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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 |
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2017-09-13 20:06:17
First BIT question & AC in one go! Using Range update and point query Last edit: 2017-09-13 20:07:56 |
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2017-09-10 12:43:11
Segment tree + lazy propagation would do. |
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2017-08-28 11:36:24
AC by ST + lazy |
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2017-08-23 16:49:13
O(1) query simple array modification |
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2017-08-16 13:15:36
Can be done with BIT and range updates. Thanks kartik sir for the tutorial. |
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2017-07-23 08:07:45
BRUTE FORCE +OPTIMIZATION==AC Last edit: 2017-07-23 08:07:57 |
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2017-06-17 21:11:40
Solved it by segment tree + lazy propagation |
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2017-06-16 12:39:26
Solved with both prefix sum and BIT |
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2017-06-09 10:57:16
Prefix sum !! geeksforgeek helped me :) learn something new :) |
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2017-05-14 15:19:14 prabodh prakash
Good question. Recommended to solve w/o use of any trees etc. |