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
hide comments
iharsh234:
2016-08-01 16:56:46
simple BIT. |
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nonushikhar:
2016-07-28 20:24:00
easy and fun! |
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square1001:
2016-07-25 05:16:04
@jainromil26
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Romil Jain:
2016-07-16 08:44:24
Solved without using bit. O(n) solution.
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shubiks:
2016-07-12 18:20:21
My 100th on SPOJ! Used BIT. ;) |
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ranjanakash166:
2016-06-27 14:12:30
no need of segment tree!!! and BIT
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ranjanakash166:
2016-06-27 14:09:06
yeee....USE FAST I/O scanf(),printf() in c... |
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anuj0503:
2016-06-18 19:54:22
Learnt something new !!! |
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KD :
2016-05-27 13:10:04
learned something new :P |
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ajay_5097:
2016-05-18 17:29:12
TLE when using Segmented Tree + lazy propagation !
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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 |