Unfair Contest

Observation: For a particular magic number $m_j$ , among two participant with same $X_i$ and $L_i \leq m_j \leq R_i$ for both of them, the participant with higher $L_i$ will reach there first. If $L_i$ if same for both of them, the one with lower index will reach there first.

As $L_i$ is divisible by $X_i$ we can iterate for all distinct values that can be $X_i$just once. When we iterate for a specific $X$, let’s consider a specific point $p$ within the iteration. We will consider the latest participant i.e the participant with maximum $L_i$ till now. If there is two or more participants with maximum $L_i$ we will prioritize the one with lower index. And, we will eleminate the ones with lesser $R_i$ than $p$. This can be done using a stack. Then we will update the result by counting the number of minimum rounds are needed to reach $p$ by the participants with this $X$. We can precalculate it for all possible distinct $X$ for all possible magic numbers.

As all the iteration will be done for just once for a distinct number the complexity will be, $O(nlog(n))$.