10445: USACO 2024 February Contest, Gold Problem 3. Quantum Moochanics
Description
For her latest experiment, Bessie has placed an even number $N$ ($2 \leq N \leq 2 \cdot 10^5$) of these particles in a line. The line starts with a mootrino on the left and then alternates between the two types of particles, with the $i$-th particle located at position $p_i$ ($0 \leq p_1 < \cdots < p_N \leq 10^{18}$). Mootrinos initially move right while antimootrinos initially move left, and the $i$-th particle moves with a constant speed of $s_i$ units per second ($1 \leq s_i \leq 10^9$).
Bessie makes observations at the following times:
- First, $1$ second after the start of the experiment.
- Then $2$ seconds after the first observation.
- Then $3$ seconds after the second observation.
- ...
- Then $n + 1$ seconds after the $n$-th observation.
This experiment may take an extremely long time to complete, so Bessie would like to first simulate its results. Given the experiment setup, help Bessie determine when (i.e., the observation number) she will observe each particle disappear! It may be shown that all particles will eventually disappear.
Input
Each test case consists of three lines. The first line contains $N$, the second line contains $p_1,\dots,p_N$, and the third line contains $s_1\dots,s_N$.
It is guaranteed that the sum of all $N$ does not exceed $2\cdot 10^5$.
Output
Constraints
- Input 3 satisfies $N = 2$.
- Input 4 satisfies $N \leq 2000$ and $p_i \leq 10^4$ for all cows.
- Inputs 5-7 satisfy $N \leq 2000$.
- Inputs 8-12 satisfy no additional constraints.
Sample 1 Input
4
2
1 11
1 1
2
1 12
1 1
2
1 11
4 6
2
1 11
4 5
Sample 1 Output
9 9
11 11
1 1
3 3
- The mootrino (initially moving right) appears at positions $2 \rightarrow 0 \rightarrow 3 \rightarrow -1 \rightarrow 4 \rightarrow -2 \rightarrow 5 \rightarrow -3$.
- The antimootrino (initially moving left) appears at positions $10 \rightarrow 12 \rightarrow 9 \rightarrow 13 \rightarrow 8 \rightarrow 14 \rightarrow 7 \rightarrow 15$.
For the second test, the antimootrino starts $1$ additional unit to the right, so the two particles meet at position $6.5$ half a second before observation $11$.
Note that we only care about observation numbers, not times or positions.
Sample 2 Input
2
4
1 3 5 8
1 1 1 1
4
1 4 5 8
1 1 1 1
Sample 2 Output
1 1 3 3
7 2 2 7
- The two leftmost particles meet at position $2$ right at observation $1$.
- The two rightmost particles meet at position $6.5$ half a second before observation $3$.