaneously invert the bits in the prefix ($$$0$$$ changes to $$$1$$$ and $$$1$$$ changes to $$$0$$$) and reverse the order of the bits in the prefix.
For example, if $$$a=001011$$$ and you select the prefix of length $$$3$$$, it becomes $$$011011$$$. Then if you select the entire string, it becomes $$$001001$$$.
Your task is to transform the string $$$a$$$ into $$$b$$$ in at most $$$3n$$$ operations. It can be proved that it is always possible.
The first line contains a single integer $$$t$$$ ($$$1\le t\le 1000$$$) — the number of test cases. Next $$$3t$$$ lines contain descriptions of test cases.
The first line of each test case contains a single integer $$$n$$$ ($$$1\le n\le 1000$$$) — the length of the binary strings.
The next two lines contain two binary strings $$$a$$$ and $$$b$$$ of length $$$n$$$.
It is guaranteed that the sum of $$$n$$$ across all test cases does not exceed $$$1000$$$.
For each test case, output an integer $$$k$$$ ($$$0\le k\le 3n$$$), followed by $$$k$$$ integers $$$p_1,\ldots,p_k$$$ ($$$1\le p_i\le n$$$). Here $$$k$$$ is the number of operations you use and $$$p_i$$$ is the length of the prefix you flip in the $$$i$$$-th operation.
5 2 01 10 5 01011 11100 2 01 01 10 0110011011 1000110100 1 0 1
3 1 2 1 6 5 2 5 3 1 2 0 9 4 1 2 10 4 1 2 1 5 1 1
In the first test case, we have $$$01\to 11\to 00\to 10$$$.
In the second test case, we have $$$01011\to 00101\to 11101\to 01000\to 10100\to 00100\to 11100$$$.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length $$$2$$$, which will leave $$$a$$$ unchanged.
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