Idea: Vladosiya
Tutorial
Tutorial is loading...
Solution
def solve():
n, m = map(int, input().split())
x = input()
s = input()
for i in range(6):
if s in x:
print(i)
return
x += x
print(-1)
for _ in range(int(input())):
solve()
Idea: Gornak40
Tutorial
Tutorial is loading...
Solution
for _ in range(int(input())):
a, b, c = sorted(map(int, input().split()))
if a == b and b == c:
print('YES')
elif b % a == 0 and c % a == 0 and (b // a - 1) + (c // a - 1) <= 3:
print('YES')
else:
print('NO')
Idea: myav, MikeMirzayanov, Vladosiya
Tutorial
Tutorial is loading...
Solution
#include <bits/stdc++.h>
#define all(arr) arr.begin(), arr.end()
using namespace std;
const int MAXN = 1010;
int n;
string A[MAXN];
int solve() {
int ans = 0;
for (int i = 0; i * 2 < n; ++i)
for (int j = 0; j * 2 < n; ++j) {
vector<char> M {A[i][j], A[n - 1 - j][i], A[n - 1 - i][n - 1 - j], A[j][n - 1 - i]};
char c = *max_element(all(M));
for(char e: M)
ans += c - e;
}
return ans;
}
int main() {
int t; cin >> t;
while (t--) {
cin >> n;
for (int i = 0; i < n; ++i)
cin >> A[i];
cout << solve() << endl;
}
}
Idea: myav
Tutorial
Tutorial is loading...
Solution
#include<bits/stdc++.h>
using namespace std;
const int maxv = 1000000;
void add_divs(int x, map<int, int>&divs){
int i = 2;
while(i * i <= x){
while (x % i == 0){
divs[i]++;
x /= i;
}
i++;
}
if(x > 1) divs[x]++;
}
bool solve(){
int n;
cin >> n;
vector<int>a(n);
map<int, int> divs;
for(int i = 0; i < n; i++) {
cin >> a[i];
add_divs(a[i], divs);
}
for(auto e: divs){
if(e.second % n != 0) return false;
}
return true;
}
int main(){
int t;
cin >> t;
while(t--) {
cout << (solve() ? "YES" : "NO") << "\n";
}
}
Idea: MikeMirzayanov
Tutorial
Tutorial is loading...
Solution
def solve():
n = int(input())
a = [int(x) for x in input().split()]
dp = [n + 1] * n
def get(pos):
if pos > n:
return n + 1
if pos == n:
return 0
return dp[pos]
dp[-1] = 1
for i in range(n - 2, -1, -1):
dp[i] = min(dp[i + 1] + 1, get(i + a[i] + 1))
print(dp[0])
for _ in range(int(input())):
solve()
1881F - Minimum Maximum Distance
Idea: senjougaharin
Tutorial
Tutorial is loading...
Solution
#include<bits/stdc++.h>
using namespace std;
int n;
vector<vector<int>> g;
void dfs(int v, int p, vector<int> &d){
if(p != -1) d[v] = d[p] + 1;
for(int u: g[v]){
if(u != p){
dfs(u, v, d);
}
}
}
int main(){
int t;
cin>>t;
while(t--){
int k;
cin>>n>>k;
g.assign(n, vector<int>(0));
vector<int> marked(k);
for(int &e: marked) cin >> e, --e;
for(int i=1;i<n;i++){
int u, v;
cin >> u >> v;
--u, --v;
g[u].push_back(v);
g[v].push_back(u);
}
if(k==1){
cout<<0<<"\n";
continue;
}
vector<int> d1(n);
dfs(marked[0], -1, d1);
int mx = marked[0];
for(int e: marked){
if(d1[e] > d1[mx]) mx = e;
}
vector<int> d2(n);
dfs(mx, -1, d2);
mx = marked[0];
for(int e: marked){
if(d2[e] > d2[mx]) mx = e;
}
cout << (d2[mx] + 1) / 2 << "\n";
}
return 0;
}
1881G - Anya and the Mysterious String
Idea: Gornak40
Tutorial
Tutorial is loading...
Solution
#include <iostream>
#include <string>
#include <set>
#include <cstring>
#define int long long
using namespace std;
const int L = 26;
const int MAXN = 200200;
int n, m;
string s;
set<int> M2, M3;
int fen[MAXN];
void fenadd(int i, int x) {
x = (x % L + L) % L;
for (; i < n; i |= (i + 1))
fen[i] = (fen[i] + x) % L;
}
int fenget(int i) {
int ans = 0;
for (; i >= 0; i = (i & (i + 1)) - 1)
ans = (ans + fen[i]) % L;
return ans;
}
void relax(int l, int r) {
l = max(l, 0LL);
r = min(r, n);
for (int i = l; i + 1 < r; ++i) {
int c1 = fenget(i);
int c2 = fenget(i + 1);
if (c1 == c2) M2.insert(i);
else M2.erase(i);
if (i + 2 >= r) continue;
int c3 = fenget(i + 2);
if (c1 == c3) M3.insert(i);
else M3.erase(i);
}
}
void build() {
M2.clear();
M3.clear();
memset(fen, 0, n * sizeof(int));
fenadd(0, s[0] - 'a');
for (int i = 1; i < n; ++i) {
fenadd(i, s[i] - s[i - 1] + L);
}
for (int i = 0; i + 1 < n; ++i) {
if (s[i] == s[i + 1]) M2.insert(i);
if (i + 2 < n && s[i] == s[i + 2]) M3.insert(i);
}
}
void update(int l, int r, int x) {
fenadd(l, x);
relax(l - 5, l + 5);
fenadd(r, L - x);
relax(r - 5, r + 5);
}
bool query(int l, int r) {
auto it = M2.lower_bound(l);
if (it != M2.end() && *it + 1 < r) return false;
it = M3.lower_bound(l);
if (it != M3.end() && *it + 2 < r) return false;
return true;
}
signed main() {
int t; cin >> t;
while (t--) {
cin >> n >> m >> s;
build();
while (m--) {
int tp, l, r; cin >> tp >> l >> r, --l;
if (tp == 1) {
int x; cin >> x;
update(l, r, x);
} else {
cout << (query(l, r) ? "YES" : "NO") << '\n';
}
}
}
}
I just have a question: Why is the algorithm for F correct? What's the intuition behind it?
Its like diameter of the tree and you want to get a point in the middle of it you can say as a centroid
Yes, I know, but we essentially don't care about which nodes are marked and which ones are not.
Assume the tree is rooted at a marked node. If you are at the deepest marked node on a branch and move further down that branch it will increase the distances to all marked nodes by 1 (meaning the value of that node will be greater than the marked node you came from). You can therefore remove any nodes past the deepest marked node on each branch. From this it follows that you have a tree where all the leaves are marked and at that point it's the standard algorithm for finding the largest distance from a leaf.
can't we do it with rerooting?
I actually did it with rerooting, with lazy segment tree. You can check out my submission 227869909 .
There is no need for a lazy segment tree to do the rerooting.
My rerooting with prefix and suffix sum. It is really a bit too more complicated, so I missed the first AK in div 3 (got AC after the contest 12min later, because of a little afraid of TLE by NodeJS). https://codeforces.com/contest/1881/submission/227927645
The first DFS will calculate 3 kinds of values,
dp[u], the max length from u to a marked node under u's subtree
prefix[u][j], the max length from u to a marked node under u's [0,j] children's subtree
suffix[u][j], the max length from u to a marked node under u's [j,] children's subtree
The second DFS is rerooting. For each u, its answer can be calculated by two parts,
dp[u], to nodes under u's subtree
out, to nodes not under u's subtreeWhen rerooting from
utov,outis updated by 3 parts. All of them should increase 1 if necessary.u's
outu itself
max(prefix[u][j-1], suffix[u][j+1]), with v is u's jth child
Hope above can help you improve understanding of rerooting, even this problem is not the best application to be solved.
What is the time complexity of below code for problem D written by me..I'm little bit confused about the TC of my code..
include <bits/stdc++.h>
using namespace std;
define fastio() ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
//#To find min divisor of N int min_divisor(int n){
for(int i=2; i<=round(sqrt(n)); i++){ if(n%i==0) return i; } return n; }
int main(){ fastio();
int t; cin>>t; while(t--){ int n; cin>>n; unordered_map<int,int> m; for(int i=0; i<n; i++){ int x; cin>>x; int min_div = min_divisor(x); while(min_div!=x){ // what is the TC of this while m[min_div]++; // is it simply O(logn) or O(logn*sqrt(n)) x /= min_div; // or something else please tell me!! min_div = min_divisor(x); }if(min_div!=1) m[min_div]++; } bool flag = true; for(auto &pr : m){ if(pr.second%n!=0) flag = false; } cout<<(flag?"YES\n":"NO\n"); }}
F can also be done in a similar fashion to Tree Distances —
Submission for Tree Distances
Solution for CF: 227871270
Exactly! Problem F can be considered as the special case of this one. 227869702 Here's mine with a slightly different approach.
I mean this finding longest chain is quite common(at least i saw some of them before in div 3) Also I first found out the longest chain trick here(hard problem): https://oj.uz/problem/view/CEOI21_newspapers
"the author is lazy"
Pretty obvious from the way this editorial is written
For problem 1881F - Минимальное максимальное расстояние , can someone please tell the proof why it is always true to consider the longest distance between the labelled vertex and then divide it by 2 then take ceil of it.
After considering some examples it becomes somehow convincing but not very intutive about why that condition will always be true.
Define $$$dist(i,j)$$$ as the distance between node $$$i$$$ and node $$$j$$$.
Assume that there are $$$3$$$ labelled vertices $$$a,b,c$$$ and $$$(a,b)$$$ is the pair that creates longest distance from two labelled vertices. To find the node position that makes the minimum value of maximum distance to either $$$a$$$ or $$$b$$$, ofc we choose the node (call this node $$$d$$$) that resides in the middle of the path from $$$a$$$ to $$$b$$$ (since if without loss of generality we assume $$$dist(d,a) \leq dist(d,b)$$$, then $$$dist(d,b) = \bigg\lceil\frac{dist(a,b)}{2}\bigg\rceil$$$ is the maximum from both and this is the min value of $$$dist(d,b)$$$).
Now, suppose $$$dist(d,c) > \max(dist(d,a),dist(d,b))$$$. Then, the longest distance from two labelled vertices must be the pair $$$(c,a)$$$ or $$$(c,b)$$$ (since $$$dist(d,c)+dist(d,b) > dist(d,a)+dist(d,b)$$$ and $$$dist(d,c)+dist(d,a) > dist(d,b)+dist(d,a)$$$) which results in contradiction.
Thank you very much HexShift for your time and effort for clearing out the doubt.
C was the coolest one and it's not even close
bruh I felt that G was much more interesting than all the problems(despite me tryping 11 attempts to AC G)
In problem F how can we be sure that ceil(n/2) is the minimum answer compared to other nodes? Can someone please express their ideas on this?
https://codeforces.com/blog/entry/121327#comment-1076496
Thank you very much!
Starters were bad(ABC) but the main course was good(DEF) but i am not able to understand the bill(editorial)
tru, i dont really enjoy ABC this round. DE was kinda meh for me. FG was quite interesting to play around with
can you explain to me why in E we are updating
dp[i]asdp[i]=min(dp[i+1]+1,dp[i+a[i]+1])and not asdp[i]=min(dp[i+1]+1,dp[i+a[i]+1+j]+j)for alljsti+a[i]+1+j<nindexing from0.Depends on how you define your DP. My design was that dp[i] = minimum removals I can have after considering all the values from [i, n — 1].
Case 1: We don't take the current index.
That means that we will need to remove this index. So, dp[i] = min(dp[i], dp[i + 1] + 1).
Case 2: Take the current block.
That means that we will need to remove all the values from [i, i + a[i] + 1]. So, dp[i] = min(dp[i], dp[i + a[i] + 1]). Need to make sure it is not out of bound. You can only end up at i + a[i] + 1, nowhere else.
I am saying that when you take the current block, it means you will take a[i] number of elements with you and then consider the next block. So I am saying that we pop some more values before taking the a[i] values and it can land us an even better solution. Thats y i am checking for all j possible
From the definition of my DP state, we know that dp[i] is the minimum removals after considering [i, n — 1] so that means that even if you check all j's, it has been it has been considered in dp[i+a[i]+1] already
Task D
If we assumed 1s can calculate 10^8 times, but in the worst condition(all a are equal to 999983), it will judge near 2000(t)*10^4(n)*10^3 times. Would it TLE?
I am a novice and thank you for your generous advice and point out my mistakes.
No, because the sum of n over all testcases does not exceed 10^4. So it's 10^4*10^3.
oh,oh i misundertand it!thanks
https://ideone.com/lyUPEZ
problem F c++ BFS code: why RTE on test 3 ???
Ok let me show some potential improvements of your code:
Why you are wrong: You missed out the corner case (n=1) where deg[1]=0 meaning your "start" is undefined and therefore shows runtime error. Meanwhile, the solution is wrong(after I tested it with the fixed code).
Data Structures used: You dont need to push the k values into a set instead use a boolean vector or smt. set requires O(nlogn) which means you will waste precious time if you are using set.
Your code checks if the root is in the set and the adjacent elements as well, you can generalise it by putting it after int node=q.front(); q.pop(); if(bla bla bla) mdis.push_back(...). You can save some lines of code.
You can try learning how to write functions such as function<void(int)> bfs=[&](int root){ code here}; this allows you to write the function inside your main fucntion where you dont need to pass all the arrays and stuff. You can access it inside the function. This should save you some implementation time as well.
Your code is not very clean (for a guy like me), therefore hard to read. I saw some parts of the code that has strange spacing. I used to be a template monster with around 400 lines of templates, but I find it hard to debug if I write a code with lots of shortforms(for some reason). I recommend spending sometime either looking at other people's code and learning a neater way of writing it. I'm not pinelizing you, just giving some personal suggestions
Neater and fixed code(although it is wrong): Sorry that I deleted the front part of your code.
code is here: https://ideone.com/fN9WEY
thank U for ur time your advice will be considered
Its a request for the author that do consider this point too that editorial is also referred by the beginners so they should make this editorial keeping in mind that beginners should also understand the intuition behind, making this platform truly helpful for the beginners as well!
I lost the top 150 due to hacking task D :(
same got TLe . because i am too dum dum and used frequency array instead of a map
now instead of + 40 delta its a negative delta :(
the predictor showed me +140, and 136 was missing before blue, but now only +50. And it would also be my best result on codeforces
sad :( now we learned something from this silly mistake at least.
yep
there's a predictor ?
yes, there is a browser extension https://chrome.google.com/webstore/detail/cf-predictor/ocfloejijfhhkkdmheodbaanephbnfhn?utm_source=ext_app_menu
this is a better one imo
https://chrome.google.com/webstore/detail/carrot/gakohpplicjdhhfllilcjpfildodfnnn
I didn't think about map either :(
You don't really need the map. You just need a way to make the rollback/reset $$$O(1)$$$ amortized. This could be done by storing what was updated in a separate stack/queue/whatever.
yes, but is it easier than just using map? In fact, if I had received a tle during the contest, I would have quickly rewritten everything on the map and what happened would not have happened.
At least it's easier than rewriting, I should tell you. It's basically just
stk.push_back(k)whenever you update index $$$k$$$, andwhile(!stk.empty()){arr[stk.back()]=0;stk.pop_back()}for the reset.https://codeforces.com/contest/1881/submission/228412567 is this reset considered o(1)? as i didn't use stack to remember what i pushed and i didn't get TLE
No, but the number of primes is not too big. Try finding out the value of $$$(\text{number of primes below }10^5) \times 2000$$$
yea its O(T) right thank you :D
map is good enough here lol
Simple and insightful Video Editorials for A-E.
Why is my code for incorrect? I feel like it closely resembles the intended idea. It's prob D. Please help me My code
I suppose in the end of the function "check" you should add max1 = max(max1, a) as it could be still the biggest prime number
yes, i did it in my code.
Well, the everything I have done is just change array to map and it worked...
https://codeforces.com/contest/1881/submission/228020694
whyyyyyy ? :(
Well, I think I've found smth. I've made the array bigger and it managed to go through the test number 4. But now it gets TL on test 16...
memset doesn't work how you think (the size is in bytes, so you only reset 1/8th of the array). Having said that using the correct length will cause a TLE.
Well, interesting. So in this case should we just use vector instead of array?
Nah, it doesn't work too. Well, therefore is there any solution without using map? I take it there isn't.
227869532 i found jiangly's code.
i solved it without maps https://codeforces.com/contest/1881/submission/228412567 but i hardcoded the prime count to 10power6 (i think i could counted it using seive)
can someone explain to me why map would be better?
because in 1 test case, you must check for 79000 primes. And with map, you simply need to check the prime factors. I think that
For F, It's always better to re-construct the tree like this.
It's related to tree diameter (I guess there's no formal proof but this blog will help Blog)
Can anyone please explain how does D solution works? Can't able understand how counting frequency of prime divisors gives our answer.
Let's use two ideas:
1) Every number can be represented as a product of prime numbers (= prime divisors). This is known as factorization.
2) If two numbers are equal, they have the same set of prime divisors.
So, the operation in the problem allows you to "take" one prime divisor from a[i] and "give" it to a[j].
If, after doing this operation several times, we can rearrange the divisors so that every number in the array has the same set of divisors, then the answer is YES.
To check that this is possible, we can simply count the number of occurrences of each divisor in the products.
For example, consider test case 4:
4
30 50 27 20
30 = 2 * 3 * 5
50 = 2 * 5 * 5
27 = 3 * 3 * 3
20 = 2 * 2 * 5
We have four occurrences of "5", four occurrences of "3" and four occurrences of "2". By doing the operation several times, we can turn each number into 2*3*5.
Thanks! Understood
Damn!I wish I can saw your explaination earlier
Only AC 1 problem... f**k me
youll get better
If someone prefers video explanations or interested in knowing how to think in a live contest or reach to a particular solution.
Here is my live Screencast of solving problems [A -> E] (with detailed commentary in Hindi language).
Can G solved with sqrt-decomposition?, I'm getting WA2 Code
I think it shoud be fine, WA means your code has some errors not TLE. But im pretty sure if your code has a large constant factor with sqrt-decomposition, your code might TLE
Yup I know, I just wanted tye TLE verdict not the WA one to make sure that my idea is correct.
Erm i also got many wa2(you can see from my submissions) I had many WA2 because I even pushed l into my valid answers when s[l-2]==s[l] but actually its invalid. If l is in the set meaning s[l]==s[l+2]. So check if your program has valid answers correctly. Also similarly for r with r+1 and r+2. r-1 with r+1.
It's TLE now, the solution was right but I forgot a case $$$(i+1,i-1)$$$ when jumping by $$$sqrt$$$ size 228099350
Yeah I also missed the cases and kept on getting wa for like 10 times
the editorial is short and to the point!!
.
Problem F solutions are based on the tree diameter, It takes some greedy proofs but in the end you can convert the problem to a direct tree diameter problem. Here's a blog
What does question C mean? please help me
erm C is actually straightforward. Turning 90 degrees clockwise is not hard to understand. One operation actually just mean that you turn one cell's values to the next alphabet (a->b).
I know What you mean. But I don't see example 4.Why 5?
so the test case is
baaa abba baba baab
now you can change a on the (1,4) cell to b
baab abba baba baab
now for the b on the (3,1) cell we will have to change the (1,2) , (2,4) and (4,3) cells to b
bbab abbb baba babb
lastly change the a at the (3,2) to b
bbab abbb bbba babb
so in total 5 operations
I know all about it.Thank you very much.
operation codeforces
Any one Feels that this contest was not for Div.3 ?
It seems to be Harder :(
One more opportunity to be filled with shame.
How do I solve 1881G - Аня и таинственная строка with Segment tree using Lazy propogation? This is what I've done 227967016, I haven't used Lazy propogation and it's TLE. Thank you in advance.
I guess its pretty obvious but let me just repeat the obervations:
You only need to keep track of palindromes of size 2 and 3 only. pairs exist like this: (i,i+2), (i,i+1). Now since we do +x to every element in a range, we need to do range add(I used fenwick tree but lazy segment tree is also fine).
Hence we can store possible elements that has pairs (i,i+1), (i,i+2). Note : We store i not i+1, or i+2.
We only need to update (l-1,l) (l-1,l+1), (l-2,l), (r-1,r+1), (r,r+1) and (r,r+2) since the elements in range [l,r] is not changed.
Store the indexes in a set, binary search to find if right end<=r
Now you have finished the problem~~~
Why this code doesn't work for problem D?
The product of the whole input is in the order of $$$O(A^n)$$$ which is at most $$$10^{60000}$$$ :unamused:
No, product of every integer input per test case would be (10^10). correct me if I'm wrong.
No man that's the sum not the product 💀💀💀
LOL True
Could someone explain me the E? How to think the dp which from right to left
Let us say we have block: 2 3 4 5 2 1 1
We can see that if we choose a[0], we continue with array 5 2 1 1
Meaning, its possible to make it into smaller subtasks: use dynammic programming.
We can see that if we want the minimum value for dp[0], we need to find dp[0+a[0]+1]=dp[3]
But we know, we cannot choose a block of 5. Hence, dp[3]=dp[4]+1.
Then we can see that either dp[i]=min(dp[i+1]+1,dp[i+a[i]+1])
Does that make sense? Ill explain further if that is not clear enough
i have a small question i used to solve dp by thinking bottom up and extending the problem (for example the subset sum problem [(Previous SUM)(Extended Number)] and now i can solve it by checking the previous state) however in this problem i failed to find a bottom up solution :(
you should start with the whole problem then break it down (Top Down Approach)
the question: are there certain problems which can be solved by only one of the 2 approaches?
I actually understand dp as a breakingdown of subtasks, where you have the dp states as the stuff you need to record down for that each subtask.
Tbh, I actually dont really think in the form of top down or bottom up? I would just think how the subtask can come from. The number of states of the dp would be determined by yourself, so its all good as long as you define it right.
I feel it was really a nice contest for anyone starting out on cp in the sense that it offers a good mix of adhoc and dsa
How to find adhoc problems? I really don't understand it.
f can be solved using dp Let define d(v) as the max distance from vertex v to a labeled vertex in its subtree and define g(v) as the max distance from vertex v to a labeled vertex which is not in the subtree of v then d(v)=max(d(child of v)+1), g(v)=max(g(parent of v)+1,d(sibling of v) + 2) then f(v)=max(d(v),g(v))
can u plz explain the intuition and the process as well as the implementation...plzz !!
F is similar to
https://cses.fi/problemset/task/1132
The only change is to fix the end points of the diameter as RED nodes.
Help please, why my solution has TL?(problem D) 228167450
I did rerooting for F, would've gotten top 100 or something but I had to have a typo that I didn't see
I have a question: why is such brutal decomposition with every integer working with Problem D It was my idea in the first place during the contest but I gave up cause I thought it would be to slow. So I took a binary search approach My code and resulted in a TLE.
Why is tutorial's code fast enough or why is my code too slow?(I learned after a better approach of precomuputing the smallest prime decomposition of every number But still my original code confuses me.)
C++ Solution for A 227837666
C++ solution for B 227857937
C++ solution for C 227893269
C++ solution for D 228002421
C++ solution for E 228148743
C++ solution for F 228240931
In F, "Let's run a breadth-first traversal from any labeled vertex v1 and find the farthest other labeled vertex v2 from it.".
Why cant we take a "non marked" vertex v1 and find farthest marked v2 ?
Even if we take non marked vertex also it's working. Not sure why the author intended to do like this ?
Is there a more efficient way to solve A??
I have two nicer(?) solutions than the official ones.
B: Suppose everything has length $$$l$$$ in the end. Then $$$l \mid A$$$, $$$l \mid B$$$, $$$l \mid C$$$ (proof: run the splitting apart in reverse). This is also sufficient, and uses $$$\frac{A}{l} + \frac{B}{l} + \frac{C}{l} - 3$$$ operations. The number of operations is minimised when $$$l$$$ is maximum, so $$$l := \gcd(A, B, C)$$$; to solve the problem just check if the minimum is at most 3. 228383697
(I saw a lot of people at the top of unofficial standings iterating over $$$(A + B + C) / l$$$ and testing each of them using this condition; it's quite strange to me that they didn't use GCD instead...)
G: There is no need for Fenwick trees or segment trees, you can check palindromes using just the difference array (and maintain bad positions using
std::setas in the tutorial). 228383431Can someone Please help me with the problem F . My code 228434471 is giving wrong answer at Test 4.
Take a look at Ticket 17107 from CF Stress for a counter example.
Problem G with sets and vectors is also possible submission
Nice Contest, Thanks for the editorial.
Can someone please explain the intuition behind problem F editorial? I knew it's right but I don't think next time I saw a similar problem I can just figure it it's about the tree diameter.
Can anybody can tell me whether my solution in F can use when it have the edge-weight?Can it be called DP?228766385
Can someone explain what does this mean ~
Notice that palindromes do not appear or disappear inside a segment, but they can appear or disappear at its boundariesI did the same thing for A( 1881A—Don't Try to Count), but I got hacked.
Problem F: for anyone wondering about the intuition behind the solution, here's a short rundown of the logic, based on the dfs algorithm for finding the diameter of a tree.
Assuming k>=2 (otherwise the solution is trivial):
-> Assume A and B to be the two marked nodes that are the furthest away from one another in the tree. The first observation we need to make is that the maximum path from any and all nodes in the tree to a marked node will always lead to A or B, i.e. either A or B (or both) will be the furthest marked nodes from any node in the tree. This can be visualized by rooting the tree on an arbitrary node on the path from A to B.
-> Now we will consider all nodes at once and using the assumption that each node's maximum distance path will lead to A or B, whichever is furthermost from said node, we can observe that the minimum distance maximum path cannot start from a node outside the path from A to B, because it would be larger than that of a node on that path.
-> This confines the possible candidates for the node(s) which give the minimum maximum distance on the path from A to B. If we conceptually flatten this path to a line, we can see that the answer has to be given by the node(s) at its center, because they represent the minimum of the function max(distance to A, distance to B) on that line. Thus, the answer has to be (distance from A to B)/2, meaning the length of the maximum path from the node at the center of the path to either A or B (which are equidistant). In case the length is an odd number, then the center node will be slightly skewed towards A or B, increasing the length of the path by 1, hence the need to round the number d/2 up.
-> The problem is now reduced to finding the two furthermost marked nodes on the tree and halving their distance, which can be solved with a simple tweak of the standard tree diameter algorithm mentioned above.
My implementation: 232911915 Hope this helps!
G can also be solved only with segment tree. In each node we keep track of the first two characters in the interval as well as the last two (or just one character if its a leaf). Then when we merge two intervals we just keep check if a palindrome is formed at the edges. Updates can also be handled very simply by just applying the shift operation to the characters we keep track of. AC with this idea: https://codeforces.com/contest/1881/submission/240931303
how is the time complexity O(2e5⋅n⋅m) of problem a? thanks in advance .
b/a−1+c/a−1≤3 question no B . what's the reason behind this logic? someone pls help me to understand
can you explain to me why in E we are updating
dp[i]asdp[i]=min(dp[i+1]+1,dp[i+a[i]+1])and not asdp[i]=min(dp[i+1]+1,dp[i+a[i]+1+j]+j)for alljsti+a[i]+1+j<nindexing from0you can just use recursion if iteration is feeling hard:
Each number will have two choice weather we should remove it or not.
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