This question is to see, and then they think of what ...... DP digital then found the answer in a community, then I think of the violence.
First, if each are located on a variable, according to the flapper law, you can only have 1e7 entries contribute to answer, and the answer to this condition are combined within a community, which inspired us to think about violence.
First of all nonbasic variables is of no use, regardless of their election and not vote, the last to be removed.
Therefore, only the basic variables contribute.
Obviously the more basic variables run slower, this time set up base variables \ (b \) months, it is clear that you can get a rough answer to the lower bound
\[ \frac{\sum_{i=0}^{2^{b-1}} i^k}{2^{b-1}} \]
The above is a first prefix and a power of nature, is clearly a \ (k + 1 \) polynomial terms, the influence brought by ignoring constant, the denominator is removed, so that \ (X = 2 ^ B \) , available
\ [X ^ k \ 2 ^ {63} \]
That is, $ b \ leq \ log_2 \ sqrt [k] {2 ^ {63}} $, readily available \ (k \ geq 3 \) when violence is enumerated groups.
Then only \ (k = 1 \) or \ (2 \) a.
Consider \ (k = 1 \) , as long as the bit to consider the contribution.
Consider \ (K = 2 \) , for the squared sums the products after only two bits for each value of the same embodiment. So we need to consider two, consider these two are \ (0 \) or \ (1 \) , and then consider the stops.
So it is still a simple question.
But I did not write ...... I wrote the code to make up, by the way where wrong will change