Submit | All submissions | Best solutions | Back to list |
DQUERY - D-query |
English | Vietnamese |
Given a sequence of n numbers a1, a2 ... an and a number of d-queries. A d-query is a pair (i, j) (1 ≤ i ≤ j ≤ n). For each d-query (i, j), you have to return the number of distinct elements in the subsequence ai, ai+1 ... aj.
Input
- Line 1: n (1 ≤ n ≤ 30000).
- Line 2: n numbers a1, a2 ... an (1 ≤ ai ≤ 106).
- Line 3: q (1 ≤ q ≤ 200000), the number of d-queries.
- In the next q lines, each line contains 2 numbers i, j representing a d-query (1 ≤ i ≤ j ≤ n).
Output
- For each d-query (i, j), print the number of distinct elements in the subsequence ai, ai+1 ... aj in a single line.
Example
Input 5 1 1 2 1 3 3 1 5 2 4 3 5 Output 3 2 3
Added by: | Jimmy |
Date: | 2008-10-26 |
Time limit: | 1s-1.5s |
Source limit: | 50000B |
Memory limit: | 1536MB |
Cluster: | Cube (Intel G860) |
Languages: | All except: ERL JS-RHINO NODEJS PERL6 VB.NET |
Resource: | Minesweeper |
hide comments
|
|||||
2025-02-28 17:18:13
The problem can be reduced to KQUERYO (but instead of elements strictly greater than k like in KQUERYO, you want to find elements strictly less than k) by making new array prev where prev[i] represents the previous index arr[i] occurred at. In the case where it's the first occurrence of arr[i] in the array, you can denote prev[i] with a negative number. Then, it becomes a problem of finding the number of elements in [l, r] such that prev[i] < l, or in other words, you want to count only the first occurrence of each number in [l, r]—If prev[i] > l, then another occurence of it must exist in [l, r], meaning it's a duplicate that shouldn't be counted. This sort of problem is straightforward to solve using a wavelet tree or persistent segment tree. Alternatively, you can also use Mo's algorithm to solve the queries offline. KQUERYO: https://www.spoj.com/problems/KQUERYO/ |
|||||
2025-02-03 03:20:58
莫队和树没有关系 但这题可以用树 不信去luogu.com看 |
|||||
2024-12-15 13:55:27
Using map gives TLE but using a vector of frequencies gives AC. |
|||||
2024-09-27 12:50:19
One of the worst problems I faced. After repeated submissions, I was facing TLE. Was optimising the code for no reason, when ultimately it turned out to be an issue of cin vs scanf. Really strict time limit for no reason at all. |
|||||
2024-08-10 11:56:28
simple MO algorithm |
|||||
2024-03-08 23:14:32
AC IN 1E9+7 LET'S GOOOOOOOOOOOOOOOO! |
|||||
2023-07-27 12:41:32
莫队和树没有关系吧(雾 |
|||||
2023-03-15 06:07:16
just use president Tree to solve this |
|||||
2023-03-14 15:47:38
Me too! AC with MO algorithm Tree Is also ok!! They both are tarjin algorihtm Last edit: 2023-03-14 15:48:01 |
|||||
2022-12-03 21:27:43
Solved in one go using your mom |