Spaces:
Running
Running
File size: 1,202 Bytes
7a8878c |
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 |
USER: https://codeforces.com/contest/1375/problem/C
BOT: Thanks! Now could you provide a link to the editorial.
USER: https://codeforces.com/blog/entry/79731
BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)?
USER: editorial
BOT: The answer is iff $$$a_1 < a_n$$$. Let's find out why.
When $$$a_1 < a_n$$$, we can repeatedly use this algorithm while the permutation contains more than one element:
Find the smallest index $$$r$$$ such that $$$a_1 < a_r$$$.
Choose $$$a_r$$$ and the element comes right before $$$a_r$$$ and delete the element before $$$a_r$$$.
Repeat step 2 until $$$a_r$$$ is adjacent to $$$a_1$$$.
Choose $$$a_1$$$ and $$$a_r$$$, and delete $$$a_r$$$.
When $$$a_1 > a_n$$$, we have some observations:
The leftmost element is non-decreasing. That is because if we want to remove the old leftmost element $$$a_1$$$, we need to pair it with $$$a_2 > a_1$$$, and that will result in the leftmost element increasing.
Likewise, we can use the same argument to show that the rightmost element is non-increasing.
Problem C author — Kuroni
**Problem D**
Let me know if you have any questions.
|