contest link I got a TLE in G of this contest.
Let me brief you on the question.
There is an array a with a length n(2 <= n <= 1e5)(1 <= ai <= n) There are T queries(1 <= T <= 1e5) For every query, there are four integers l, r, p, q(1 <= l <= r <= n, 1 <= p < q <= n) Then, change the array. Keep the elements equal to p and q. Remove other elements. Output the inverse number of the array after changing. After every query, the array becomes to the initial one.
I came up with a divide and conquer algorithm.
Let me describe my solution.282174326
The key point lies in how to get the answer of the big problem(the range of [l, r]) after getting the answer of the small problem(the range of [l, (l + r) / 2] and the range of [(l + r) / 2 + 1, r]).
The inverse number of range[l, r] equals to the inverse number of range[l, (l + r) / 2] plus the inverse number of range[(l + r) / 2 + 1, r] plus the amount of q in range[l, (l + r) / 2] multiply the amount of p in range[(l + r) / 2 + 1, r], because p < q.
I can get the amount of an element x in range[l, r] in O(logn) using binary search.
Of course, I need to get the ID vector for each element before all the queries.
my code:
include <bits/stdc++.h>
using namespace std; using ll = long long; vector f[100005]; ll cnt(int l, int r, int tar) { return upper_bound(f[tar].begin(), f[tar].end(), r) - lower_bound(f[tar].begin(), f[tar].end(), l); } ll solve(int l, int r, int p, int q) { if (l == r) return 0LL; int m = l + r >> 1; return solve(l, m, p, q) + solve(m + 1, r, p, q) + cnt(l, m, q) * cnt(m + 1, r, p); } int main() { ios::sync_with_stdio(false), cin.tie(nullptr); int n, T, a; cin >> n >> T; for (int i = 1; i <= n; i++) { cin >> a; f[a].push_back(i); } for (int ttt = 1; ttt <= T; ++ttt) { int l, r, p, q; cin >> l >> r >> p >> q; cout << solve(l, r, p, q) << '\n'; } return 0; }
