When we do bitwise or of the values a_i we can get one of the following numbers:
1101 = 13
1100 = 12
1011 = 11
1010 = 10
1001 = 9
1000 = 8
0111 = 7
0110 = 6
0101 = 5
0100 = 4
0011 = 3
0010 = 2
0001 = 1
0000 = 0

That is, when we do bitwise or of numbers from set A, we can get all kinds of numbers ranging from 0 to K.

So, our first guess is to go through all of these 14 numbers and collect the numbers a_i which fit the pattern:

for (int pattern : {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13})
{
   set<int> numbers_fitting_pattern;

    for (int a : A)
      if (a & pattern == a)  
          numbers_fitting_pattern.insert(a);
}

Now we can optimize this algorithm. Let's say S₀₁₁₀ is the set of {a_i} such that theirs bitwise or gives 0110. Namely, S₀₁₁₀ = {0000, 0010, 0100, 0110}.

Now let's look at S₀₁₁₁ = {0000, 0001, 0010, 0100, 0011, 0110, 0111}.
As you can see S₀₁₁₀ ⊂ S₀₁₁₁. Even more — the set S₀₁₁₁ contains all the sets S₀₀₀₀, S₀₀₀₁, S₀₀₁₀, S₀₀₁₁, S₀₁₀₀, S₀₁₀₁, S₀₁₁₀.

That means we can use only S₀₁₁₁ and skip the sets resulting from patterns 0000, 0001, 0010, 0011, 0100, 0101, 0110. After this skipping our original loop will look like this:

for (int pattern : {7, 8, 9, 10, 11, 12, 13})
{
   set<int> numbers_fitting_pattern;

    for (int a : A)
      if (a & pattern == a)  
          numbers_fitting_pattern.insert(a);
}

This has improved our loop almost twice, but we can get rid of a few more patterns in this loop.

Do you understand now?

According to chokudai's tweet, the affected person was only seven people (Hec, hogloid, kawatea, logicmachine, nuip, E869120, TangentDay), and all of them would gain rating if it was rated. Though seeing this tweet, he proved the existence of people who affected by the mistake which is not getting AC for this problem. That's why he decided to be unrated.

Be lucky.

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