MULTQ3  Multiples of 3
There are N numbers a[0], a[1]... a[N  1].
Initially all are 0. You have to perform two types of operations :
1) Increase the numbers between indices A and B (inclusive) by 1.
This is represented by the command "0 A B"
2) Answer how many numbers between indices A and B (inclusive) are divisible by 3.
This is represented by the command "1 A B".
Input
The first line contains two integers, N and Q. Each of the next Q lines are either of the form "0 A B" or "1 A B" as mentioned above.
Output
Output 1 line for each of the queries of the form "1 A B" containing the required answer for the corresponding query.
Sample
Sample Input : 4 7 1 0 3 0 1 2 0 1 3 1 0 0 0 0 3 1 3 3 1 0 3 Sample Output : 4 1 0 2
Constraints
1 <= N <= 100000
1 <= Q <= 100000
0 <= A <= B <= N  1
hide comments
drago_codes:
20230201 06:36:05
Use shift operators for multiplying and dividing by 2.


tomatim:
20221126 06:53:06
Is lazy optimization necessary? getting TLE on testcase 8 

rebel_roar:
20220124 15:03:33
Same solution on codechef giving TLE. Are you guys encountering same problem or it just me ? 

jrseinc:
20200731 14:28:09
after 7 WA, 2TLE, and a single silly mistake finally AC!


silent_1:
20200616 13:45:19
https://www.codechef.com/problems/MULTQ3 

manish_thakur:
20200527 07:59:51
Use fast input output in c++ 

jenishmonpara:
20200519 19:04:35
AC in one go


fighter_4:
20200425 17:36:58
guys there is nothing tricky in 9th test case, you may have done any small mistake if you are getting WA,, took me 4 hours to debug,,,, but finally AC :D, my 20th. 

perdedor_9:
20200414 10:16:36
naive solution with optimization,AC in one go 

aditya15:
20200402 12:23:53
Test cases upto case 9 are simple sum queries without any updates, so even a wrong code may pass cases till 9. Just use Segment Trees + lazy propagation as mentioned below. No need to even use Fast IO 
Added by:  Varun Jalan 
Date:  20100912 
Time limit:  1s 
Source limit:  50000B 
Memory limit:  1536MB 
Cluster:  Cube (Intel G860) 
Languages:  All except: NODEJS OBJC VB.NET 
Resource:  own problem 