We hope you enjoyed the contest!

Idea: SlavicG

**Tutorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
void solve()
{
string s, c = "codeforces";
cin >> s;
int ans = 0;
for(int i = 0; i < 10; i++)
{
if(s[i] != c[i])
{
ans++;
}
}
cout << ans << endl;
}
int32_t main(){
int t = 1;
cin >> t;
while (t--) {
solve();
}
}
```

Idea: mesanu

**Tutorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
void solve()
{
int n;
cin >> n;
int a[n];
int cnt = 0, ans = 0;
for(int i = 0; i < n; i++)
{
cin >> a[i];
if(a[i] == 0)
{
cnt++;
}
else
{
ans = max(ans, cnt);
cnt = 0;
}
}
cout << max(ans, cnt) << endl;
}
int32_t main(){
int t = 1;
cin >> t;
while (t--) {
solve();
}
}
```

Idea: SlavicG

**Tutorial**

Tutorial is loading...

**Solution**

```
#include "bits/stdc++.h"
using namespace std;
#define ll long long
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(),v.rend()
#define pb push_back
#define sz(a) (int)a.size()
void solve() {
int n; cin >> n;
map<string, int> mp;
mp["00"] = mp["01"] = mp["10"] = mp["11"] = 1e9;
int ans = 1e9;
for(int i = 0; i < n; ++i) {
int x; cin >> x; string s; cin >> s;
mp[s] = min(mp[s], x);
}
if(min(mp["11"], mp["10"] + mp["01"]) > (int)1e6) {
cout << "-1\n";
} else {
cout << min(mp["11"], mp["10"] + mp["01"]) << "\n";
}
}
int32_t main() {
ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
int t = 1;
cin >> t;
while(t--) {
solve();
}
}
```

Idea: flamestorm

**Tutorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
bool ok(int n, int m) {
if (n == m) {return true;}
else if (n % 3 != 0) {return false;}
else {return (ok(n / 3, m) || ok(2 * n / 3, m));}
}
void solve() {
int n, m;
cin >> n >> m;
cout << (ok(n, m) ? "YES" : "NO") << '\n';
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int tt; cin >> tt; for (int i = 1; i <= tt; i++) {solve();}
// solve();
}
```

Idea: mesanu

**Tutorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
#define startt ios_base::sync_with_stdio(false);cin.tie(0);
typedef long long ll;
using namespace std;
#define vint vector<int>
#define all(v) v.begin(), v.end()
#define MOD 1000000007
#define MOD2 998244353
#define MX 1000000000
#define MXL 1000000000000000000
#define PI (ld)2*acos(0.0)
#define pb push_back
#define sc second
#define fr first
#define endl '\n'
#define ld long double
#define NO cout << "NO" << endl
#define YES cout << "YES" << endl
int n, m;
bool vis[1005][1005];
int a[1005][1005];
int dfs(int i, int j)
{
vis[i][j] = true;
int ans = a[i][j];
if(i != 0 && a[i-1][j] != 0 && !vis[i-1][j])
{
ans+=dfs(i-1, j);
}
if(i != n-1 && a[i+1][j] != 0 && !vis[i+1][j])
{
ans+=dfs(i+1, j);
}
if(j != 0 && a[i][j-1] != 0 && !vis[i][j-1])
{
ans+=dfs(i, j-1);
}
if(j != m-1 && a[i][j+1] != 0 && !vis[i][j+1])
{
ans+=dfs(i, j+1);
}
return ans;
}
void solve()
{
cin >> n >> m;
for(int i = 0; i < n; i++)
{
for(int j = 0; j < m; j++)
{
vis[i][j] = false;
cin >> a[i][j];
}
}
int ans = 0;
for(int i = 0; i < n; i++)
{
for(int j = 0; j < m; j++)
{
if(!vis[i][j] && a[i][j] != 0)
{
ans = max(ans, dfs(i, j));
}
}
}
cout << ans << endl;
}
int32_t main(){
startt
int t = 1;
cin >> t;
while (t--) {
solve();
}
}
```

Idea: flamestorm

**Tutorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
const int MAX = 200007;
const int MOD = 1000000007;
void solve() {
int n, m;
cin >> n >> m;
int cnt[n + 1];
for (int i = 1; i <= n; i++) {
cnt[i] = 0;
}
for (int i = 0; i < m; i++) {
int u, v;
cin >> u >> v;
cnt[u]++;
cnt[v]++;
}
map<int, int> cnts;
for (int i = 1; i <= n; i++) {
cnts[cnt[i]]++;
}
vector<int> v;
for (auto p : cnts) {
v.push_back(p.second);
}
sort(v.begin(), v.end());
if (v.size() == 3) {
cout << v[1] << ' ' << v[2] / v[1] << '\n';
}
else {
cout << v[0] - 1 << ' ' << v[1] / (v[0] - 1) << '\n';
}
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int tt; cin >> tt; for (int i = 1; i <= tt; i++) {solve();}
// solve();
}
```

Idea: flamestorm

**Tutorial**

Tutorial is loading...

**Solution**

#include <bits/stdc++.h>
using namespace std;
long long ans[2000007];
long long a[1500][1500] = {}, curr = 1;
void solve() {
int n;
cin >> n;
cout << ans[n] << '\n';
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
for (int i = 1; i < 1500; i++) {
for (int j = i - 1; j >= 1; j--) {
a[j][i - j] = a[j - 1][i - j] + a[j][i - j - 1] - a[j - 1][i - j - 1] + curr * curr;
ans[curr] = a[j][i - j];
curr++;
}
}
int tt; cin >> tt; for (int i = 1; i <= tt; i++) {solve();}
// solve();
}

Idea: SlavicG

**Tutorial**

Tutorial is loading...

**Solution**

```
#include "bits/stdc++.h"
using namespace std;
#define ll long long
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(),v.rend()
#define pb push_back
#define sz(a) (int)a.size()
const int mod = 1e9 + 7;
void solve() {
int n, x; cin >> n >> x;
vector<int> a(n + 1);
vector<vector<int>> dp(n + 1, vector<int>(1 << 6, 0));
for(int i = 1; i <= n; ++i) {
cin >> a[i];
for(int mask = 0; mask < (1 << 6); ++mask) {
dp[i][mask] += dp[i - 1][mask];
if(dp[i][mask] >= mod) dp[i][mask] -= mod;
dp[i][mask & a[i]] += dp[i - 1][mask];
if(dp[i][mask & a[i]] >= mod) dp[i][mask & a[i]] -= mod;
}
dp[i][a[i]] = (dp[i][a[i]] + 1) % mod;
}
int ans = 0;
for(int mask = 0; mask < (1 << 6); ++mask) {
if(__builtin_popcount(mask) == x) {
ans = (ans + dp[n][mask]) % mod;
}
}
cout << ans << "\n";
}
int32_t main() {
ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
int t = 1;
cin >> t;
while(t--) {
solve();
}
}
```

thanks for the speedy editorial.

wow what speedy editorial！

In F , according to the constraints , if m = 1 , then how x and y could be both greater than 1 such that x*(y+1) = m = 1 ?

It is guaranteed that this graph is a snowflake graph.

So m=x+xy>2

m>=1 doesn't means that m can be 1.

In fact, $$$m=n-1$$$

Problem H can be solved in

`O(k * 2^k + n)`

using AND Convolution, as described here: https://mirror.codeforces.com/blog/entry/115438Implementation: https://mirror.codeforces.com/contest/1829/submission/204840516

Wow, dude, this is really nice! Because of commutativity of AND, this is possible! I thought that I could solve it as well with convolutions, but didn't see the commutativity and ended up solved it using the standard

`n * 2^k`

dp.This is very interesting! However, I have a major doubt: How did you realize that applying the inverse SoS on the amount of subsets whose AND contain mask is the same as the amount of subsets whose AND is EXACTLY mask? In the linked blog the example is given for pairs, but I fail to realize how to connect these ideas formally. Did you prove it? Can you share your insight?

Maybe I would understand it if I understood why it works for pairs, lol.

Thanks for the round!!

I liked problem G(only problem which I couldn't AC :sadge:) and the intended solution idea.

Wow. Loved G. Though was not able to solve it. But didn't even notice that just rotating it will make it a standard 2-D prefix sum problem. Lol struggled the whole 1hr during the contest.

Can you please explain it?

I am not able to get the tutorial.

Maybe it's because you don't know that technique called 2-D prefix sum. Google it and study it.

The pattern was [1,3,6..][2,5,9..].

Imagine this as the 2-D grid.

G was a really good problem, it reminded me of http://www.usaco.org/index.php?page=viewproblem2&cpid=416.

Why I always get tle?

use ios:sync_with_stdio(false); command

It doesn't work.

thank you, I have solved it.The array size is a little small.

Didn't get H

It's a standard dp bitmask problem.

G is possible by simply using sum of squares of N natural numbers for each row and is easy implementation too.

https://mirror.codeforces.com/contest/1829/submission/204823035

I think problem D amounts to checking whether the equation

`2^a * n == 3^b * m`

has a solution where`b >= a`

. I used two-pointers to search for a and b.Implementation: https://mirror.codeforces.com/contest/1829/submission/205007790

Yup, I noticed that too.

can you explain more about your approach?

think like that any number a can be divided to the form a/3 , 2a/3 which further divide into a/9 ,2a/9 and 4a/9 and so on so the number m should be of latter form . so find how much is the differences of maximum power of 3 say y and divide the first number by that . and then try checking for all powers of 2 if u can get to number m before finishing the tries u have which is y

thanks sir!

in problem F what is the output of this case :

$$$m$$$ can't be $$$1$$$. (https://mirror.codeforces.com/blog/entry/115896?#comment-1027861)

There is an additional constraint in the inputs.

Lol 30 minutes for trying to get suitable solution for case like that

Tonight's problems' interesting and not typical, thank the writer and codeforces for such a great contest!

Have anyone solved E using BFS?

Yes. 204779800

Thanks. Actually I am very new to Graph. I missed this --> "The idea is to consider each cell of the grid as a potential starting point for a lake, and explore all the cells reachable from it by only moving up, down, left or right, without stepping on any cell with depth 0"

My guess is that most people solved it using BFS

I solved it with DFS. However, the code was in Go, and I needed a few submissions to optimize memory usage. Strange thing...

Code link : https://mirror.codeforces.com/contest/1829/submission/205309024

please help me,it is giving Runtime error in test-12

If you open the submission details, it says that it's "Stack overflow" error. I don't know Java, but you should either increase the stack size or rewrite it using bfs

I solved by union find.

I solved it through Recursion

I did

Yes

https://mirror.codeforces.com/contest/1829/submission/204820630

i solved 5/8. Great contest everyone!!

Alt solution to G:

From the given $$$n$$$, try to go upwards to the left block $$$l_n$$$, and to the right block $$$r_n$$$.

If we can visit both ways, there some blocks will be taken twice. So we find the common blocks between $$$l_n$$$ and $$$r_n$$$ and subtract them from the answer. If the common block exists, it's either $$$l_{r_n}$$$ or $$$r_{l_n}$$$.

There is a recursive relation, until the blocks exist. It can be either of:

Implementation: 204825154

Do you mean to say like this: My Submission I guess it's easier this way.Not sure!

I also coded it this way, but if you look closely, this is actually just the 2d prefix sum solution!

It is!

This is just a different way of looking at it, without all the modifications. I feel this is more intuitive.

this is 2023 per test case right?

originally, this solution was right on the edge of passing :( and was supposed to not pass. But setters remembered its a div4.

It's actually $$$10^6$$$ for all test cases since you can do memoization.

Also you do not need all of the $$$2023$$$ rows.

My Submission my logic is to go upwards in the form of an inverted tree until the base of the inverted tree is greater than the number of cans in that level, after which all the cans above this level will fall, and if the can lies on the left or right boundary then only the boundary cans fall, but it seems to fail for large testcases can you please explain why?

New Submission

previous submission was a built faulty this is better.

For problem F, if you see:

`Wrong answer on test 2`

`wrong answer 65th numbers differ`

Then it is the

`x == y+1`

edge case mentioned in editorial.Someone please help me spot the error in my solution to D I spent almost the entire contest trying to fix it:

Here is a failing test case:

Can you see why this happens?

Ohhhhhhhhh, thank you so much.

Yeah I see now.

27 // 3 is 9 -> 9 > 8, so my solution tries to split up 9 to get 8 -> outputs NO.

And a fix be to use return (solve(n//3, m) or solve(n-n//3, m)) right?

Yeah, this should fix the problem.

did u solved it?

yup, here is final submission:

There is a no brainer dp solution to C.

DP

basically if we have S as a set of skills we took then xor(s)=3 ATP.

we have to find a set such that cost is minimum

204848665

my solution for (E) The lakes is clearly an O(mn) solution, but was exceeding limit, can anyone explain where it can be optimised?

204858437 is logically same as my solution 204848665 , but the first solution was accepted but mine got TLE, both are written in python

Currently you are adding the same squares into the queue multiple times: when adding a new square to the queue you're only checking if you've been to the square before. Importantly, you're not checking if the square has already been added to the queue. This starts actually growing exponentially and the time complexity is definitely not $$$O(nm)$$$. It can be fixed with a small modification — changing the place where some operations are done. This ensures that no square will be added to the queue more than once and the complexity is now truly $$$O(nm)$$$.

204888301

understood, thanks for the explanation

Actually an even simpler modification gets AC:

Before adding the next squares into the queue, just make sure that

`grid[i][j] > 0`

. Now no exponential growth will happen and each square will get added to the queue at most 4 times.204889157

Use visited array

I tried Solving E using Standard DFS in python Python Code. But I was continuously getting Runtime Error at TC 6. C ++ did the trick though (for the same) C++ code. Can someone tell me why I was getting Error in Python for the same Implementation ?

Python's default stack limit is 1000 which can be changed with

`sys.setrecursionlimit(big number)`

, on cases where you are forced to recurse deeper then 1000 layers (imagine a spiral pattern of values separated by 0's) you will run time error. (Be careful python recursion is very slow so you may run into issues with time limit)Changing to c++ fixed the issue because c++ does not have this issue with low default stack limit.

Thanks a lot. That solve a lot of my problems with python. Been getting these kinda errors frequently.

for beginners, how large can the "big number" can be in-case of O(N) required space complexity ? Will 10^4 do?

Ps Update: Just tried using it. It didn't make any difference in the outcome of the submission (RE Again ;_;). But thanks again, It was Informative.

Since the grid can be of size 1000 * 1000, you would need the stack limit to be that big as well (ie 1e6+ a bit). Though you might lead to TLE since recursion is quite slow in python. I would suggest if you do get tle (and can't switch languages) to try programming the flood fill as an iterative bfs as although the time complexity is the same the constant factor should be significantly faster.

yes, thankyou.

204849479

Why am I getting runtime error in this code?

Python's default stack limit is 1000 which can be changed with

`sys.setrecursionlimit(big number)`

, on cases where you are forced to recurse deeper then 1000 layers (imagine a spiral pattern of values separated by 0's) you will run time error.Got it! Thanks

WoW Fast Editorial...

wow.

so speedyJUST CAME BACK TO VISIT A TAYLOR SWIFT ROUND. OMG!!

come back...be here (another Taylor's song)

message in a bottleDear Reader, This is me trying to come back be here at this holy ground. and i almost do, but everything has changed and story of us looks like a tragedy now, so long live, had a marvellous time ruining everything. People in cf are the lucky one and they should stay stay stay.

It's time to go - Yours slayor twift

Can anybody hack this solution for TLE ?? https://mirror.codeforces.com/contest/1829/submission/204874235

Problem F: Observe that the starting vertex is the ONLY vertex that is not a leaf AND also has no leaf neighbours. 204824064

Or you can go with that whenever there is a level greater than 2 then that is surely not our central node of snowflake.204843315

for problem D, why we used this formula for master theorem: T(n) = 2T(n/3) + O(1) , but it's really T(n/3)+T((2*n)/3)+O(1) why we can Ignore that coeff (2*) ? thank you

Usually for time complexity we don't consider coefficients so we just say it is 2 * T(n/3), since the order is the same.

Thank you!

I thik this is wrong. Recurrence T(n)=T(n/3)+T(2n/3) +c has an O(n) solution for T(n). In fact, T(n) is Omega(n). This can be proved by induction easily. That is, you can prove that T(n) >= kn for some k and for every n sufficiently large.

Another approach to solve E (https://mirror.codeforces.com/contest/1829/submission/204823633)

Have anybody solved problem E without using bfs and DFS???

Here you can find another approach -> (https://mirror.codeforces.com/contest/1829/submission/204823633)

I just used dsu

$$$H$$$ can be solved in $$$(maxa_i^2 * t + ∑n)$$$ where you can just keep track of how many ways each number between $$$0-63$$$ inclusive can be constructed. Submission: 204886301 (binpower is not needed, we can just pre-calculate powers of $$$2$$$).

task E is the best

wow, fast editorial <3

that was a good contest, thanks :)

Amazing

There is a much faster solution to D

Each time you either multiply by 1/3 or 2/3, so the final multiplier must be 2^a / 3^b where a <= b.

Then just check whether the ratio m/n can be expressed in that form.

Code (gets AC):

204904968

There is an even more cheesy solution, where once you realize the 2^a / 3^b and a<=b part, you precompute a list of all such fractions of that form within 10^7, and for each test case, just check whether m/n is equal to some fraction in that list:

(also AC)

One thing I noticed on F:

Each node is in the 2nd layer of nodes if and only if its degree is $$$1$$$. That means we can perform the following:

my submission

I used similar things too during the contest.

The number of the 1st layer's vertices is (the sum of vertices — the number of the 2nd layer's vertices — 1), and we can use that to calculate the answer.

my submission during the contest

my strat for F was to take a node with degree 1, and go to the next node. the degree of this node is y + 1. then go to a node which degree is greater than one from this node, and the degree of this node is x.

question E was interesting. loved solving it.

Can someone explain in problem c why did we use >1e6 for the -1 case.

Doesn't really matter you just have to take a number which is larger than any input number which are given less than or equal to 10^5 so any number greater than 10^5 will work.

One interesting Observation in F is that all leaf nodes have degree 1 so count all degree 1 lets say total and then other are degree >1 which means they would be x+1 as one node with degree >1 would be central vertex so answer would be x and total/x.

For E, my dfs solution gives TLE which has the same logic as the solution given. When I changed the visited array and input array to global variables, it got accepted. Why is that so?

because you define a new 2d vector after each test, and that consumes much time

Got it, thanks :)

Can someone tell me why my H problem is giving TLE with $$$dp[n][64]$$$ 204872960 but accepted with space optimization 204874966?

Hi, I wonder is this round rated for all who has a rating < 1400? If so, why my ratings didn't change? I'm new to codeforces so please forgive me for asking these questions. Thanks!

System testing is going on, after some time ratings will change.

Thanks!

The round is not rated for anyone

I submitted the question (C) 2 min before the contest end , last time it was showing it green now its showing it red ?? WHY?

Some of the hacks are very... strange looking.

First

Second

I solved G in a similar way. Let $$$dp_i$$$ be the answer for $$$i$$$ and $$$x_i$$$ be the row for $$$i$$$. Now we know that $$$dp_1 = 1$$$ , and $$$dp_0 = 0$$$, and for each $$$1 \leq i \leq N$$$ we have two cases:

$$$i - x_i$$$ only lies on the previous row $$$x_i - 1$$$ , then $$$dp_i = i^2 + dp_{i-x_i}$$$

$$$i - x_i + 1$$$ only lies on the previous row $$$x_i - 1$$$ , then $$$dp_i = i^2 + dp_{i-x_i+1}$$$ both $$$i - x_i$$$ and $$$i - x_i + 1$$$ lie on the previous row $$$x_i - 1$$$. Here we can't simply do $$$dp_i = i^2 + dp_{i-x_i} + dp_{i-x_i+1}$$$ because $$$dp_{i-x_i}$$$ and $$$dp_{i-x_i+1}$$$ have common numbers that were added to both of them. We can see that they have different answers except the one they took from $$$dp_{i-2(x+1)}$$$ so it becomes $$$dp_i = i^2 + dp_{i-x_i} + dp_{i-x_i+1} - dp_{i-2(x+1)}$$$

I approached F the following: let be the number of outermost nodes l.

One can easily find this "l" from the input. It turns out that l = x*y, also, the total number of nodes is n = 1 + x + x*y. So we have due independent equations which yields to x and y values

alt for D, we can just check wether we can get from m to n if we can do the operations below

m := 3*mm := 3*m/2 (if m mod 2 = 0)So we need to check if there exist integer pair

xandy(x >= y and m mod 2^y = 0)s.t.m * 3^x / 2^y = nwe can do it in

O(log2(n))How could H be solved with larger values? for example (a[i] <= 1e6 or 1e9) and of course a corresponding K.

I am Expert now thanks to this round, SpeedForces forever xD

That's a huge jump orz

What about this?

I dont know why my submission 204795848 in the contest using C++ 17 giving the TLE but it's Accepted if I submit it with C++20 204978167

I set the recursion limit and I still get Runtime Error: 204981121 Can anyone help?

Thanks for a great round! I solved 5/8 problems, which is the highest ever (usually no more than 3). I solved the first 4 problems in 30 minutes, and then an hour later I solved F, which seemed to me easier than E, the idea of which I did not understand.

/* this question is basically of precomputation + dp in this question we will first precompute the matrix then make a loop from 1 to 1e6 and calculate the answer for all the elements and then for each query we will give answer in O(1)

/ /dp of current element will be the x^2 (dp[k]=(vec[i][j]*vec[i][j])) + dp of the two elements above it if exists i.e. { if(i-1>=0 && j-1>=0) dp[k]+=dp[vec[i-1][j-1]]; if(i-1>=0 && j<vec[i-1].size()) dp[k]+=dp[vec[i-1][j]]; }*/ // do a dry run you will get by yourself

/* <--------------------------------- THANK YOU -----------------------------> */

https://mirror.codeforces.com/contest/1829/submission/205005182

My Python implementation of the editorial solution passes only first 5 tests, then gets Runtime Error on test 6. https://mirror.codeforces.com/contest/1829/submission/205011680

Can someone help find the problem?

I got the same problem. After some trials (to me) it seems impossible to pass the problem in python with recursion. I used bfs to solve this. (dfs is also not ok for it will cause MLE)

Here's my submission: 204988219

Tip: check each cell if it is visited or is 0 before bfs to optimize the time

C hack has been evil U_U

Just an implementation detail about the tutorial solution for problem 1829H - Don't Blame Me: if you are using GNU C++20 compiler, then you do not have to use the built-in function

`__builtin_popcount`

. You can use the standard library function std::popcount instead. Also, as the next-state at any index $$$i$$$ depends only on the previous state at index $$$i-1$$$, it is sufficient to compute the answer using two vectors only instead of using a matrix of $$$n+1$$$ vectors.Accepted SolutionNice problems ,i have got +156 points (:

I had different idea for G. Keep start and end positions of cups in a line then go up and use math formula for the sum of n squares: 205064632 O(1) space and O(n) run

Hello! Does anyone know how to solve H using combinatorics?

Can someone post DP solution of G?

See the comment above.

This is my solution, I just calculate how many numbers appear once. Then i subtract that number from number of nodes and i subtract -1 to get rid of the first node and from there i get x.

void solve() { int n,m; cin >> n >> m; map<int,int> mp;

}

int main() { ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);

}

Can anyone please explain the tutorial for F (Forever Winter)? I am unable to understand the logic behind it. I only coded a TLE solution which was accepted at the time of the contest, then it changed to TLE.

EDIT:- Nevermind. I understood with the help of other's submissions.

UPD : solved

Nice tutoriel

## include

## include

## include

using namespace std; typedef long long ll;

bool solve1(ll n, ll x, vector& dp) { if (n == x) { return true; } if (x > n || n % 3 != 0) { return false; } if (dp[n] != -1) { return dp[n]; } bool ans = solve1(n / 3, x, dp) || solve1(2 * n / 3, x, dp); dp[n] = ans ? 1 : 0; return dp[n]; }

void solve() { ll n, m; cin >> n >> m; vector dp(n + 1, -1); if (solve1(n, m, dp)) { cout << "YES" << endl; } else { cout << "NO" << endl; } }

int main() { cin.tie(0); cin.sync_with_stdio(0); cout.tie(0); cout.sync_with_stdio(0);

} Why this code is giving Time complexity in Problem D, ;Gold Rush?

Editor seems to be fan of Taylor Swift

In problem G, I was using bfs for solution but it didnt seem to give any answer for the last test of test case 1. Can anyone tell why.

## include<bits/stdc++.h>

## include <ext/pb_ds/assoc_container.hpp>

## include <ext/pb_ds/tree_policy.hpp>

using namespace std; using namespace __gnu_pbds;

## define ll long long

## define pb push_back

## define msort(v) sort(v.begin(),v.end())

## define loop(ii,n) for(long long ii = 0; ii < n; ++ ii)

## define pii pair<int,int>

## define ff first

## define ss second

// #define ordered_set tree<int, null_type,less, rb_tree_tag,tree_order_statistics_node_update>

## define ordered_set tree<pair<int,int>, null_type,less<pair<int,int>>, rb_tree_tag,tree_order_statistics_node_update>

## define o_key order_of_key

## define o_set ordered_set

bool cmp(pair<int,int>& i1, pair<int,int>& i2) { return (i1.first < i2.first); } bool prime[1000005]; void sieve() { memset(prime, true, sizeof(prime));

} ll len(ll a) { int ans=0; while(a>0){a/=10; ans++;} return ans; } map<int,int> mprime; void pf(int n) {

} long long binpow(long long a, long long b) { long long res = 1; while (b > 0) { if (b & 1) res = res * a; a = a * a; b >>= 1; } return res; } ll cnt(ll i) { ll ans=0; while(i>0) { ans+=(i%10); i/=10; } return ans; } /////////////////////////////////////////////////////////////////////////// //////////////////// DO NOT TOUCH BEFORE THIS LINE //////////////////////// ///////////////////////////////////////////////////////////////////////////

int main() { ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);

}

Alternate approach for G: https://mirror.codeforces.com/contest/1829/submission/207013133

Basically whenever we move up the tower, we make the left top and right top fall down. But the sum of a single element is not simply the sum of its parents and itself, this solution deals with that by choosing the right upper part only for the right parent, hence minimizing any overlap.

Examples: If we were to calculate for n=13, it would go the following way-