# PROBLEM LINK:

Contest

Practice

**Setter:** Priyanshu Vats

# DIFFICULTY:

Easy

# PROBLEM:

Problem setters are busy with their PS these days and have a pending report submission tonight. For setters to not lose marks and get ‘A’ in PS, they have asked you to solve this problem for them.

They have N special cakes from D-Mess, each of same size X. i-th of them is divided into i equal pieces. Size of each piece is rounded down to integer so that it can be processed easily, though it will generate some waste cake. One piece as a sample is taken from each special cake for testing purpose. Find the sum of sizes of all sample pieces. (These are actually reserved pieces in mess :P)

X, N are given.

# EXPLANATION:

There are N cakes and one sample piece is taken from each cake. Therefore, there will be N sample pieces. Also, the size of i-th piece will be x / i (i-th cake is divided into i pieces).

So, answer will be the sum : x / 1 + x /2 + x / 3 .... + x / N (Integer division).

**Brute force method:** Running a for-loop from 1 to N will give **TLE** as N <= 10^9.

**Better approach:** For a better time complexity, we can do the following:

To start with let N = x because for all N > x, sizes further x-th cake will be 0.

We can find the sum into 2 parts : x / 1 to x / \sqrt x and x / \sqrt x to x / x.

**First part** can be done by simply using a for-loop from 1 to \sqrt x. **For part 2**, we will calculate its value in reverse order i.e. x / x to x / \sqrt x.

So, from x / x to what value will the result be equal to 1?

It would be upto x / (x / 2).

Therefore, from x / (x / 1) to x / (x / 2) result will be 1 (notice that we covered x / 2 operations from brute force method in 1 operation),

from x / (x / 2) to x / (x / 3) the result will be 2, and so on…

We continue this reverse part upto x / (x / \sqrt x) as x / (x / \sqrt x ) = x / \sqrt x. There will be some edge cases where x is a perfect square, so we will need to add/subtract to adjust our calculation.

Now, since we have calculated the sum x / 1 to x / x. We can similarly calculate the sum x / x to x / (n + 1) separately(as we did in **part 2**) and subtract it from previously found sum.

# TIME COMPLEXITY:

In both parts, we are traversing upon \sqrt x terms **at max**. So, the overall time complexity will be O(min(N, \sqrt x))

# SOLUTION:

Author’s Solution : PS lite - solution

Feel free to Share your approach, if you want to. Suggestions are welcomed as always had been.