Editorial for TLE '16 Contest 7 P3 - NOR


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Author: d

It is not necessary to use a segment tree or a sparse table to solve this problem. These approaches are overkill and run quite slowly. Binary search is also not necessary, and it increases execution time.

The first observation is that \text{anything} \operatorname{NOR} 1 = 0, where \text{anything} can be any expression. In particular, if A_y = 1, then the answer to the query x\ y is always 0.

To solve a query, it is only necessary to get the largest k that satisfies both k \le y and A_k=1. In case A_1 \dots A_y are all 0's, let k=0. This information can be preprocessed and stored in an array in \mathcal{O}(N) time (linear time). Now there are 3 cases to consider:

  • k<x: In this case, all integers are equal to 0.
  • k=x: In this case, only the first integer is equal to 1. The rest are equal to 0.
  • k>x: In this case, (A_x \operatorname{NOR} A_{x+1} \operatorname{NOR} \dots \operatorname{NOR} A_k) = 0, because A_k=1 and \text{anything} \operatorname{NOR} 1 = 0.

A query can be solved in \mathcal{O}(1) (constant time). Make sure to print the correct answer for each case.

Time Complexity: \mathcal{O}(N+Q)


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