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UPDATEIT - Update the array ! |
You have an array containing n elements initially all 0. You need to do a number of update operations on it. In each update you specify l, r and val which are the starting index, ending index and value to be added. After each update, you add the 'val' to all elements from index l to r. After 'u' updates are over, there will be q queries each containing an index for which you have to print the element at that index.
Input
First line consists of t, the number of test cases. (1 <= t <= 10)
Each test case consists of "n u", number of elements in the array and the number of update operations, in the first line (1 <= n <= 10000 and 1 <= u <= 100000)
Then follow u lines each of the format "l r val" (0 <= l, r < n, 0 <= val <=10000)
Next line contains q, the number of queries. (1 <= q <= 10000)
Next q lines contain an index (0 <= index < n)
Output
For each test case, output the answers to the corresponding queries in separate lines.
Example
Input: 1
5 3
0 1 7
2 4 6
1 3 2
3
0
3
4
Output:
7
8
6
Added by: | Pandian |
Date: | 2013-10-15 |
Time limit: | 1s |
Source limit: | 50000B |
Memory limit: | 1536MB |
Cluster: | Cube (Intel G860) |
Languages: | All except: ASM64 |
Resource: | Own |
hide comments
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2017-02-04 19:07:47
simple BIT my 100th... |
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2017-01-20 08:29:57
why timit limit exceeded?? |
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2016-11-24 14:44:41 Chinmay Kousik
Bit doesn't seem to make sense here |
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2016-11-23 04:26:50
python always time exceed |
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2016-11-23 04:25:35
Any python ACs? |
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2016-11-14 12:12:49
TLE in c++ using cin, cout |
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2016-10-30 15:44:31
easy prefix sum :) |
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2016-10-22 15:12:36 m.orazow
Any Java ACs? |
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2016-10-06 10:20:13
solved without using any kind of trees... all updates done together in O(n) all queries answered in O(1)... AC in a go :) |
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2016-09-10 15:54:40 munjal
my first BIT :) |