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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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2019-11-10 14:42:05
use printf scanf as cin cout slows you down!::-) |
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2019-10-09 09:38:49
AC in one go!! Last edit: 2019-10-09 09:39:15 |
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2019-09-25 22:31:36
just use your idea to get red of TLE |
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2019-08-10 07:18:44
Difference Array !! |
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2019-06-13 14:25:20
use scanf and printf |
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2019-03-17 10:04:09
BIT took 0.13 secs Diff Array took 0.10 secs |
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2019-01-08 12:00:04
learned difference array use Last edit: 2019-01-08 12:00:12 |
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2018-12-20 16:24:04
Simply use a difference array :) |
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2018-12-15 06:36:10
AC in no goes! :) |
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2018-10-14 23:43:03
ac in one go :) done with normal cumulative sum array |