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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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2024-06-14 05:16:18
The Fenwick tree can work here, but it is easier and simpler to prefix the sum on segments, for O(N) - go AC! |
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2022-11-28 22:57:13
prefix sum? |
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2022-01-11 06:31:34
simple BIT |
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2021-11-19 19:21:05
Range update, point query from BIT |
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2021-06-22 02:53:09
What a waste question ... |
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2021-06-07 19:28:12
basic question on BIT |
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2020-03-28 07:13:58
has anyone done it using bitmasks?? |
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2019-12-31 19:52:39
i solved it in O(n) using difference array still getting tle is java |
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2019-11-20 20:51:00
I always sement tree because i only know segment tree ! |
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2019-11-10 14:42:29
use printf scanf as cin cout slows you down!::-) |