RACETIME - Race Against Time
As another one of their crazy antics, the N (1 ≤ N ≤ 100,000) cows want Farmer John to race against the clock to answer some of their pressing questions.
The cows are lined up in a row from 1 to N, and each one is holding a sign representing a number, Ai (1 ≤ Ai ≤ 1,000,000,000). The cows need FJ to perform Q (1 ≤ Q ≤ 50,000) operations, which can be either of the following:
- Modify cow i's number to X (1 ≤ X ≤ 1,000,000,000). This will be represented in the input as a line containing the letter M followed by the space-separated numbers i and X.
- Count how many cows in the range [P, Q] (1 ≤ P ≤ Q ≤ N) have Ai ≤ X (0 ≤ X ≤ 1,000,000,000). This will be represented in the input as a line containing the letter C followed by the space-separated numbers P, Q, and X.
Of course, FJ would like your help.
The first line gives the integers N and Q, and the next N lines give the initial values of Ai. Finally, the next Q lines each contain a query of the form "M i X" or "C P Q X".
Print the answer to each 'C' query, one per line.
Input: 4 6 3 4 1 7 C 2 4 4 M 4 1 C 2 4 4 C 1 4 5 M 2 10 C 1 3 9 Output: 2 3 4 2
FJ has 4 cows, whose initial numbers are 3, 4, 1, and 7. The cows then give him 6 operations; the first asks him to count the how many of the last three cows have a number at most 4, the second asks him to change the fourth cow's number to 1, etc.Warning: large input/output data.
very good problemLast edit: 2018-01-10 19:23:58
@mathecodician I am gonna say you that my O(10e9 * NQ) passed. :) B)
Here people are talking about N*log^2(N) not passing but my NQ passed lol
Is there any tutorial for this problem?
Easy problem. My complexity is N + q * sqrt(n) * log(n).
I couldn't make my N*log^2(N) pass, I used segment tree with treap in each node and added fast input but still TLE :(
Junior Andrade [UNIFEI]:
N + q*sqrt(n)*log(n) works! :)
How come N*sqrt(N) gets AC, and N*lg^2(N) gets TLE?
how can an algorithm of complexity (N+Q)*log(N+Q) be TLE??!