All problems were invented MikeMirzayanov and developed by me Supermagzzz and Stepavly.

**Editorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
void solve() {
int w, h, n;
cin >> w >> h >> n;
int res = 1;
while (w % 2 == 0) {
w /= 2;
res *= 2;
}
while (h % 2 == 0) {
h /= 2;
res *= 2;
}
cout << (res >= n ? "YES\n" : "NO\n");
}
int main() {
int t;
cin >> t;
while (t--) {
solve();
}
}
```

**Editorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
void solve() {
int n;
cin >> n;
int cnt1 = 0, cnt2 = 0;
for (int i = 0; i < n; i++) {
int c;
cin >> c;
if (c == 1) {
cnt1++;
} else {
cnt2++;
}
}
if ((cnt1 + 2 * cnt2) % 2 != 0) {
cout << "NO\n";
} else {
int sum = (cnt1 + 2 * cnt2) / 2;
if (sum % 2 == 0 || (sum % 2 == 1 && cnt1 != 0)) {
cout << "YES\n";
} else {
cout << "NO\n";
}
}
}
int main() {
int t;
cin >> t;
while (t--) {
solve();
}
}
```

**Editorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
void solve() {
int n;
cin >> n;
vector<int> a(n);
for (int &x : a) {
cin >> x;
}
vector<int> dp(n);
for (int i = n - 1; i >= 0; i--) {
dp[i] = a[i];
int j = i + a[i];
if (j < n) {
dp[i] += dp[j];
}
}
cout << *max_element(dp.begin(), dp.end()) << endl;
}
int main() {
int tests;
cin >> tests;
while (tests-- > 0) {
solve();
}
return 0;
}
```

**Editorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
void solve() {
int n;
cin >> n;
vector<int> v(n);
for (int &e : v) {
cin >> e;
}
sort(v.rbegin(), v.rend());
ll ans = 0;
for (int i = 0; i < n; i++) {
if (i % 2 == 0) {
if (v[i] % 2 == 0) {
ans += v[i];
}
} else {
if (v[i] % 2 == 1) {
ans -= v[i];
}
}
}
if (ans == 0) {
cout << "Tie\n";
} else if (ans > 0) {
cout << "Alice\n";
} else {
cout << "Bob\n";
}
}
int main() {
int t;
cin >> t;
while (t--) {
solve();
}
return 0;
}
```

**Editorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
using pii = pair<int, int>;
struct man {
int h, w, id;
};
bool operator<(const man &a, const man &b) {
return tie(a.h, a.w, a.id) < tie(b.h, b.w, b.id);
}
struct my_min {
pii mn1, mn2;
};
vector<pair<int, my_min>> createPrefMins(const vector<man>& a) {
vector<pair<int, my_min>> prefMin;
my_min curMin{pii(INT_MAX, -1), pii(INT_MAX, -1)};
for (auto x : a) {
if (x.w < curMin.mn1.first) {
curMin.mn2 = curMin.mn1;
curMin.mn1 = pii(x.w, x.id);
} else {
curMin.mn2 = min(curMin.mn2, pii(x.w, x.id));
}
prefMin.emplace_back(x.h, curMin);
}
return prefMin;
}
int findAny(const vector<pair<int, my_min>> &mins, int h, int w, int id) {
int l = -1, r = (int) mins.size();
while (r - l > 1) {
int m = (l + r) / 2;
if (mins[m].first < h) {
l = m;
} else {
r = m;
}
}
if (l == -1) {
return -1;
}
auto mn1 = mins[l].second.mn1;
auto mn2 = mins[l].second.mn2;
if (mn1.second != id) {
return mn1.first < w ? mn1.second + 1 : -1;
}
return mn2.first < w ? mn2.second + 1 : -1;
}
void solve() {
int n;
cin >> n;
vector<man> hor, ver;
vector<pii> a;
for (int i = 0; i < n; i++) {
int h, w;
cin >> h >> w;
hor.push_back({h, w, i});
ver.push_back({w, h, i});
a.emplace_back(h, w);
}
sort(hor.begin(), hor.end());
sort(ver.begin(), ver.end());
auto horMins = createPrefMins(hor);
auto verMins = createPrefMins(ver);
for (int i = 0; i < n; i++) {
auto[h, w] = a[i];
int id = findAny(horMins, h, w, i);
if (id == -1) {
id = findAny(verMins, h, w, i);
}
cout << id << " ";
}
cout << endl;
}
int main() {
int tests;
cin >> tests;
while (tests-- > 0) {
solve();
}
return 0;
}
```

**Editorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
void solve() {
int n, m;
cin >> n >> m;
map<int, int> v;
for (int i = 0; i < m; i++) {
int x, y;
cin >> x >> y;
v[y] |= (1 << (x - 1));
}
const int FULL = 3;
v[2e9] = FULL;
int hasLast = 0, lastColor = 0;
for (auto[x, mask] : v) {
if (mask != FULL && hasLast) {
int color = (x + mask) % 2;
if (lastColor == color) {
cout << "NO\n";
return;
} else {
hasLast = false;
}
} else if (mask == FULL && hasLast) {
cout << "NO\n";
return;
} else if (mask != FULL) {
lastColor = (x + mask) % 2;
hasLast = true;
}
}
cout << "YES\n";
}
int main() {
int t;
cin >> t;
while (t--) {
solve();
}
return 0;
}
```

**Editorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
vector<int> calcDist(vector<vector<int>> const &g) {
vector<int> dist(g.size(), -1);
dist[1] = 0;
queue<int> pq;
pq.push(1);
while (!pq.empty()) {
int u = pq.front();
pq.pop();
for (int v : g[u]) {
if (dist[v] == -1) {
dist[v] = dist[u] + 1;
pq.push(v);
}
}
}
return dist;
}
void dfs(int u, vector<vector<int>> const &g, vector<int> const &dist, vector<int> &dp, vector<bool> &used) {
used[u] = true;
dp[u] = dist[u];
for (int v : g[u]) {
if (!used[v] && dist[u] < dist[v]) {
dfs(v, g, dist, dp, used);
}
if (dist[u] < dist[v]) {
dp[u] = min(dp[u], dp[v]);
} else {
dp[u] = min(dp[u], dist[v]);
}
}
}
void solve() {
int n, m;
cin >> n >> m;
vector<vector<int>> g(n + 1);
for (int i = 0; i < m; i++) {
int u, v;
cin >> u >> v;
g[u].push_back(v);
}
vector<int> dist = calcDist(g);
vector<int> dp(n + 1);
vector<bool> used(n + 1);
dfs(1, g, dist, dp, used);
for (int i = 1; i <= n; i++) {
cout << dp[i] << " ";
}
cout << endl;
}
int main() {
int tests;
cin >> tests;
while (tests-- > 0) {
solve();
}
return 0;
}
```

How the editorial posted 2 years ago? If the contest is running.

2 "Hours" ago.

You can write blogs and not publish them. In this case, it is so the author can get a fast editorial out — by pre-writing the editorial but not making it public yet.

You wrote "years" instead of "hours":)

problem D was almost same as this one.

Can someone please explain why in 1472G - Переезд в столицу you can't go in first sample:

so there is exactly one step of kind 2.

To calc the distances you need to consider the direction of the edges.

You use road with d[2] -> d[1] twice. It's not allowed

I use different roads. Are you talking about same distances change?

Yes, you use road with this distances twice

Oh, I got where is my mistake. I mixed up two kinds. Actually [2]->[1] is second kind of move. I thought it is first kind. shame.

Your move from 6->2 is a move of the second kind (dist 2 to dist 1) The move from 5->1 is also a move of the second kind (dist 2 to dist 0).

Therefore, you should stop at city 2 — a distance of 1 from the capital.

yeah,I have the same question.

DIV 3 PROBLEM C VIDEO TUTORIAL LINK https://www.youtube.com/watch?v=7FFzXk-brRU I have tried my best to give you guys proper intuition of this dp problem. Hope you guys will enjoy this video !!!

how can editorial be posted if hacking phase is still ongoing

That's ok.

anyone please explain d

For D you only have to justify to yourself why taking the biggest number is always the best choice to make on your turn. The editorial actually does a pretty job of this, but think of it like this — either you take the biggest item left, and it matches your parity (i.e. add it to your score), or it doesn't (your opponent doesn't gain from that item on their turn — also good for you). It's simple to see that therefore a player should ALWAYS take the biggest item (i.e. sort the array and check each move).

thanks got it : )

Thank you, really simple explanation, even i wasn't able to understand editorial's solution.

Thanks again !!

In the code though I saw that the if condition simply added or minused from result based on it's eveness or oddness. Shouldn't it be based on whose turn it was instead? Cause Alice can take an odd number and result wouldn't change.

the index takes care of alice/bob turn...after sorting it in reverse order...the even index corresponds to alice and the odd index corresponds to bob

Oh yeah, i misread it

Actually the solution was so simple that I spent 20 mins trying to prove it but still submitted this method without proving it because I couldn't find any other better method

Can you please tell me for which test case I am getting WA by using the above same approach? https://codeforces.com/contest/1472/submission/172212393

Another way to look at it:

At each point the player wants to maximise the distance between the scores of one another.

You may be tempted to look for parity while picking the numbers, but if the goal is to simply maximise the distance then you increasing your score is equal to hindering the others progress.

therefore at each point you pick The largest number which has a potential to reduce the difference between two scores.

I hope this makes sense.

Thank you much

Can someone please explain the approach for B (non DP approach)?

h

There are two cases

in second case if there is no zeroes then and only then it cant be divided other wise its always YES

Can you explain the odd+odd part?

if both parts are odd then there must be atleast two onez their to make both parts equal.

For ex: 6=3+3 if it doesn't consist of any both part can't be equal (it will be like (2,4) ) because by only 2's we can't make a odd number.

Odd means even+odd So, dividing it by odd+odd It means if sum/2 is odd, there has to be atleast 1 ones in the array. As, 2 is even

how dp is used i dont understand

`Simple solution — let's iterate through how many candies of weight 2 we will give to Alice, then the remaining weight should be filled by candies of weight 1. If there are enough of them, then we have found a way of division.`

How is this true ? If we take the case of 1 1 2 2 2, then Alice will get 2 2, while Bob will get 2 ,1 and 1.Hello jiggly_puff Lets consider the test case 1 1 2 2 2 then As you have writted by yourself Alice will get {2,2} and bob will get {1,1,2} now total weight of alice candies = 4 toatl weight of bob candies = 4 , hence it will return YES as alice_weight = Bob_weight

Not the best solution, but probably might be easier to understand:

https://codeforces.com/contest/1472/submission/103314992

So you can use it as a base for a further optimization.

Approach for problem B.

Let the number of 1's is

one& the number of 2's istwo.Let total sum = one + 2 x two.

Casework:Case 1Case 2Case 3Case 4.1Case 4.2You can also see my solution here : 103195164

Lets say s represents the sum of all elements, If s is odd then its totally impossible to split it into two equal halves

Otherwise if s / 2 odd then you need to have atleast one 1 in the array to make the odd half

so the answer is no if (you have a odd total) or (a even total but odd half and no ones with you), else answer is yes

I used a method based on some observations. You can have a look at my code here. If the sum is odd, obviously there's no way of distributing them in equal halves. Also, if you have no 1's and an odd number of 2's, you again can't do it. In all other cases, you can always come up with a solution, just try out some test cases yourself!

Problem B is really similar to Div-2A Kitahara Haruki's Gift (https://codeforces.com/problemset/problem/433/A)

Great Contest Overall

Can someone explain

lasttestcase of Problem G.For node 3 , shouldn't the ans be 1

we can go

3 (2) --> 4 (3) ---> 2 (1)

value in brackets is the corresponding d values

3--> 4 is of type 1

and

4--> 2 is of type 2

Distance to 4 is equal to 2 by 1>5>4, not 3.

thanks

Can someone explain why 6->2->5->1 isn't a valid path for the first testcase of the sample input for problem G? I'm pretty confused cause we go 6(2) -> 2(1) -> 5(2) -> 1(0).

Thank you very much, now I can solve E, F and G tasks!

A way to avoid repeating the algorithm in E, is to always have H < W for every i. So just swap H and W for some person, if H > W.

Could you explain a little bit more ? thanks in advance.

Now I got it. the condition for the person i to be placed in front of the person j can be rewritten like this : min(h_i, w_i) < min(h_j, w_j) and max(h_i, w_i) < max(h_j, w_j) This means you can compare like that.

this makes question very easy. thanks

`Simple solution — let's iterate through how many candies of weight 2 we will give to Alice, then the remaining weight should be filled by candies of weight 1. If there are enough of them, then we have found a way of division.`

How is this true ? If we take the case of 1 1 2 2 2, then Alice will get 2 2, while Bob will get 2 ,1 and 1.How is this not true?

Here are my attempts to explain solutions in video form after messing almost everything up (this seems to happen more and more often recently...)

on G,can someone explain why (dist[u] must <dist[v]),my code doesn't add this condition and i got WA. Thanks!

if (!used[v] && dist[u] < dist[v]) { dfs(v, g, dist, dp, used); } (In the DFS function)

Because you can use only the first option from the statement as much times, as you want. We build the new graph, with edges that satisfy this condition. The second option we check by DP.

Hi, This is my Dfs function. Where did I perform the second operation twice?

mark: stores the final answer.d: stores the distance from root.Adding that conditon you mentioned, gives me an AC. Can you please tell, where am I going wrong? Thanks

I am stuck in the same situation. Even without dist[u] < dist[v] condition, we seem to be ensuring that 2nd condition is only counted once because

Thumbs up to the guys above. I also got stuck over this condition. For anyone interested, here's a graph exposing this bug:

In short, without the

`dist[u] < dist[v]`

condition, your DFS may update answer for some vertex`x`

(`dp[x]`

) by using`dp[y]`

at a moment when`y`

is still being processed by DFS somewhere up the tree. In this case,`dp[y]`

isn't finalized, it's just not ready. In the graph above,`x`

is 5 and`y`

is 3.Takeaway: DFS doesn't like when you close loops; better let the old DFS finish and only then spawn the new DFS tree to process

`x`

.Thanks man!

The editorial for G is very smooth. Great work Supermagzzz and Stepavly.

I don't know how people solve problem like today's D so fast !! sed lyf :((

Donald trump complaining even on someone solving a problem faster. Not surprising.

I am not Donald Trump..see again

Problem B Video Editorial link :https://www.youtube.com/watch?v=ZVLnPd0_3BU I have tried my best to explain with intuition and corner case.I hope you will enjoy this video !!!

What a fast editorial! Really helpful; Thank you!

I wonder with B they did not include the alternate solution with DP

Could anyone please help me to resolve

MLEin this solution for Problem CYour dp size is 20000 instead of 200000

you're passing the vector by value, so each time you call

`solve()`

a new copy is created. Pass it by reference or make it global.Thanks :-)

Why this dp solution gives WA on B.I thought it is equal sum partition. https://codeforces.com/contest/1472/submission/103310451

I had two use greedy approach for this but I am willing to know it's do solution and fault in my code. Thanks in advance.

Thanks bro.

I came across an ID in this round which was submitting solution with wrong answer for a particular corner case. FOR eg if n==123234 then print "NO" , this was done so that it could be hacked by another IDs. :p

Name of the ID please.

Charon_souls_carrier

Can someone please help me why my code gets a WA(I get why it might get TLE but no WA): 103300499

wrong answer jury found the answer, but participant didn't (test case 11)

This is a comment of your code, so you can look for test case 11(it is visible) and see why it doesn't work. EDIT: you're only looking for elements with smaller X but a correct answer could also be element whose X is larger Y is smaller and so when it is laid sideways than it could fit in the element that you are checking. So your for loop J should go from 0 to n (i.e. check all elements) P.S. even if you did this you would get TLE

Can anyone please explain why my solution is showing WA. My logic is obvious. One can easily get that after seeing my code. Submission

Thanks in advance for helping :)

I think its because you didn't sort the arrays. You should always take largest number, and even if you divide numbers in odd and even you should sort the array and look for larges numbers. I used a similar approach 103227720. Hope this helps.

Yeah.. Thanks I get it. I didn't sort the array (T_T) Literally if you see my commented solution below it i had sorted the array. But i missed while writing new one. I thought i have sorted the arrays. Why these things happen to me (T_T)

I didn't actually understand your code, but I think the approach is similar. So hopefully this will be helpful!Explanation for DLet's observe first, when it's Alice's turn, she either blocks(selects the biggest odd number so the maximum point that Bob can take decreases) or takes the biggest even number.

It's also similar for Bob, he either tries to block Alice by selecting the biggest even number(so the maximum point that Alice can take decreases) or takes the biggest odd number.

Why this is optimal?

Let's take this case as an example: $$$ \text {N = 4, } \ $$$ $$$\text{A: {5 3 4 6}}$$$

1-Alice starts, now there are 2 options, blocking 5 or taking the 6.1.A- If she blocks the Bob, then Bob will also block her and she'll lost 1 points because of herself. And it'll stay at "Tie" when it's Alice's turn(3rd turn) again, thing is, if Alice had choosen to take 6, she would be on the lead now(check1.B)1.B- If she takes the 6, then Bob will take the 5, and now it's Alice's turn(3rd) again, and now she is on the lead because of her decision.2-It's the 3rd turn, Alice 1 — Bob 0, and the array is {3,4}2.A-Alice can choose taking the 4 or blocking the 3, if she blocks the Bob, then Bob will also block her in the next round, so the final points would be (1 — 0)(It's true but not optimal, check

2.B)2.B-Now if Alice chooses taking the 4, Bob will take the three in the next round and the final points would be (5 — 3)Because the difference is higher,

2.Bis more optimal for Alice. But, because of Bob has no action over the 1st turn of Alice, it's also optimal for him too.I hope this helps. Now for the coding phase,

Code partMy code for D

Basically what this code does is, if $$$\text{i is even}$$$ it means it's Alice's turn, and if the current biggest value is even, she takes it, if it's odd, she blocks it.

Opposite case for Bob.

Note: I wrote this 5 minutes before I went to sleep, so if there are mistakes, forgive me and correct me please.Thanks for the help. I get it now :)

Man thanks it really helped !

Can anyone explain problem G? Why do we have to change the graph? Why can't we apply dp on the graph with cycles?

I have applied dp only my friend. You can see my submission. Reply here if you didn't get the logic.

I enjoyed the last dp problems of this contest :)

Does anyone else feel that E was too much implementation heavy?

I thought about exactly this solution but with 30 mins remaining.

Then dropped the idea thinking that it might have some better segment tree solution instead of so much implementation. lol

Actually, it is not (at least for me).

It could be implemented using a segment tree in about 30 ~ 40 lines of code.

G was cool, E took me a while but it was fun too. Overall good problems!

Is it good to write an editorial while the hacking phase is still on?? I don't think so.

It was a bit tough contest

Are we allowed (yet) to ask inquiries regarding hacks for this round?

Can anyone tell me how ans of this test case of ques d is TIE.

3 2 1

I guess here bob should win. First alice chooses 2, then bob goes with 3 . Then at end alice chooses 1 and removes it. So bob wins here(since score of bob is 3 and alice score is 2).

How is it draw?

IF Alica pick 3 first, then if Bob pick 1 Alica wiil win because in next turn she will pick 2(Bob_score = 1, Alica_score = 2), if Bob picked 2 insted of 1, Alica will be left with 1 (Bob_score = 0, because he piced even number, Alica_score = 0, because she picked odd number). Then answer is TIE.

Can C be done by top-down approach?

Yes , link to my submission https://codeforces.com/contest/1472/submission/103216440

Got it, thanks!!

Why is this Solution got hacked? It is almost same as the Editorial. Please let me know as I don't wanna to repeat the same mistake again. Thanks in advance !

Yes, I was wondering the same thing as well. My solution (and many other Java solutions of a similar format) were also hacked (and ultimately received TLE), but I cannot find any apparent inefficiencies.

I heard Java sort uses quick sort. Quick sort in a worst case takes $$$O(N^2)$$$ complexity in an anti quick sort test.

That explains it. Thanks!

Is this Dp forces ?

I have two implementations for E which I think are easier than the one in the editorial.

A : With comments, Without comments

B : With comments, Without comments

Thankx Bro ! your A with comments was easy to understand!

no problem

Frustrating , Frustrating .... Can't believe it i wasted more than 1 hour , to solve D and never noticed variable was crossing integer range , So guys always check if range is crossed or not :(

In the first test case of the given example of question E,

Test case:

3

3 4

5 4

3 3

Ans: -1 3 -1

Why is it -1 for the first pair? 3,3 can be placed in front of 3,4 right? Am I missing something?

because the height and weight have to be strictly less than the person behind. 3 is less than 4, but 3 is not less than 3.

Oh, silly mistake

Very Simple solution using Priority Queue, C++.

DPContest :)There is only 2 problems I solved with dp and I think this is enough. Problems which have dp tag also have some other solutions this contest

cf 693 1st question please help me , on what test case my code fails because I think that my code is right Its humble request thanks in advance

include <bits/stdc++.h> using namespace std;

long long count(int n){ long long c=0; while(n%2==0){ n=n/2; c++; } return c; }

int main() {

int t; cin >> t; while (t-- > 0) { int w, h; long long n; cin >> w >> h >> n;

}

return 0; }

U multiplying count (w/h) by 2. Rather, make it like this-

long long x1/x2 = pow (2,count (w/h))

And in, a1/a2 u r multiplying it.. i don't know if it works or not.. but u can give it like this

long long a1= pow (2,count (w)+count (h))

then, just check if a1>=n

Reason- suppose u can cut 3 times, u can make 8 papers. But your answer gives 6 papers

My code give correct answer on the test cases given by codechef .

In this I use 5 condition 1st if n==1 if n is one then ans is yes 2nd if w and h is odd and n is not 1 then ans is NO 3rd if w is even and h is odd then I count how many times w is divided by 2 ,and finally multiply by 2 For example w= 4 and h =7 then max=4 4th case is vice versa of 3rd 5th case if both w and h is even then I count how many times w and h is divided by 2 ,and and multiply both by 2 For example w=4 and h=8 then max=24

But when I submit the code it is saying wrong output on test cases 2 Please help

I made the changes that I told you to make and it is giving me AC.

Here you go

Thank you so much Finnaly accepted I need one suggestion from you Suppose I gave a codeforces contest and solve 2 question and I got stuck in 3rd question Then should I only have to see the 3rd question editorial or how many questions should I have to up solve I mean that how I improve myself Any tip from your side

I literally did the same thing for problem D — 103294044

Can anyone help me how to optimise this, because I got tle

Quoting some lines in your code :

Finding max in an array takes O(n) time, and you are doing that operation for n times, that means your code is effectively getting to a complexity of O(n^2). But the constraints on n is

`(1 ≤ n ≤ 2 * 10^5)`

, so for O(n^2) your code would have to do`(2 * 10^5) * (2 * 10^5) = 4 * 10^10`

operations per second. But normal CPUs can do only about`10^8`

operations per second. Hence you are getting a TLE.Oh ryt! So would it be better if I first sort and then use them? or is there any other way? which is even better than sorting and using

First Sorting and then using would work fine. I am also linking my solution to the problem so that you may take reference (Although it's pretty much the same what editorial says) — 103483268. If my answer helped you in any way, please do vote a like!

Thanks a lot! I liked it too XD

I took a different approach for A, first I reduced the width and height to minimum it could be. Thus I have the width and height of the smallest card. I can use it find the area. Now I divide the area of the original card with the area of the smallest card. Now I have the number of cards I can create.

For D first test case my code is giving right output in my pc whereas on codeforces it gives wrong output 103274282

else if (o[j]>=e[i] && !fl) { sumb+=**o[i];** --j; fl=!fl; }

Maybe this is causing some problem

I was given similar problem as D in one of my interviews and I couldn't think of greedy approach, could only come with DP approach and was rejected. Now I solved it with so much ease in comfortable environment. Hate interview pressure.

Can someone please explain E?

Have a look at the video by galen_colin: https://www.youtube.com/watch?v=p5FNuXJoLDk&t=1054s

In problem D (language used : JAVA 11) , I am getting TLE in testcase 10 with this snippet of code

while(t-->0){ int n = ni();

and getting AC with this snippet of code

while(t-->0){ int n = ni();

Can anyone help me to know what is the cause?

Exactly what you are not able to understand?

In first code I sort the array in non decreasing order and I traverse the array from last for chosing the number for alice and bob

and in second I sort the array in non increasing order and I traverse the array from starting for chosing the number for alice and bob

Both are about same approaches nearly while first one is getting TLE and second one is AC

Ok, I found the reason: It's because you're using Wrapper class (Integer class) in the second case. When we use wrapper classes, we are basically sorting "Objects" instead of "Values", and Arrays.sort() handles them differently. It implements "Merge Sort" while sorting Objects (Worst case time complexity: O(N*log(N))) but implements "Quick sort" while sorting Values ((Worst case time complexity: O(N^2))). So, if you use Wrapper class in the first case instead of primitive type, it would work fine as well.

Accepted Submission: https://codeforces.com/contest/1472/submission/103375034

For detailed explanation, check out the reference: https://www.geeksforgeeks.org/java-handling-tle-while-using-arrays-sort-function/

Thank you so much

Learn a lot from above

Please use ruffle sort (Randomized quick sort to avoid quick sort worst case)-: My Code in case it helps-: `package Div3_693;

import java.io.BufferedReader; import java.io.InputStreamReader; import java.util.Arrays; import java.util.Random;

public class D {

// Arrays.sort(arr); ruffleSort(arr); long ascore=0,bscore=0;

} `

https://codeforces.com/contest/1472/submission/103236341 this is my solution of D main idea is what person must bring max(his parity, opposite parity(it means not give this element to enemy))

Can problem E be solved with monotone stack? Think the array to be circular, then for each element find the next element so that (h1 < h2 and w1 < w2) or (w1 < h2 and h1 < w2)

anyone can provide better sol for Problem E or any video

I wrote my solution here.

Hi, I am facing some problems with D. Even-Odd Game. I have read the editorial as well as the solution provided. I think I am doing the exact same thing given in the solution: always selecting the biggest available number for either player and adding or subtracting it from a global sum depending on who the player is and the parity. However, I am getting wrong answer on test case 3. ~~~~~

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

using namespace std;

int main() { int t, n; cin >> t;

} ~~~~~

Make

scorelong longThanks! It worked.

Always take care of data type you are using and what max data program can generate.

Thanks for this round. I like problem G very much, it is the classic Codeforces' style task. Quality of this round was much better than previous five or even more.

In problem D i have used erase for deleting the element and it has failed, but i see solution with same logic but using pop_back() instead of erase getting their solution accepted. I want to know that whether the time complexity of erase more than pop_back() or are they same in c++ ?

See this and this.

I have a different solution for F, where I keep a offset between the top and bottom pointer, and use some case analysis. Time is O(mlogm).

As the problem D has a trival solution, I have faced a false plagarism accusion on it.

"Your solution 103217882 for the problem 1472D significantly coincides with solutions lit2019039/103212550, whohet/103217882."

This question has a really trival solution and hence is clearly a coincidence. Please help and remove this plagarism mark. This is literally a coincidence, as the problem really have same solution. My solution link https://codeforces.com/contest/1472/submission/103217882

Match solution link https://codeforces.com/contest/1472/submission/103212550

Please have a look as problem really have a trival solution

How does the code which was giving RE on testcase 4 with a slight modification in the comparator function (used for sorting) gives AC(aka happy new year). The change simply is using "<" instead of "<=". any input in this regard is highly appreciated

code with AC VERDICT:https://codeforces.com/contest/1472/submission/103375747

code with RE on test4 :https://codeforces.com/contest/1472/submission/103375782.

I bealive comparator function should be unique. Meaning if A > B then B < A. In your case with <= you are saying that at that if A == B then a could be before or after B which should not be possible because then sorting is not unique. I.E. if your functions returns A < B it should also return !(B < A)

I believe when you say "A<=B" that would mean that if A and B are equal then A should be before B , Although I understand what you are trying to infer, and I think that it could be right but that would depend on the sorting algorithm, furthermore if it was the case that A and B are repetitively compared with each other, shouldn't it result in a TLE instead of RE?

I bealive it works that if you compare A and B, than comparator automaticly compares B with A, and it should yield same results. In your case it gives A before B in first, and B before A in second which is contradictory. And so it gives an error.

alright that makes sense. thanks a ton. have been struggling with it for a while now. thanks!

I felt E was harder than F. I spent wayy too long on E. Shouls have attempted F first.

Why is there a "Tie" in the second test at problem D?

Each player takes the largest number so their opponent can't take it. So Alice takes 3, then Bob takes 2, than Alice takes 1, after that score is 0:0. This is optimal for both players. If Alice takes 2 in first turn then Bob will take 3 and win, so she has to take 3 herself. Same logic aplies after first move.

Thanks

can anybody please explain C ?

could you be a bit more specific as to what you didn't understand?

Can someone please help me out for question D. My submission is showing runtime error on testcase 1 but it works fine on my ide

yeah, just select g++ 17 instead of whatever variant you're selecting right now, also your logic fails the further test cases.

incase you need to see the correct code similar to yours, you can go through mine :https://codeforces.com/contest/1472/submission/103348877 however even this is a too slow code and can be made better by using pop_back instead of erase. but anyway it passes the test cases. :)

My logic did not fail! Thank you so much for the help though :)

You pass

`odd.size() - 1`

as a parameter —`size()`

(in many implementations) returns an unsigned 64-bit value, so if it's 0 the result after subtracting will be the highest representable value. However, that is converted to a signed`long long`

, and since the value is greater than can be represented in a`long long`

, the value is implementation-defined.You can cast the result from

`size()`

to a signed type first.Thank you so much! It has worked now!

Is there any way that I can uphack my solution to G. It's linked here 103279910. I have tested it on custom invocation on CF and I know there is definitely a case that makes it TLE. I waited out the hack phase since I wanted to keep my rating, but now I don't see the option to hack it anymore.

EDIT: I think you have to be 1900+ to uphack. Here is my generator that would cause my solution to TLE. https://pastebin.com/uYxwMnYz

Any1 who is eligible to uphack, go for it :)

Link1 Link2

Problem E.

Link1 solution is accepted wherease link2 is not. The only diff between the 2 solutions is that for pair(hi,wi) whose hi is least is being calculated differently:-

In link1: let the main loop take care of everything

In link2: comparing all (hj,wj) with (hi,wi) for second condition [wj < hi && hj < wi = > ans[i] = j] [Since first condition can never be satisfied]

Why does link2 solution fail?

can anyone link a good video solution for E? thanks

https://codeforces.com/contest/1472/submission/103375747 this happens to be a self explanatory code.

for me it is always hard to understand other people's code, can you please elaborate how did you solve it? did you follow editorial's method or your own different method?

same here, anyway I guess the code they've provided is too complicated and involves a lot of things like binary search etc. i have rather written a much simple code. here is how I did it

make a vector having 3 fields -> height, width and index

before populating the vector sort the height width in decreasing order OR just swap height with width if width>height.

now sort the vector populated above, such that the first field (height) is the dominating factor, in case 2 elements have same height then compare width.

After step 3 you'd be having a pretty much sorted vector. Now the only thing that remains is going through the sorted vector(left to right), and keeping a check of a the smallest width encountered till now (as heights are already sorted they need not be checked), however one edge case needs to be verified, the minimum value of width could be of a element having same height as the current element, hence i've used another variable for it,

for every element having index(3rd field) 'i'if above condition is met by a previously encountered element having index 'j' then save ans[i]=j; else ans[i]=-1;

I guess I might have missed a few/many things (lemme know if you know any further explanation).

What's the DP state/solution for D no. problem? I can just feel the greedy approach. But problem tag shows DP too.

Another way to think about E:

We can think about the pairs $$$(h_i, w_i)$$$ and $$$(w_i, h_i)$$$ as

points on a 2D plane.Then the problem becomes: for each point $$$P(h_i, w_i)$$$, find if there is any point that lies

completely insidethe rectangle that is formed by the origin $$$O(0, 0)$$$ as its bottom left corner and $$$P$$$ as its upper right corner.I've uploaded my post-contest stream to YouTube: https://youtu.be/nCSLlTCDnDs

Can someone please point out what is the mistake in this solution 103260792 for problem D? I hope this is the correct place to ask... Thanks!

Use

`long long`

instead of`int`

for storing the sum as the range of`int`

is small for storing sum in worst case.Can someone please explain why my code is giving TLE for 1472E — Correct Placement.

I have swapped w and h whenever w>h while storing the inputs in a vector. I have then sorted the vector, found index of hj for every hi such that hj<hi using binary search. I have stored the index of the smallest value of w from indexes 0 to i in the sorted vector in an array b. At last, I choose w for every hj from the array in O(1) after checking the required condition.

For D how can this input be a tie?

`3 2 1`

Alice will pick 3 in his first move(because he picks 2, then bob will pick 3 and wins thereafter), Then bob will pick 2(same reason for 1). Then alice will pick 1 at end. Both will have 0pts at the end.

Ok, got it, basically, it means when the game ends, both of their scores is 0.

Yes

Thanks!

https://codeforces.com/contest/1472/submission/103513913 Can anyone tell me what is wrong with this? It is for the Problem G.

Could someone who solved D explain thought process/mental steps leading to solution?

Is there any general approach to similar problems?

I found pretty nice explanation here https://youtu.be/p5FNuXJoLDk?t=665

Could anyone please help me,I Am getting different answer in different online compilers. (Problem B) (link to my solution: https://codeforces.com/contest/1472/my) Thanks.

first of all you used a variable 'z' for size of array , and also you used it as variable to count number of 2 present , and also after correcting this , your code will not give right answer for test case — {2,2,2,1,1}

Thanks.

problem E VIDEO TUTORIAL LINK : https://www.youtube.com/watch?v=M8jqjbJjoxA I HAVE TRIED BY BEST TO EXPLAIN WITH PROPER INTUITION(twopointer + dp)SO THAT YOU CAN UNDERSTAND EASILY . I HOPE YOU GUYS WILL ENJOY THIS VIDEO !!!

DIV 3 PROBLEM C VIDEO TUTORIAL LINK https://www.youtube.com/watch?v=7FFzXk-brRU I have tried my best to give you guys proper intuition of this dp problem. Hope you guys will enjoy this video !!!

Can someone help me find the complexity of this code:103390550? It just barely passed the 4s time limit.

Can anyone explain to me why my problem C solution is having a memory exceeded limit? What should i change to make it more memory efficient? 103327344 — Here i thought i used dp to max it faster but my other solution(103317010) without dp was quicker. Can anyone explain why this is?

Problem E from my view:

Always assume that w<h, (if found w > h, we want to swap them), sort the array of (w, h)s in ascending order, we may want to save index (for tracking indexes)

Traverse the sorted array from left to right, for each i-th pair (w, h), i.e. (wi, hi), the problem becomes searching for j (j = 1..i-1) where wj < wi and hj < hi Now that we have wj <= wi for any j in 1..i-1, we only needs to binary search for upper bound of j which

wjnot exceed wi-1, let's call the bound is kThe only problem is we want to search for some j in 1..k where hj < wi. Let's construct a min array in such a way that min[k] = min { hj, 1 <= j <= k }, we also want to maintain 'argmin' array to store index which contains min

Then, result of finding j where hj < wi is j = argmin[k]

can someone please explain the DP approach from problem B?

SpoilersFirst, you can sort; it does not change the result. Second, observe that for equal numbers like 1 1 1 1; you only need to increase the last number (maintain increasing property). Third, let dp[i][0] and dp[i][1] be the answer for the first i elements where we do not increase ai or we increase ai. You build the DP relations for dp[i] and dp[i-1],

I cant stand the fact that the dude on problem E invite n friends 1<= n <= 2 * 10^5 to celebrate new year! Guys we are in corona virus pandemic we should be more carefull! Imagine what will happen after 14 Days if there was someone with corona in the party! Be more carefule stay home and code, stay safe :D

Simpler solution for F: 103672111

for problem G: can someone explain on what basis the second kind kind of action is taken?

for example :

// if there are cities "A", "B" and "C" of distance 3 having a road to city "D" of distance 4, and with 2nd kind of action city "D" can become of distance 1 making all cities "A", "B", "C" and "D" of distance 1.

// we have city "A" of distance 3, has a road to city "E" of distance 4 that can be reduced to distance 0 with 2nd kind of action, making "A" and "D" of distance 0.

which one should be chosen??

for 1472D - Четно-нечётная игра it shows a dp tag , but I am not able to solve with dp if anyone has solved with dp please post ur solution

sorry I am a noob...but what will be the complexity of this algorithm??

Why there is a graph tag in problem C?

i need help..! submission to D.

Please find what is wrong with this code Your text to link here.... i am not able to find any error.! it is giving runtime error.!

Does anyone know a possible reason for MLE in Problem G?

I am getting run time error for 12th test case in problem G for the following code in python 3, can someone please help me out?

import sys t = int(input())

def dfs(i, adj, d, ans, vis): vis[i] = 1 for k in adj[i]:

def solve(): input() n, m = map(int, input().split()) adj = [[] for i in range(0,n+1)] d = [sys.maxsize]*(n+1) ans = [0]*(n+1) for i in range(0,m): u, v = map(int, input().split()) adj[u].append(v) queue = [1] vis = [0]*(n+1) vis[1] = 1 d[1] = 0

for i in range(0,t):

QC: Long jumps Why

`i`

starts from n — 1 in this loop? for (`int i = n - 1; i >= 0; i--`

) { dp[i] = a[i]; int j = i + a[i]; if (j < n) { dp[i] += dp[j]; }## include <bits/stdc++.h>

using namespace std; typedef long long int lli; typedef long double lld;

## define FOR(i,l,u) for(lli i=l;i<=u;i++)

int main() { lli t;cin>>t; FOR(k,1,t) { lli n;cin>>n; lli A[n+1]; FOR(j,1,n) { cin>>A[j]; } lli score[n+1]={0},maxi=0; FOR(j,1,n) { if(score[j]==0) { lli i=j; score[j]+=A[i]; i+=A[i]; while(i<=n) { score[i]=-1; score[j]+=A[i]; i+=A[i]; } if(score[j]>maxi) maxi=score[j]; } } cout<<maxi<<"\n";

}

why am I getting time limit exceeded? In the above code no element is traversed more than twice, so time complexity is-O(n) for each test case.

Technically ,your solution is O(n^2).

I've written a Python solution of F, with my own thinking process inside of it. Not a solution that I could expect to achieve within the time frame of the contest, but hope it helps!

Problem B: Fair Division is tagged as

`dp`

. Is there also a DP solution for that? If yes, can anyone tell how to solve the problem using DP?Can anybody please explain to me why the same DP approach in problem c doesn't work if we consider 1 based indexing!

problem C i don't know but why maximum score in the second test case is 6 :) 1+2 --> 3; a1=2; 3+2 --> 5; a3=2; 2+2=6??