[contest link](https://mirror.codeforces.com/gym/105173)↵
I got a TLE in G of this contest.↵
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Let me brief you on the question.↵
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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.↵
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I came up with a divide and conquer algorithm.↵
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Let me describe my solution.[submission:282174326]↵
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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]).↵
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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.↵
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I can get the amount of an element x in range[l, r] in O(logn) using binary search.↵
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Of course, I need to get the ID vector for each element before all the queries.
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.[submission: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.
