The Wu
Contest: Codeforces Round 502 (in memory of Leopoldo Taravilse, Div. 1 + Div. 2)
Difficulty rating: 1900
Tags: bitmasks, brute force, data structures

Childan is making up a legendary story and trying to sell his forgery — a necklace with a strong sense of "Wu" to the Kasouras. But Mr. Kasoura is challenging the truth of Childan's story. So he is going to ask a few questions about Childan's so-called "personal treasure" necklace.

This "personal treasure" is a multiset $$$S$$$ of $$$m$$$ "01-strings".

A "01-string" is a string that contains $$$n$$$ characters "0" and "1". For example, if $$$n=4$$$, strings "0110", "0000", and "1110" are "01-strings", but "00110" (there are $$$5$$$ characters, not $$$4$$$) and "zero" (unallowed characters) are not.

Note that the multiset $$$S$$$ can contain equal elements.

Frequently, Mr. Kasoura will provide a "01-string" $$$t$$$ and ask Childan how many strings $$$s$$$ are in the multiset $$$S$$$ such that the "Wu" value of the pair $$$(s, t)$$$ is not greater than $$$k$$$.

Mrs. Kasoura and Mr. Kasoura think that if $$$s_i = t_i$$$ ($$$1\leq i\leq n$$$) then the "Wu" value of the character pair equals to $$$w_i$$$, otherwise $$$0$$$. The "Wu" value of the "01-string" pair is the sum of the "Wu" values of every character pair. Note that the length of every "01-string" is equal to $$$n$$$.

For example, if $$$w=[4, 5, 3, 6]$$$, "Wu" of ("1001", "1100") is $$$7$$$ because these strings have equal characters only on the first and third positions, so $$$w_1+w_3=4+3=7$$$.

You need to help Childan to answer Mr. Kasoura's queries. That is to find the number of strings in the multiset $$$S$$$ such that the "Wu" value of the pair is not greater than $$$k$$$.

Input format:
The first line contains three integers $$$n$$$, $$$m$$$, and $$$q$$$ ($$$1\leq n\leq 12$$$, $$$1\leq q, m\leq 5\cdot 10^5$$$) — the length of the "01-strings", the size of the multiset $$$S$$$, and the number of queries.

The second line contains $$$n$$$ integers $$$w_1, w_2, \ldots, w_n$$$ ($$$0 \le w_i \le 100$$$) — the value of the $$$i$$$-th caracter.

Each of the next $$$m$$$ lines contains the "01-string" $$$s$$$ of length $$$n$$$ — the string in the multiset $$$S$$$.

Each of the next $$$q$$$ lines contains the "01-string" $$$t$$$ of length $$$n$$$ and integer $$$k$$$ ($$$0\leq k\leq 100$$$) — the query.

Output format:
For each query, print the answer for this query.

Constraints:
Time limit: 2.0 s
Memory limit: 256.0 MB

Note:
In the first example, we can get:

"Wu" of ("01", "00") is $$$40$$$.

"Wu" of ("10", "00") is $$$20$$$.

"Wu" of ("11", "00") is $$$0$$$.

"Wu" of ("01", "11") is $$$20$$$.

"Wu" of ("10", "11") is $$$40$$$.

"Wu" of ("11", "11") is $$$60$$$.

In the first query, pairs ("11", "00") and ("10", "00") satisfy the condition since their "Wu" is not greater than $$$20$$$.

In the second query, all strings satisfy the condition.

In the third query, pairs ("01", "11") and ("01", "11") satisfy the condition. Note that since there are two "01" strings in the multiset, the answer is $$$2$$$, not $$$1$$$.

In the fourth query, since $$$k$$$ was increased, pair ("10", "11") satisfies the condition too.

In the fifth query, since $$$k$$$ was increased, pair ("11", "11") satisfies the condition too.

Example input:
2 4 5
40 20
01
01
10
11
00 20
00 40
11 20
11 40
11 60

Example output:
2
4
2
3
4

Example input:
1 2 4
100
0
1
0 0
0 100
1 0
1 100

Example output:
1
2
1
2

Editorial:
We can regard a $$$01-string$$$ as a binary number. $$$$$$ Wu(s,t) = \sum_{i=0}^nw_i[s_i = t_i] = f(s \oplus t) $$$$$$ where $$$\oplus$$$ is bit-wise XOR and we can preprocess $$$f$$$.
Notice that $$$n \le 12$$$, so $$$\frac{n}{2} \le 6$$$, so we can do something like meet-in-the-middle, split the numbers into higher $$$6$$$ bits and lower $$$6$$$ bits:
$$$f[S_1][S_2][j]$$$ count the number of binary numbers with higher bits equal to $$$S_1$$$ and $$$f((\text{lower bits}) \oplus S2) = j$$$.
Then one can easily get $$$g[S][k]$$$ stores the answer and then answer queries in $$$O(1)$$$ time.
If you don't understand, see the code.
Complexity: $$$O(|S|2^{\frac{n}{2}}+2^{\frac{3n}{2}}+q)$$$.