1624G — MinOr Tree

Here are some good problems from codeforces problemset that can be solved using binary lifting

1. 191C — Fools and Roads

2. 1328E — Tree Queries

3. 1702G2 — Passable Paths

4. 832D — Misha, Grisha and Underground

Hi, You are not checking for dead nodes when you are picking the nodes that are at the least possible distance from the root in the priority queue. Here's the correct segment of the code :

	while (!q.empty()) {
		ll d = q.top().first;
		int node = q.top().second;	
		q.pop();
                if(distance[node] < d) continue ;
		for (auto i : adj[node]) {
			int it = i.first;
			int wt = i.second;
			if (d + wt < distance[it]) {
				distance[it] = d + wt;
				q.push({distance[it], it});
			}
		}
	}

The author has stated that a, and b will be provided in the test cases such that it always leads to a valid x. I will quote the statement — "It is guaranteed that a, b are the two largest divisors for some number 1≤x≤10^9. "

Congratulations, your progress is appreciable.

Can someone give any hints about how to solve D1

B > C > A

I managed to solve up to E during the contest and F afterward. The problems were nice, they were from various topics and I enjoyed solving them.

yash_daga sir orz !..

Nice contest, problems were from various topics, and their difficulty and overall time to solve were adequate

I have a doubt about my submission for F. My code is similar to the one provided in the editorial, the only difference being that I simply moved the iterator to the next element in the set after first setting the iterator to the lower bound of l. But it gives TLE on test case 18. To my knowledge, wouldn't it be better to just move the iterator to the next element in the set as it is already sorted rather than using a binary search to get the next valid index at every iteration? Here's my code -> https://codeforces.me/contest/1791/submission/192143264