Overview

For each request specifying multiple resources, find the permission bit string \(K\) that grants access to all of them simultaneously, minimizing \(K\) as a binary number.

Analysis

When the required permission for resource \(i\) is \(S_i\), the access condition is: \(K \mathbin{\&} S_i = S_i\) This means “every bit that is 1 in \(S_i\) must also be 1 in \(K\).”

If a request wants to access the resource set \(\{c_1, c_2, \dots\}\), then \(K\) must satisfy all of them:

• The 1-bits of \(S_{c_1}\) must be 1 in \(K\)
• The 1-bits of \(S_{c_2}\) must also be 1 in \(K\)
• …

In the end, any bit required by at least one resource must be 1 in \(K\). This is the bitwise union, expressed by the following formula: \(K \supseteq S_{c_1} \lor S_{c_2} \lor \cdots\)

Furthermore, to minimize \(K\) as a binary number, it is optimal to set all unnecessary bits to 0. Therefore, the minimum \(K\) is:

\[K = S_{c_1} \lor S_{c_2} \lor \cdots\]

Concrete Example

Let \(L=5\), and suppose we want to access both: - \(S_1 = 00101\) - \(S_3 = 10000\)

Then setting all required bits gives: - \(K = 00101 \lor 10000 = 10101\)

This is the minimum. To make it any smaller, we would need to turn some 1 into 0, but that would prevent access to the corresponding resource.

A naive search for \(K\) satisfying the conditions would have \(2^L\) candidates (up to \(2^{64}\)), which is infeasible. However, from the observation above, each query can be answered by simply taking the OR.

Algorithm

1. Read each \(S_i\) (a bit string of length \(L\)) as an integer (binary number) and store it in an array.
   • Since \(L \le 64\), Python’s int handles this safely.
2. For each request, sequentially OR the integer values of the specified resources to build val:
   • val |= S[idx]
3. Convert val to a binary string of length \(L\) (zero-padded on the left) and output it.

Complexity

• Time complexity: \(O\!\left(N + \sum_{j=1}^{Q} M_j\right)\) (Each query only performs OR for the specified number of resources)
• Space complexity: \(O(N)\) (Storing each resource’s bit string as an integer)

Implementation Notes

• Fast input: Since \(N, Q, \sum M_j\) can be up to \(10^5\), reading all at once with sys.stdin.buffer.read().split() is reliable.

• Zero-padded output: Using format(val, "0{}b".format(L)) pads the result to length \(L\).

• Rather than performing bit operations on strings each time, converting to integers and using OR is both simpler and faster.

  Source Code

import sys

def main():
    data = sys.stdin.buffer.read().split()
    it = iter(data)

    N = int(next(it))
    L = int(next(it))
    Q = int(next(it))

    S = [0] * (N + 1)
    for i in range(1, N + 1):
        S[i] = int(next(it).decode(), 2)

    fmt = "0{}b".format(L)
    out_lines = []

    for _ in range(Q):
        m = int(next(it))
        val = 0
        for _ in range(m):
            idx = int(next(it))
            val |= S[idx]
        out_lines.append(format(val, fmt))

    sys.stdout.write("\n".join(out_lines))

if __name__ == "__main__":
    main()

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This editorial was generated by gpt-5.2-high.