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Mancunian And Multiplicative Queries
Tag(s):

Medium, Sqrt-Decomposition

Problem
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Mancunian is in a hurry! He is late for his date with Nancy. So he has no time to deal with a long and complex problem statement.
You are given two arrays, A and \(FUNC\). Each element of either array, \(A_i\) or \(FUNC_i\), is a positive integer not greater than C. Given Q queries, each of the form L R, the answer for each query is \begin{equation} \prod_{i=1}^{C} FUNC[cnt_i] \end{equation} where \(cnt_i\) is the frequency of the integer i in the range L R.
The answer to the problem is the product of the answers of all the individual queries modulo \(10^9+7\).

Input format

The first line contains three integers N, C and Q denoting the size of the array, the range of allowed values for each array element and the number of queries respectively.
The second line contains N positive integers, denoting the elements of the array A. The \(ith\) (1-indexed) of these represents \(A_i\).
The third line contains \(N+1\) positive integers, denoting the elements of the array \(FUNC\). The \(ith\) (0-indexed) of these represents \(FUNC_i\).
The \(ith\) of the next Q lines contains two integers, L and R for the \(ith\) query.

Output format

Print a single integer, which is the answer to the given problem.

Constraints

  • \(1 \le N, Q \le 50,000\)
  • \(1 \le C \le 1,000,000\)
  • \(1 \le L \le R \le N\)
SAMPLE INPUT
5 5 3
1 2 2 3 3
1 1 2 3 4 5
2 4
1 2
4 5
SAMPLE OUTPUT
4
Explanation

In the range specified by the first query, there are 0 occurrences of 1, 2 occurrences of 2, 1 occurrence of 3 and 0 occurrences of both 4 and 5. So the answer for the first query is \(FUNC[0]\) * \(FUNC[2]\) * \(FUNC[1]\) * \(FUNC[0]\) * \(FUNC[0]\) \(=\) 2.

In the range specified by the second query, there is 1 occurrence of 1, 1 occurrence of 2, 0 occurrences of 3 and 0 occurrences of both 4 and 5. So the answer for the second query is \(FUNC[1]\) * \(FUNC[1]\) * \(FUNC[0]\) * \(FUNC[0]\) * \(FUNC[0]\) \(=\) 1.

In the range specified by the third query, there are 0 occurrences of 1, 0 occurrences of 2, 2 occurrences of 3 and 0 occurrences of both 4 and 5. So the answer for the third query is \(FUNC[0]\) * \(FUNC[0]\) * \(FUNC[2]\) * \(FUNC[0]\) * \(FUNC[0]\) \(=\) 2.

Hence the final answer is 2 * 1 * 2 \(=\) 4.

Time Limit: 5.0 sec(s) for each input file.
Memory Limit: 256 MB
Source Limit: 1024 KB
Marking Scheme: Marks are awarded when all the testcases pass.
Allowed Languages: C, C++, C++14, Clojure, C#, D, Erlang, F#, Go, Groovy, Haskell, Java, Java 8, JavaScript(Rhino), JavaScript(Node.js), Julia, Kotlin, Lisp, Lisp (SBCL), Lua, Objective-C, OCaml, Octave, Pascal, Perl, PHP, Python, Python 3, R(RScript), Racket, Ruby, Rust, Scala, Swift, Visual Basic

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This Problem was Asked in

HackerEarth

Challenge Name

August Easy '16

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