Ok write smth here.
If the total number of occurrences of some character $$$c$$$ is not a multiple of $$$n$$$, then it is impossible to make all $$$n$$$ strings equal — because then it is impossible for all $$$n$$$ strings to have the same number of $$$c$$$.
On the other hand, if the total number of occurrences of every character $$$c$$$ is a multiple of $$$n$$$, then it is always possible to make all $$$n$$$ strings equal. To achieve this, for every character $$$c$$$ we move exactly ((the total number of occurrences of $$$c$$$) / $$$n$$$) characters $$$c$$$ to the end of each string, and by the end we will have all $$$n$$$ strings equal each other.
We can easily check if the condition satisfies by counting the total number of occurrences of each character $$$c$$$ and check its divisibility by $$$n$$$. The final complexity is $$$O(S \cdot 26)$$$ or $$$O(S)$$$ where $$$S$$$ is the sum of lengths of all strings.
1397B - Степенная последовательность
First of all, the optimal way to reorder is to sort $$$a$$$ in non-decreasing order. Proof: the cost to transform $$$a_i$$$ to $$$c^i$$$ is $$$\lvert a_i - x^i \rvert$$$, and $$$\lvert a_i - x^i \rvert + \lvert a_j - x^j \rvert \le \lvert a_j - x^i \rvert + \lvert a_i - x^j \rvert$$$ when $$$i < j$$$ and $$$a_i \le a_j$$$, thus it is optimal to have $$$a_i \le a_j$$$ for each $$$0 \le i < j < n$$$.
From now on, we assume $$$a$$$ is sorted in non-decreasing order.
Denote $$$a_{max} = a_{n - 1}$$$ as the maximum value in $$$a$$$, $$$f(x) = \sum{\lvert a_i - x^i \rvert}$$$ as the minimum cost to transform $$$a$$$ into $$${x^0, x^1, \cdots, x^{n-1}}$$$, and $$$c$$$ as the value where $$$f(c)$$$ is minimum.
Note that $$$c^{n - 1} - a_{max} \le f(c) \le f(1)$$$, which implies $$$c^{n - 1} \le f(1) + a_{max}$$$.
We enumerate $$$x$$$ from $$$1, 2, 3, \dots$$$ until $$$x^{n - 1}$$$ exceeds $$$f(1) + a_{max}$$$, calculate $$$f(x)$$$ in $$$O(n)$$$, and the final answer is the minimum among all calculated values. The final complexity is $$$O(n \cdot max(x))$$$.
But why doesn't this get TLE? Because $$$f(1) = \sum{(a_i - 1)} < a_{max} \cdot n \le 10^9 \cdot n$$$, thus $$$x^{n - 1} \le f(1) + a_{max} \le 10^9 \cdot (n + 1)$$$. When $$$n = 3, 4, 5, 6$$$, $$$max(x)$$$ does not exceed $$$63245, 1709, 278, 93$$$ respectively; so we can see that $$$O(n \cdot max(x))$$$ comfortably fits in the time limit.
1396A - Кратные длине - If $$$n = 1$$$
- Else
Let us denote $$$S$$$ as the current total number of stones.
Consider the following cases:
- Case A: There is a pile that has more than $$$\lfloor \frac{S}{2} \rfloor$$$ stones.
The first player ($$$T$$$) can always choose from this pile, thus he ($$$T$$$) is the winner.
- Case B: Every pile has at most $$$\lfloor \frac{S}{2} \rfloor$$$ stones, and $$$S$$$ is even.
It can be proven that the second player ($$$HL$$$) always wins.
Proof 1: Let us prove by induction:
When $$$S = 0$$$, the second player obviously wins.
When $$$S \geq 2$$$, consider the game state after the first player moves. If there is a pile that now has more than $\lfloor \frac{S}{2} \rfloor$ stones, then we arrive back at case A where the next player to move wins. Otherwise, the second player can choose from any valid pile (note that the case condition implies that there are at least two non-empty piles before the first player's move). Now $$$S$$$ has been reduced by $$$2$$$, and every pile still has at most $$$\lfloor \frac{S}{2} \rfloor$$$ stones.
Proof 2: The condition allows us to assign a perfect matching of stones, where one stone is matched with exactly one stone from a different pile.
A greedy way to create such a matching: Give each label $$$0, 1, \dots, S - 1$$$ to a different stone so that for every pair of stones with labels $$$l < r$$$ that are from the same pile, stones $$$l + 1, l + 2, \dots, r - 1$$$ are also from that pile; then match stones $$$i$$$ with $$$i + \frac{S}{2}$$$ for all $$$0 \le i < \frac{S}{2}$$$.
For every stone that the first player removes, the second player can always remove its matching stone, until the first player can no longer make a move and loses.
- Case C: Every pile has at most $$$\lfloor \frac{S}{2} \rfloor$$$ stones, and $$$S$$$ is odd.
The first player ($$$T$$$) can choose from any pile, and we arrive back at case B where the next player to move loses.
So the first player ($$$T$$$) wins if and only if there is a pile that has more than $$$\lfloor \frac{S}{2} \rfloor$$$ stones or $$$S$$$ is odd. This can be easily checked in $$$O(n)$$$.
In this problem, it is useful to note that when the boss only has $$$1$$$ hp left, just use the pistol because it has the least reloading time. So there are 3 strategies we will use when playing at stage $$$i$$$ $$$(1 \le i \le n)$$$:
- Take $$$a_i$$$ pistol shots to kill first $$$a_i$$$ monsters and shoot the boss with the AWP.
- Take $$$a_i + 1$$$ pistol shots and move back to this stage later to take another pistol shot to finish the boss.
- Use the laser gun and move back to this stage later to kill the boss with a pistol shot.
Observation: We will always finish the game at stage $$$n$$$ or $$$n - 1$$$. Considering we are at stage $$$i$$$ $$$(i \le n - 1)$$$ and the boss at both stage $$$i$$$ stage $$$i - 1$$$ has $$$1$$$ hp left, we can spend $$$2 * d$$$ time to finish both these stages instead of going back later, which costs us exactly the same.
Therefore, we will calculate $$$dp(i,0/1)$$$ as the minimum time to finish first $$$a_i - 1$$$ stages and 0/1 is the remaining hp of the boss at stage i. The transitions are easy to figure out by using 3 strategies as above. The only thing we should note is that we can actually finish the game at stage $$$n - 1$$$ by instantly kill the boss at stage $$$n$$$ with the AWP so we don't have to go back to this level later.
Answer to the problem is $$$dp(n, 0)$$$. Time complexity: $$$O(n)$$$.
