Thank you all for participating in the round.
All the problems were created by BiNARyBeastt during our classes in winter.
There are 2 observstions needed to solve the problem:
Let's define the numbers of vowels by $$$a_0, \cdots, a_4$$$ and assume we have fixed them. Obviously, $$$a_0 + \cdots + a_4 = n$$$.
At first, let's not consider the empty string as it doesn't change anything. Then, the number of palindrome subsequences will be at least $$$A = 2^{a_0} + \cdots + 2^{a_4} - 5$$$ (every subsequence consisting of the same letter minus the five empty strings). Now, notice that if we put the same characters consecutively then the answer would be exactly $$$A$$$, and that would be the best possible answer for that fixed numbers (there cannot be other palindrome subsequences because if the first and last characters are the same, then all the middle part will be the same as well).
Now, we need to find the best array $$$a$$$. To do this, let's assume there are 2 mumbers $$$x$$$ and $$$y$$$ in the array such that $$$x + 2 \leq y$$$. Then, $$$2^x + 2^y > 2 \cdot 2^{y-1} \geq 2^{x+1} + 2^{y-1}$$$. This means, that replacing $$$x$$$ and $$$y$$$ with $$$x+1$$$ and $$$y-1$$$ will not change the sum of the array $$$a$$$ but will make the number of palindrome subsequences smaller. We can do this replacing process until no two numbers in $$$a$$$ have difference bigger than $$$1$$$. Actually, there is only one such array (not considering its permutations) and it contains only $$$(n / 5)$$$-s and $$$((n / 5) + 1)$$$-s.
#include <bits/stdc++.h>
using namespace std;
const string VOWELS = "aeiou";
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
int t; cin >> t; // test cases
while (t--) {
int n; cin >> n; // string length
vector<int> v(5, n / 5); // n - (n % 5) numbers should be (n / 5)
for (int i = 0; i < n % 5; i++) v[i]++; // and the others should be (n / 5) + 1
for (int i = 0; i < 5; i++) for (int j = 0; j < v[i]; j++) cout << VOWELS[i]; // output VOWELS[i] v[i] times
cout << "\n";
}
}
B1 — The Strict Teacher (Easy Version), B2 — The Strict Teacher (Hard Version)
For both versions let's assume the array $$$b$$$ is sorted in increasing order.
For the easy version, there are three cases and they can be considered separately.
Case 1: David is in the left of both teachers. In this case, it is obvious that he needs to go as far left as possible, which is cell $$$1$$$. Then, the time needed to catch David will be $$$b_1-1$$$.
Case 2: David is in the right of both teachers. In this case, similarly, David needs to go as far right as possible, which is cell $$$n$$$. Then, the time needed to catch David will be $$$n-b_2$$$.
Case 3: David is between the two teachers. In this case, David needs to stay in the middle (if there are two middle cells, it doesn't matter which one is picked as the middle) of two teachers, so they both have to come closer to him simultaneously. So, they will need the same amount of time, which will be $$$(b_2-b_1) / 2$$$. Notice, that David can always go to the middle cell not depending on his cell number.
For this version, there are three cases, too. Case 1 and Case 2 from the above solution are still correct, but the last one should be changed a bit because now it is important between which two consecutive teachers David is. To find that teachers, we can use binary search. After finding that David is between teachers $$$i$$$ and $$$i+1$$$, the answer is $$$(b_{i+1}-b_i) / 2$$$, just like the easy version.
#include <bits/stdc++.h>
using namespace std;
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
int t; cin >> t; // test cases
while (t--) {
int n, m, q; cin >> n >> m >> q;
vector<int> a(m);
for (int i = 0; i < m; i++) cin >> a[i];
sort(a.begin(), a.end());
for (int i = 1; i <= q; i++) {
int b; cin >> b;
int k = upper_bound(a.begin(), a.end(), b) - a.begin(); // finding the first teacher at right
if (k == 0) cout << a[0] - 1 << ' '; // case 1
else if (k == m) cout << n - a[m - 1] << ' '; // case 2
else cout << (a[k] - a[k - 1]) / 2 << ' '; // case 3
}
cout << "\n";
}
}
A greedy approach doesn’t work because the end of one string can connect to the beginning of the next. Try thinking in terms of dynamic programming.
Before processing the next string, we only need to know which letter we reached from the previous selections.
Let's loop through the $$$n$$$ strings and define $$$dp_i$$$ as the maximal answer if we are currently looking for the $$$i$$$-th letter in the word "Narek". Initially, $$$dp_0 = 0$$$, and $$$dp_1 = \cdots = dp_4 = -\infty$$$. For the current string, we brute force on all five letters we could have previously ended on. Let’s say the current letter is the $$$j$$$-th, where $$$0 \leq j < 5$$$. If $$$dp_j$$$ is not $$$-\infty$$$, we can replicate the process of choosing this string for our subset and count the score difference (the answer). Eventually, we will reach to some $$$k$$$-th letter in the word "Narek". If reaching $$$dp_k$$$ from $$$dp_j$$$ is bigger than the previous value of $$$dp_k$$$, we update $$$dp_k$$$ by $$$dp_j + counted\texttt{_}score$$$.
Finally, the answer is $$$dp_i - 2 \cdot i$$$. This is because if $$$i$$$ is not $$$0$$$, then we didn’t fully complete the entire word (the problem states that in this case, these letters are counted in the GPT’s score, so we subtract this from our score and add it to the GPT’s).
Note: Updating the array $$$dp$$$ is incorrect, because we may update some $$$dp_i$$$ for some string, and then use that updated $$$dp_i$$$ for that same string. To avoid this, we can use two arrays and overwrite one to the other for each string.
Time complexity: $$$O(n*m*5)$$$ or $$$O(n*m*25)$$$, depending on the implementation.
#include <bits/stdc++.h>
using namespace std;
const string narek = "narek";
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
int t; cin >> t; // test cases
while (t--) {
int n, m; cin >> n >> m;
vector<string> s(n);
for (int i = 0; i < n; i++) cin >> s[i];
vector<int> dp(5, int(-1e9)), ndp;
dp[0] = 0;
for (int i = 0; i < n; i++) {
ndp = dp; // overwriting dp
for (int j = 0; j < 5; j++) {
if (dp[j] == int(-1e9)) continue;
int counted_score = 0, next = j;
for (int k = 0; k < m; k++) {
int ind = narek.find(s[i][k]);
if (ind == -1) continue; // if s[i][k] is not a letter of "narek"
if (next == ind) { // if s[i][k] is the next letter
next = (next + 1) % 5;
counted_score++;
}
else counted_score--;
}
ndp[next] = max(ndp[next], dp[j] + counted_score);
}
dp = ndp; // overwriting dp back
}
int ans = 0;
for (int i = 0; i < 5; i++) ans = max(ans, dp[i] - 2 * i); // checking all letters
cout << ans << "\n";
}
}
Let's define by $$$M$$$ the maximal value of $$$a$$$.
Try to find how to check whether it is possible to get some fixed $$$\gcd$$$-s for $$$a$$$ and $$$b$$$ (i.e. fix the $$$\gcd$$$-s and try to find sufficient range to swap).
Choosing a range is equivalent to choosing a prefix and a suffix. Not all prefixes and suffixes are needed to be checked to find the answer.
There are only few different values in the $$$\gcd$$$-s of prefixes and suffixes. Try to find a limit for them.
Let's define $$$p_i$$$ and $$$q_i$$$ as the $$$\gcd$$$-s of the $$$i$$$-th prefixes of $$$a$$$ and $$$b$$$, respectively. Then, for each index $$$i$$$, such that $$$1 < i \leq n$$$, $$$p_i$$$ is a divisor of $$$p_{i-1}$$$, so either $$$p_i=p_{i-1}$$$ or $$$p_i \leq p_{i-1} / 2$$$ (the same holds for $$$q$$$). This means, that there are at most $$$\log(M)$$$ different values in each of $$$p$$$ and $$$q$$$. The same holds for suffixes.
Let's assume we have fixed the $$$\gcd$$$-s of $$$a$$$ and $$$b$$$ after the operation as $$$A$$$ and $$$B$$$. Now, let's consider the shortest of the longest prefixes of $$$a$$$ and $$$b$$$ having $$$\gcd$$$ $$$A$$$ and $$$B$$$, respectively. If the swapped range has intersection with that prefix, we can just not swap the intersection as it cannot worsen the answer. The same holds for the suffix. This means, that the swapped range should be inside the middle part left between that prefix and suffix. But the first and last elements of the middle part have to be changed, because they didn't allow us to take longer prefix or suffix. So, the part that has to be swapped is that middle part.
Then, we can see that the only sufficient lengths of the longest prefixes and suffixes are the places where one of them (i.e. in array $$$a$$$ or $$$b$$$) changes (i.e. $$$i$$$ is a sufficient prefix if $$$p_i \neq p_{i+1}$$$ or $$$q_i \neq q_{i+1}$$$, because otherwise we would have taken longer prefix). So, we can brute force through the sufficient prefixes and suffixes (the number of each is at most $$$\leg(M)$$$). All we are left with is calculating the $$$\gcd$$$-s of the middle parts to update the answer, which can be done with sparse table.
Time complexity: $$$O(n*\log(n)*\log(M) + \log(M)^3)$$$.
Memory complexity: $$$O(n*\log(n))$$$.
Note: Please read Full Solution 1 for definitions and better understanding.
Let's brute force through the sufficient prefixes. Assume we are considering the prefix ending at $$$i$$$. This means the left border of the range will start at $$$i+1$$$. Then, we can brute force the right border starting from $$$i+1$$$. In each iteration, we keep $$$\gcd$$$-s of the range and update the answer. To proceed to the next one, we only need to update the $$$\gcd$$$-s with the next numbers of $$$a$$$ and $$$b$$$.
Time complexity: $$$O(n*\log(M)^2)$$$ (this solution is $$$\approx2$$$ times faster due to low constant and memory).
Memory complexity: $$$O(n)$$$.
#include <bits/stdc++.h>
using namespace std;
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
int t; cin >> t; // test cases
while (t--) {
int n; cin >> n;
vector<int> a(n), b(n);
for (int i = 0; i < n; i++) cin >> a[i];
for (int i = 0; i < n; i++) cin >> b[i];
vector<int> pa(n), pb(n), sa(n), sb(n);
for (int i = 0; i < n; i++) { // calculating the gcd-s of prefixes and suffixes of a and b
pa[i] = gcd((i == 0 ? 0 : pa[i - 1]), a[i]);
pb[i] = gcd((i == 0 ? 0 : pb[i - 1]), b[i]);
sa[n - i - 1] = gcd((i == 0 ? 0 : sa[n - i]), a[n - i - 1]);
sb[n - i - 1] = gcd((i == 0 ? 0 : sb[n - i]), b[n - i - 1]);
}
int ans = pa[n - 1] + pb[n - 1];
for (int i = 1; i < n; i++) {
if (pa[i] == pa[i - 1] && pb[i] == pb[i - 1]) continue; // checking sufficiency of the prefix
int ga = 0, gb = 0; // current gcd-s of the range
for (int j = i; j < n; j++) {
ga = gcd(ga, a[j]); // updating gcd of the range of a
gb = gcd(gb, b[j]); // updating gcd of the range of b
int ca = gcd(pa[i - 1], gcd((j == n - 1 ? 0 : sa[j + 1]), gb)); // gcd of a after the operation
int cb = gcd(pb[i - 1], gcd((j == n - 1 ? 0 : sb[j + 1]), ga)); // gcd of a after the operation
ans = max(ans, ca + cb); // updating the answer
}
}
cout << ans << "\n";
}
}
