_Nyu_'s blog

By _Nyu_, history, 7 months ago, In English,

There is n number of elements & two types of queries. 1st query: update an element by new value 2nd query: how many numbers are greater than k in l to r range.

My approach is by using PST for every 1 to i (1<=i<=n) build a seg tree between 1 to maxValue of elements. Then in the 2nd query, I just calculate how many numbers greater than k in (l-1)'th seg tree & r'th seg tree in o(log N) & give the difference between them.

But in the 1st query, I can't do it by O(log N)......

Here is the problem link: https://www.spoj.com/problems/KQUERY2/en/ My solution: https://ideone.com/mdJeU4

please help it to do it..........

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7 months ago, # |
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You can try to use segment tree of Treaps. In this case each query will execute in: [number of segments] log(N) * log(N) [query to the Treap to get elements greater than k]