시간 제한 | 메모리 제한 | 제출 | 정답 | 맞힌 사람 | 정답 비율 |
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1 초 | 256 MB | 36 | 26 | 21 | 72.414% |
You are working on a new high-performance database engine — Instant Compression and Processing Codec (ICPC). ICPC stores user activity records. Each user activity record has an integer user identifier. The records are stored in a number of data files. Each data file is compressed and can contain records from multiple users, however ICPC has to process queries that look for a specific subsets of users. In order to do so, there has to be a way to quickly determine which data files may contain records for a specific user before attempting to decompress them, which may be a long and CPU-consuming process.
ICPC uses an algorithm called Bloom Filter. The way it is implemented in ICPC is described below. For each ICPC database the following integer parameters are chosen:
A value of the bloom filter is computed for each data file. The data file’s bloom filter is a vector of m bits. A bit number j (0 ≤ j < m) is set to one if and only if there is a record in this data file for some user identifier uk, such that for some hash function i (0 ≤ i < f) the following equality holds:
j = (uk · ai) mod m (1)
Your task is to implement ICPC filtering logic. You are given filter parameters and values for a number of data files and a set of user identifiers. Your task is determine which data files may contain record with at least one user identifier from the specified set. A data file may contain a record with a user identifier uk if and only if for all i (0 ≤ i < f) all the bits j given by equality (1) in its filter value are set to one.
The first line of the input file contains filter parameters — integer numbers m, f, and ai for 0 ≤ i < f (1 ≤ m ≤ 1000, 1 ≤ f ≤ 100, 1 ≤ ai < 231).
The second line of the input file contains an integer n — the number of data files (1 ≤ n ≤ 1000). Each of the following n lines contains bloom filter value of the corresponding file in hexadecimal form. Each value is represented by a string of ⌈m/4⌉ hexadecimal digits (one of 0123456789abcdef). The first digit of the string represents bits 0–3 of the value (stored in order from the least significant bit of a hexadecimal digit to the most significant bit), the second digit — bits 4–7, the third — 8–11, etc. When m mod 4 ≠ 0, then the last hexadecimal digit represents the last m mod 4 bits of the value in its least significant bits.
The following line of the input file contains an integer q — the number of user identifiers in a query (1 ≤ q ≤ 1000), followed by q integers uk — the set of distinct user identifiers in the query (1 ≤ uk < 231).
Write a line with the integer number s to the output file — the number of data files that may contain a record with at least one user identifier from the specified set, followed by s numbers dt (0 ≤ dt < n) — the 0-based numbers of the corresponding data files in ascending order.
23 4 3 5 7 11 3 effde7 c07902 0800c1 3 2 4 6
2 0 2
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