arc060 a

arc060_a

2^50, so it's impossible.

Is it better to loose than to determine if it matches A for 2^50 streets?

But even 2^25 is a tough call, and the worst case is 2^49.

If we take the bottom frequency table as L, the top frequency table as U, and the number of zeros as Z, then Li x Ri x 2^Z.

To be -1 because it includes cases where no one card is selected.

computational complexity

Split by sign with N

Make a frequency table with DP in N^3

Frequency table collation with N^2

Official Explanation

The move to reduce the order was to make the objective "sum to zero" by drawing A, regardless of the number of cards

I counted the "sum to zero" by comparing the two frequency tables, but the official solution is to DP them together and look at the point where the sum reaches zero.

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