# FRJUMP - Editorial

Testers: Istvan Nagy
Editorialist: Praveen Dhinwa

medium

### PREREQUISITES:

sqrt decomposition, log

### PROBLEM:

You are given an array a of size n. You have to answer following queries and update.

• In each query, you are given two numbers - i, r. Let prod = a_i \cdot a_{i + r} \cdot a_{i + 2 r} \cdot ... \cdot a_{i + k r}, s.t. i + k r \leq n and k has largest possible value satisfying the equation. You to find prod \pmod{10^9 + 7} and first digit of prod.
• In the update operation, you can set a_i to some value v.

### EXPLANATION:

Finding first digit and modulo value of product of some numbers
Finding product of some elements modulo some number is easy. Let us see how can we find the first digit of product of some elements. Note that direct multiplication could be very large and won’t fit into any typical data type. Even if you use a big type, number of digits in the product could be huge and entire calculation will be quite slow.

As we know that sum of these elements would be quite smaller, so can we somehow represent the product of some numbers in terms of sum. We can take log of the product. Let b_1, b_2, \dots, b_n be n positive numbers. Let prod = b_1 \cdot b_2 \cdot \dots \cdot b_n. We get log(prod) = log(b_1) + log(b_2) + .. \dots + log(b_n). So, given log of a number, can we find first digit of the number?
Yes, if we know the fractional part of the log value, we can get the first digit by calculating 10^{\text{fractional part}}.
Fractional part of a double x can be found by x - floor(x).

Solution based on number of divisors of a number
Let us use 0 based indexing for the problem. For the given query of an integer r, we have to find first digit and modulo value of product a_0, a_r, a_{2 * r}, a_{3 * r}, a_{k * r}. Note that the indices are all multiples (including zero) of r.
Due to this, we can realize that the update at index i by setting a_i to v, will modify the values of products for all r's which are divisors of r (0 is also counted).

Based on these observations, we can design a solution as follows.
Let prod_r = a_0 \cdot a_r \cdot a_{2 * r} \cdot a_{3 * r} \cdot a_{k * r} \pmod{10^9 + 7}.
Similarly, let logSum[r] = log(a_0) + log(a_r) + log(a_{2 * r}) + \dots + \cdot a_{k * r}.
Note that we can find the first digit of the product using logSum[r] for a given query - r as described above. In particular, it will be pow(10, logSum[r] - floor(logSum[r]).

For the given array, before processing all the queries, we can precalculate values of prod_r and logSum_r for each r from 1 to n. Note that the time taken will be \mathcal{O}(n log n).

Now let us see how to process the updates and query.

If we receive an update of setting a_r to v, we will iterate over all the divisors d of r and update the corresponding prod_d and logSum_d.
Now if we receive a query of finding answer for given a r, we can directly answer it by using values of prod_r and logSum_r.

Now, there is a small catch with the above solution. If we want to make an update at r = 0, then it will affect the answers of all the prod and logSum values. So if we directly update these values, we can take O(n) in the worst case. To handle this case, we can separately maintain the value of first element, (i.e. a_0). So, whenever we receive an update of setting a_0 to some value, instead of changing all the prod, logSum values, we will just change the value of first element we maintained. Note that we will have to change our update accordingly. For each query, we will have to additionally output the modified values of prod and logSum values respectively.

Note that this extra catch was because of the reason that r = 1 to n, all are divisors of 0. But for other numbers, this is not the case. Maximum number of divisors of a number 1 \leq n \leq 10^5 are 128.

Hence, time complexity of this solution will be \mathcal{O}(128 * n). Please note that in order to pass your solution with the given time limit, you have to implement this carefully, e.g. while update, don’t calculate inverse of a_r in the loop.

### Time Complexity:

\mathcal{O}(n \, 128)

### AUTHOR’S AND TESTER’S SOLUTIONS:

1 Like

This question was all about tackling the precision, once the algo was clear! Nice one

1 Like

Using double also works, just keep diving/multiplying by 10 so as to keep 1 digit before the decimal.

Example, store 1234.56789 as 1.23456789.
And if on division you get to something like 0.0001234, make it 1.234.

Obviously there is some loss of precision here, but some tinkering allowed a full AC.

I wonder if a case that can stop this approach exists?

4 Likes

The pre-requisites mention about Square Root Decomposition, I did not understand how we are implementing Square Root Decomposition here?

Exactly same logic as described in the editorial, still it was WA.
If any body could figure out.

One in C++ 14 gets a WA and TLE(by 0.1 sec ) and the other(in C++ 4.3.2) gets AC,these things should not happen with a programmer

Anyways what can be the reason ?

1 Like

Can anyone tell me why am i getting WA?
https://www.codechef.com/viewsolution/10512326

We can do this question using sqrt decomposition also.Divide the array into sqrt n blocks and for each block maintain an array of sqrt n size in which ith element contains information required about the product of the elements in this block due to r=i.For r>=sqrt n just calculate the answer manually(it takes O(root n))and for r< sqrt n traverse the blocks to get the information.We can update in sqrt n time by checking divisors of that particular index and update in its block. So total complexity is O(n*root n).

I’m getting WA for subtask 2. Can somebody explain why? (I know the approach is not ideal, still)
https://www.codechef.com/viewsolution/10453146

Can anyone help me to optimize my code??

https://www.codechef.com/viewsolution/10504174

If N=50,say. and we update the 12th element. Then we have to change the value of product for r=1,2,3,4,6,12. How will we do that? Do we have to iterate through each value of r?

I got 95 points in the question since my solution did not pass the first subtask. Can anyone explain why?
https://www.codechef.com/viewsolution/10511094

Implemented as in editorial still getting WA. Can someone figure it why?
https://www.codechef.com/viewsolution/10515781

I dont understand why this additional EPS was needed in the power of 10.

This problem could be solved without the prior knowledge of square-root decomposition.
https://www.codechef.com/viewsolution/10448317

I did this problem without using sqrt-decomposition…I made an array which stores the new value of friendliness at pth index during update query…I never updated the array which stores enjoyment for different values of R. During query 2, I checked whether there is an any update of friendliness int the multiples of r. In this way i updated my Answer…
See my solution here for more clarity… https://www.codechef.com/viewsolution/10384547

Editorialist solution giving TLE for last 3 testcases . Why ?

can anyone explain me how to precompute the product array in O(nlogn) ?? I am getting it in O(n*sqrt(n)) .

In the editorialist code, what is the purpose of computing modulo inverse and multiplying with existing product of r along with y?

if (type == 1) {
scanf("%d", &y);
int iv = inv(a[x - 1]);
long double nv = log10l((long double)y);
for (int j = 0; j < (int)d[x - 1].size(); ++j) {
int r = d[x - 1][j];
prod[r] = 1LL * prod[r] * iv % MOD * y % MOD;
slg[r] += nv - lg[x - 1];
}
a[x - 1] = y;
lg[x - 1] = nv;
}

Another much easier method is to pre-compute answer for 1<= R<= min( N, 300). For R> 300 compute by iterating over 0 to N- 1, which takes O( N/ 300) since we make a jump of >300 during iteration. To update just alter the pre-computed values which takes O( min( 300, N)). So overall complexity becomes O( Q*N/300).

1 Like