Here is comparatively short and clean implementation of O (n lg n) solution based on segment-tree approach passed in 0.52 seconds. solution

You should see my solution if you are not very good in segment tree . good and easy solution to understand . i used basic concept (as zscoder described power of 2 and 5) . you can easily understand hows lazy working
Solution

The Intended solution was i think uses Segment tree or similar binary data structure.

The takeaway from the problem was to learn how two lazy fields propagate down the tree during queries.
One lazy term was for the Multiplying a number to a node and then marking its children lazy(if they exist) and one lazy term was for resetting the node to a new sequence which was nothing but a factorial multiplied by a number and mark the children lazy for reset.

While propagating a reset update you can ignore the child's previous lazy terms(both), convince yourself here. For propagating "multiplication" update if you encounter a node having reset lazy field then first reset the node from its reset lazy field, mark its child lazy for reset and then do a update query on this node.

For double-lazy propagation to work, in my opinion, there should be at least one reset update of some kind, because reset update tells us that we can ignore the child previous lazy terms and set them according to the new update, irrespective of the child node history, as it is a reset update. If you don't have a reset update, then after some queries you will notice that some nodes contain both the lazy terms and you would not be able to distinguish which one to apply first and usually the order in which they are applied would matter and would yield different result, unless you suggest me some two different good operation whose order of application yields the same result(then that would make a good codechef long challenge problem :P)

REFERENCE SOLUTION

[EDIT]
A similar question that you can try is SEGSQRSS, although the test cases are very weak and you can get AC without lazy propagation, but then its just a waste of time. Do try that.

Nice. Though Sqrt Decomposition is viable here, I have an alternative solution using Lazy Propogation on Segment Tree.

The only "hard" query here is the second one. However, if you store :

1. The largest power of $2$ and $5$ that divides the product of the segment
2. A boolean denoting whether a lazy update is needed
3. The type of update that has to be applied (either a range multiplication or reset)
4. The ends of the segment $[l, r]$
5. $ex2, ex5$ denoting that we should multiply the range by $2^{ex2}$ and $5^{ex5}$
6. The endpoints of the current type $2$ query that needs to be propogated
7. $laz2, laz5$ denoting the largest power of $2$ and $5$ of $y$ in a type $2$ query.

AC Solution

codechef is very slow in updating editorials

Here is my solution which uses single segment tree. It uses oops to further simplify the problem.Solution link

Nice editorial or I would say article.

I have added my editorial/explanation to the same problem. The Update and Query range complexity for my solution is O(ln(N)). It runs in .77 seconds in java. You can see the same in my solution.

My solution is $\mathcal O(n \lg n)$ and passed in $0.47$ seconds with some preprocessing. Idea is same as @zscoder described.

I have tried to implement the Lazy Propagation on a Segment Tree to fetch the queries and update the values in O(logN). But my solution wasn't accepted. Can someone help me out with where am I going wrong? PS: My code gives the correct output for the given test case and some random test cases. Here is my code.

can anyone suggest mistakes in my link text solution please??

And Why do we need to store the ex2,ex5,lz2,lz5?? can anyone please explain??

little complex.. :)

My solution

is O(N log N) too with Segment Tree and Lazy Propagation, pass in 0.71 s.