We will hold AtCoder Beginner Contest 296.
- Contest URL: https://atcoder.jp/contests/abc296
- Start Time: http://www.timeanddate.com/worldclock/fixedtime.html?iso=20230401T2100&p1=248
- Duration: 100 minutes
- Number of Tasks: 8
- Writer: kyopro_friends, math957963, physics0523, Nyaan, nok0
- Tester: leaf1415, MMNMM
- Rated range: ~ 1999 The point values will be 100-200-300-400-500-500-600-600.
We are looking forward to your participation!
How to solve F?
Key observation $$$1$$$: Swapping $$$(i, j)$$$ in $$$A$$$ and $$$(i, k)$$$ in $$$B$$$ is equal to swapping $$$(i, k)$$$ in $$$B$$$ first followed by swapping $$$(i, j)$$$ in $$$B$$$. Therefore, we can always fix $$$A$$$ and operate on $$$B$$$ to make $$$B$$$ equals to $$$A$$$. Note that $$$(i k)(i j)$$$ could be merged into a rotation of three words, we call them $$$3$$$-rotations.
Key observation $$$2$$$: The answer is "Yes" iff $$$B$$$ is an even permutation of $$$A$$$. $$$B$$$ always swap even times, so if $$$B$$$ can reach $$$A$$$, $$$B$$$ is an even permutation. On the other hand, it is well-known that the alternative group $$$A_n$$$ (which consists of all even permutations) are generated by $$$3$$$-rotations $$$\{(123), (124), ..., (12n)\}$$$, therefore, if $$$B$$$ is an even permutation, then $$$B$$$ can convert into $$$A$$$.
Key observation $$$3$$$: If the content of $$$A$$$ and $$$B$$$ are different, it is impossible. Otherwise, if there are duplicate elements in $$$A$$$, it is always possible, as we can add a dummy swap between two elements with the same value to make an even permutation. Finally, we only need to consider the case where all elements are distinct. We can use the merge sort to calculate its inversion number in $$$O(nlogn)$$$.
Solve $$$A, B, C, E, F$$$ but WA on $$$D$$$ $$$8$$$ times. How $$$D$$$ please?
Upd: I am an idiot. I wrote something like $$$n \times n$$$ which easily overflows.
I looked up for a and b upto sqrt(m).
Can you give a hint for E?
Build a functional graph which consists of edges $$$i \rightarrow a[i]$$$. Then T can win at $$$i$$$ iff $$$i$$$ is in a circle. To find all elements not in a circle we can utilize the toposort.
Can you share your submission? I thought the element which is in circles can be reached whatever the value of k will be. Atcoder editorial is quite complex to understand.
I think he wins iff i is in a circle.
Sorry, critical typo.
what's going wrong in this 51 TestCases Passes and 5 Failed.
I faced this 5 WA too. The solution is to check if n*n is overflow.
how to fix it did you use sqrt to compare and in result did you use LLONG_MAX?
Yes, my submission
n can be 1e12. n*n can become 1e24 which is too big.
In $$$G$$$ was killed by just only one case: when query point is equal to rightmost point in polygon.
How did you approach G?
This is mostly implementation based problem without any thinking. There is algorithm for checking if point lies in convex polygon.
First we do cyclic shift on polygon such that its leftmost point becomes minumum. Then for all other point $$$atan2(y[i] - y[0], x[i] - x[0]) \in [-\frac{\pi}{2}, \frac{\pi}{2}]$$$. Then we precalc for every $$$i$$$ this $$$atan2$$$. This is increasing array.
We have a query point. We have to check in which triangle of points with indices $$$0, i, i+1$$$ does in lie. Use lower_bound to do it in $$$log{n}$$$. Now we have to check, if point is in corresponding triangle and is on corresponding segment. To remove many ifs and epsilons and make life easier, we just apply these checks to all $$$[max(0, i - 3), min(n - 1, i + 3)]$$$.
Now we only have to check, if point is on segment, and if point is in triangle. This can be done in integers.
F is similar to 1585D - Yet Another Sorting Problem / 1591D - Yet Another Sorting Problem
Also 986B - Petr and Permutations
https://atcoder.jp/contests/abc296/submissions/40253903 can someone give a example where this fails.
3 2 3 2 The answer is 2
Thanks achvanov
Submission — one of the shortest solution to problem E using binary jumps
Can you explain the approach?
If $$$i$$$ is on a cycle, then we always win.
If $$$i$$$ is outside of the cycle, then our opponent can pick $$$k_i = N$$$ and he always wins.
Pair of intersected cycles don't exist in this graph.
So, the answer is the sum of lengths of cycles.
Let's do $$$2^{20}$$$ (or $$$2^{30}$$$, or $$$2^{60}$$$ — some value, which we can choose) steps for each vertex with using Binary Jumps technique. We will finish on the cycle (always) for each vertex. No one vertex on cycle will be missed because we are doing equal number of steps. Total number of distinct results (
set.size()) is the total number of vertex on cycles — answer on the problem.How can we always win if i is on a cycle
Take such vertex on the cycle which will equal $$$i$$$ after $$$k_i$$$ steps