1/4

Most of the number theory problems involving divisors etc that have intended solutions using mobious or inclusion exclusions (I-E), do not really need these buzzwords, there is almost always an easier solution using dp. We can still say its I-E, but code is sort.

2/4

Let's calculate dp[i] where dp[i] denotes no of segments which have gcd(min, max) = i. To calculate dp[i], we can first calculate the no of segments which have gcd as multiple i, this part is easy and the same as editorial.

3/4

Then subtract dp[k*i] from dp[i] where 2<=k<=n/i. These will remove all segments which have gcd multiple of i but no exactly i. At last just print dp[1].

4/4

I forgot to mention one should find these dp values in descending order.

These comments were taken from
aryanc403 's twitter.