MULTQ3 - Multiples of 3


There are N numbers a[0],a[1]..a[N - 1]. Initally 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
ramini1996: 2018-02-03 13:48:51

Wrote 160 lines of code. AC in ONE GO !!!

amitboss: 2018-01-29 20:10:07

Ac in 1 go!
can use anything i.e, long long int or int

Last edit: 2018-01-29 20:13:06
atulbi: 2018-01-24 17:17:26

Same problem on codechef giving TLE , while AC in one go on SPOJ

shiv2111: 2018-01-20 15:56:24

easy peasy :|

shady51: 2018-01-10 12:04:32

Rectify the time limit for Java. I am using segment tree with lazy propagation with fast input/output, and still getting TLE, used the same approach in C++14 and got accepted.

Last edit: 2018-01-10 13:35:57
code_aim: 2017-09-29 16:43:28

Good one!!

vishakha14: 2017-08-17 15:08:56

Spoj is just too harsh for java users. TLE despite O(n + qlogn) + Fast I/O soln.

shubham_7616: 2017-06-14 18:00:23

Take size of array to be 10^6 and size of seg tree to be 4*(10^6).
Took 4*(10^5) and got 2 WAs.

sas1905: 2017-06-05 21:08:59

Segment+lazy..

lord_poseidon: 2017-05-29 11:44:31

AC in one go


Added by:Varun Jalan
Date:2010-09-12
Time limit:0.169s
Source limit:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Languages:All except: NODEJS OBJC VB.NET
Resource:own problem