ABC467 Participation Report
It is me.
I participated in ABC467. I solved 4 problems. I beat Iwai-seijin. I think this is the second time.
Since both E and F are far beyond the difficulty level I can solve, this looks like quite a good result. I am happy.
Problem A Obesity
The units are confusing, and it is a very untidy problem.
The answer is .
Problem B Keep the Change
The answer is the total sum of when keep.
Problem C Adjacent Sums (easy)
It strongly smells like DP, but it is not.
Since is determined once is fixed, you can:
- Run a simulation for
- Run a simulation for
The answer is the min of these two cases.
Problem D Concentric Circles
Suddenly, geometry. I wish they would stop because it is bad for my heart.
The important things are:
- If the dot product of two vectors is , they are perpendicular.
- If the cross product of two vectors is , they are parallel.
First, if P == R && Q == S, it is Yes because you can use the same circle.
After that, consider whether the perpendicular bisector of and the perpendicular bisector of intersect.
If they intersect, you can use that intersection as the common center.
However, calculating the perpendicular bisectors properly is a hassle, so first check if and are parallel.
If and are not parallel, then their perpendicular bisectors are also not parallel and will always intersect. Therefore, the answer is Yes.
So, does it mean No if and are parallel? Not necessarily, because there is a case where the perpendicular bisectors coincide.
You can determine this by checking if the vector connecting the midpoints of and is perpendicular to .
If it is perpendicular, the perpendicular bisectors coincide, so the answer is Yes; otherwise, it is No.
That is all.
Final Thoughts
I might have solved Problem E if I could have performed a binary search on ModInt.
For Problem F, the problem to be solved for each query looks like interval scheduling, but maybe I am wrong.
The rate of improvement recently has been abnormal, but I am anxious because my practice cannot keep up at all. It is scary.
See you.