Thank you for participating. I hope you enjoyed the contest!

**Tutorial**

**Solution Code**

```
#include<bits/stdc++.h>
#define endl '\n'
using namespace std;
main()
{
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int k, n, t;
cin >> t;
for(;t--;)
{
cin >> n >> k;
if(n < k)
cout << k - n << endl;
else if(n % 2 == k % 2)
cout << 0 << endl;
else
cout << 1 << endl;
}
return 0;
}
```

**Behind Story**

After 2 problems got rejected, I made this problem but it has 2D coordinate originally.

**Tutorial**

**Solution Code**

```
#include<bits/stdc++.h>
#define endl '\n'
using namespace std;
main()
{
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int t;
cin >> t;
for(;t--;)
{
int m, sum = 0, x0, x1, x2, y0, y1, y2;
cin >> x0 >> x1 >> x2 >> y0 >> y1 >> y2;
m = min(x0, y2);
x0 -= m;
y2 -= m;
m = min(x1, y0);
x1 -= m;
y0 -= m;
m = min(x2, y1);
x2 -= m;
y1 -= m;
sum += 2 * m;
sum -= 2 * min(x1, y2);
cout << sum << endl;
}
}
```

**Behind Story**

One tester suggested to swap $$$A$$$ and $$$B$$$ but I and adedalic thought it is not good idea, so the order didn't changed.

**Tutorial**

**Solution Code**

```
#import<bits/stdc++.h>
#define endl '\n'
using namespace std;
int a[100005], b[100005];
main()
{
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int t;
cin >> t;
for(;t--;)
{
int k = 0, m = 1000000000, n;
cin >> n;
for(int i = 0; i < n; i++)
{
cin >> a[i];
b[i] = a[i];
m = min(m, a[i]);
}
sort(b, b + n);
for(int i = 0; i < n; i++)
if(a[i] != b[i] && a[i] % m > 0)
k = 1;
if(k)
cout<<"NO"<<endl;
else
cout<<"YES"<<endl;
}
}
```

**Behind Story**

Most testers especially liked this problem.

1401D - Maximum Distributed Tree

**Tutorial**

**Solution Code**

```
#include<bits/stdc++.h>
#define endl '\n'
using namespace std;
typedef long long LL;
struct H{int x, y;};
int ii, n;
const int q = 1e9 + 7;
LL p[100005], pv[100005], vi[100005], w[100005];
H e[200040];
int C(H a, H b){return a.x < b.x;}
int G(LL a, LL b){return a > b;}
LL dfs(int v)
{
LL d = 1;
vi[v] = 1;
for(int i = pv[v]; i < pv[v+1]; i++)
if(!vi[e[i].y])
d += dfs(e[i].y);
w[ii] = d * (n - d);
ii++;
return d;
}
main()
{
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int t;
cin >> t;
for(;t--;)
{
int k = 0, m;
LL x = 1;
cin >> n;
for(int i = 0; i < n - 1; i++)
{
cin >> e[i].x >> e[i].y;
e[i + n - 1].x = e[i].y;
e[i + n - 1].y = e[i].x;
}
int sz = 2 * n - 2;
sort(e, e + sz, C);
for(int i = 1; i < sz; i++)
{
if(e[i].x > e[i - 1].x)
{
for(int j = e[i - 1].x + 1; j <= e[i].x; j++)
pv[j] = i;
}
}
for(int j = e[sz - 1].x + 1; j <= n + 2; j++)
pv[j] = sz;
ii = k = 0;
dfs(1);
cin >> m;
for(int i = 0; i < m; i++)
cin >> p[i];
sort(p, p + m, G);
sort(w, w + n - 1, G);
if(m < n)
for(int i = m; i < n - 1; i++)
p[i] = 1;
else
{
int i;
for(i = m - 1; i > m - n; k = i, i--)
w[i] = w[i - m + n - 1];
for(; i; i--)
w[i] = w[0];
}
int l = max(m, n - 1);
int i;
for(i = 0, x = w[0]; i <= k; i++)
x = x * p[i] % q;
for(;i < l; i++)
x = (x + w[i] * p[i]) % q;
cout << x << endl;
for(int i = 1; i <= n; i++)
vi[i] = 0;
}
}
```

**Behind Story**

This problem was inspired by 1187E - Tree Painting and 1281E - Jeremy Bearimy. Originally this was proposed as E, but after some fixing it became more easier than original one. And the modulo was new one originally, but I changed it later by myself.

**Tutorial**

**Solution Code**

```
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
struct H{int k, l, r;};
const int p = 1048576;
LL l[2100000], r[2100000], z[2100000];
H h[100005], v[200040];
int C(H a, H b){return a.k < b.k;}
int D(H a, H b){return a.l < b.l || a.l == b.l && a.r > b.r;}
LL sum(int x, int y, int d) // find the sum of range [x, y]
{
if(x <= l[d] && r[d] <= y)
return z[d];
if(x > r[d] || y < l[d])
return 0;
return sum(x, y, d << 1) + sum(x, y, (d << 1) + 1);
}
void REP(int i, LL k) // add k to i-th number
{
i += p;
for(z[i] += k; i >>= 1;)
z[i] = z[i << 1] + z[(i << 1) + 1];
}
main()
{
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
LL m, n, s = 1;
cin >> n >> m;
for(int i = 0; i < p; i++)
l[i + p] = r[i + p] = i;
for(int i = p; --i;)
{
l[i] = l[i << 1];
r[i] = r[(i << 1) + 1];
}
for(int i = 0; i < n; i++)
{
cin >> h[i].k >> h[i].l >> h[i].r;
if(h[i].r - h[i].l == 1000000)
s++;
}
sort(h, h + n, C);
for(int i = 0; i < m; i++)
{
int j = i << 1;
cin >> v[j].k >> v[j].l >> v[j+1].l;
v[j + 1].k = v[j].k;
v[j].r = 1;
v[j + 1].r = -1;
if(v[j + 1].l - v[j].l == 1000000)
s++;
}
sort(v, v + 2 * m, D);
for(int i = 0, j = 0; i < n && j < 2 * m;)
{
if(h[i].k < v[j].l || h[i].k == v[j].l && v[j].r < 0)
{
s += sum(h[i].l, h[i].r, 1);
i++;
}
else
{
REP(v[j].k, v[j].r);
j++;
}
}
cout << s;
return 0;
}
```

**Behind Story**

This problem was inspired by 1075C - The Tower is Going Home. Originally the restriction of $$$n, m$$$ was small so $$$O(n^2)$$$ solution could pass, but adedalic suggested to increase them up to $$$10^5$$$. In fact, I only knew brute forces solution until he tells me the solution using segment tree. And the size of the square was $$$10^9 \times 10^9$$$ so coordinate compression would be done additionally, but we eliminated it to avoid too many implementation.

**Tutorial**

**Solution Code**

```
#include<bits/stdc++.h>
#define endl '\n'
using namespace std;
typedef long long LL;
LL p, a[528000], ll[528000], rr[528000];
LL SUM(LL l, LL r, LL d)
{
LL x, y, z;
x = ll[d];
y = rr[d];
z = b & ~(y - x);
x ^= z;
y ^= z;
if(x > r || y < l)
return 0;
else if(x < l || y > r)
return(SUM(l, r, z << 1) + SUM(l, r, (z << 1) + 1));
else
return a[z];
}
void REP(LL i,LL k)
{
i += p;
for(a[i] = k; i >>= 1;)
a[i] = (a[i << 1] + a[(i << 1) + 1]);
}
main()
{
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int b = 0, i, n, q;
cin >> n >> q;
p = 1 << n;
for(i = 0; i < p; i++)
{
cin >> a[p + i];
ll[p + i] = rr[p + i] = i;
}
for(; i < p; i++)
ll[p + i] = rr[p + i] = i;
for(; --i;)
{
a[i] = (a[i << 1] + a[(i << 1) + 1]);
ll[i] = ll[i << 1];
rr[i] = rr[(i << 1) + 1];
}
for(; q--;)
{
int m;
cin >> m;
if(m == 1)
{
int k, x;
cin >> x >> k;
REP((x - 1) ^ b, k);
}
else if(m == 2)
{
int k;
cin >> k;
b ^= (1 << k) - 1;
}
else if(m == 3)
{
int k;
cin >> k;
b ^= 1 << k;
}
else
{
int l, r;
cin >> l >> r;
cout << SUM(l - 1, r - 1, 1) << endl;
}
}
}
```

**Behind Story**

This problem has 6 generators, and I most liked it among 6 problems.

Video Editorial of Problem C: Mere ArrayI tell you and other video editorial makers one thing honestly , if you really want to help people help in problem D , E (and F some times) . Just see the number of submissions in problem A,B,C . Almost everyone did them and those who did not would have got by now.

Yes, most of the participants in cp community belong to average level and hence need help in difficult problems (need explanation of problems above C in cf) in order to increase thier level.

Thanks , it's not difficult for any person of any level to upsolve A,B,C .Real difficulty starts from D.

To upsolve definitely not, given the fact, that tutorial essentially give you exact way how to code them. But to understand WHY those solutions are correct might not be that easy and this is much more important if you want to improve your cp skills.

Moreover no matter on your skill level it does happen, that you come up with a more complicated solution/same solution but more complicated way to understand why it works/have something like "my intuition tells me something like this should work but I have no idea how to prove it" even for a relatively simple problem, and then it might be worth it to see the simpler solution

Some people think that the difficult part starts from E or F, it's different for everybody. Yes, A, B, C tend to be easy for most people but there are others who celebrate, if they solve C and we as a community got to help each other not make fun of.

Hi @bully....maguire Why are you pulling him down? Maybe you can help the community by making video editorials of D,E,F

He is putting in efforts to give back to the community,atleast appreciate what he is doing, it may not be helpful to you, but to someone out there it may be immensely helpful.

Btw no doubt your feedback does also make business sense, because if we look at it like a CP Question (:P), the number of views that he gets on a video editorial of Question X would be = Number of people who solved Ques(X-1) — Number of people who solve Ques(X)

So for this contest, the audience for Editorial of C is 2000 whereas the audience for an editorial of D is 6000 :D

while u got a point, still this is no reason to demotivate him.

I overkilled it with DSU :/ saw that some set formation is there and quickly started coding. Now I know Div2C is not meant to write large amount of code. Nice video tutorial :)

sir if we take all numbers who are not at the right position and then take out their gcd finally check weather that gcd is equal to the minimum number so is it correct ??

[8,4,6] Their gcd is 2.

Their gcd should be divisible by the minimum number.

https://codeforces.com/contest/1401/submission/90619678 why is it wrong ??

try this array

2 3 8 4

ans should be YES

`2 4 8`

. The answer should be yes because even though 2 is at the correct place, we can still use it to swap the other values. your solution would be correct if you just check whether the gcd of the calculated gcd and the minimum is equal to minimum or not.why the fuck people are down voting this

Great problem set tlsdydaud1!

In F, is solution O(n^2 q) solution supposed to fail? strict time limit :(

No, my O(n^2 q) solution works in <400ms

why does this solution give TLE? https://codeforces.com/contest/1401/submission/90614540

Updating/Querying a BIT at position 0.

https://codeforces.com/contest/1401/submission/90613864

can someone tell the error in this solution....i think it has same idea as in editorial :(

Ummm.... I think you didn't have the code for handling the m < n-1 part.

for m<n-1 case i'm just assigning 1 to rest of the lowest visited edges ... is it correct?

Oh oooops sorry I meant the m > n-1 part ;)

till then priority queue will become empty right?? and before it i will greedily assign values?

According to the problem, "the product of all n-1 numbers should be equal to k;", which means you have to use up all numbers in p[]. But in your program it seems to leave them unused though.

thanks for replying...got the bug :|

Thanks for a good round and fast editorial!

Problems are too interesting...

Thanks a lot for the fast editorial 😀

tlsdydaud1 I really liked your round!

Editorial came quicker than most people solved the first question :P

Can someone tell me a testcase where my code for problem B is failing? https://codeforces.com/contest/1401/submission/90608794 Thankyou

absolutely wrong, try another approach

How so? I am also making pairs of (a2, b0), (a0, b2) and (a1, b0). Just like they did. Please give any testcase where it is failing. Thanks

There you go,

Answer is 0 but your program gives -2, Also you are making arrays of x1+y1+z1 length which can through Memory Exeption or TLE.

Pretests are too strong..

Swapping problem A & C would be the correct easy to hard order xD.

difficulty level according to me B==C < D <= A < E < F

You haven't solve any problem and now say that D is easier than A.........

I want your Weed

Fast editorial !!!! Great problems specially D problem . Thanks codeforces community

In problem E:

"∙ When a segment intersects with two sides (facing each other) of the square, the number of pieces increases by one."

Doesn't the number of pieces in this case increase by two?

Think about the simplest case: you only add one segment going across the square. Initially there is one piece (the whole square) and after adding the segment, there are two pieces, above and below the segment. So, the number of pieces increased by one.

Right, in my head the inital number of pieces was zero lol. Thanks!

But, suppose you then added another segment vertically touching both the opposite sides and intersecting the previous drawn line. Now,the number of peices is 4 (increased by 2). How?

Because you also have a new piece from the intersection of the two segments.

"∙ When a segment intersects with two sides (facing each other) of the square, the number of pieces increases by one."

This statement is not counting new pieces from internal intersections.

Can anyone help me?

My code for Problem D here(https://codeforces.com/contest/1401/submission/90610162)

Like editorial's method, I was trying to get answer. but i don't know why it does not work.

You take the weights modulo 10^9+7 before sorting them, so the order can come out wrong.

I made the same mistake during the contest but gave up trying to figure it out. No motivation when the contest is unrated :D

what............?!!!!!?

i have never thought about that LoL...

Thank you very much, and i will never modulo before sorting :(

thanks man i spent 1 hour during the contest to find the same bug in my code and also after contest i couldn't find that.

Can you please tell what's wrong in this code for problem d Maximum distributed index.

I think the mistake is in this part of the code:

Actually here for each edge between node i and j, the number of times that edge is traversed is sub[i]*(n-sub[i]), where i must be the child in the DFS tree representation of the tree, but in your case you are taking every i. For reference, see my submission: https://codeforces.com/contest/1401/submission/90657259

I found the bug.....I was adding the contribution of extra primes instead of multiplying....

As for what you are saying that piece of code is completely fine....as every node will be a child to some node except for the root(note I start with i=1 and don't count the root). Thanks anyway though!

Okay:)

Thanks, my code had the same mistake and I couldn't find the bug

About E: in the editorial, is it segment tree or interval tree?

It's the efficient segment tree implementation https://codeforces.com/blog/entry/18051

I meant: is it a segment tree (to do operations with numbers), or it is an interval tree (to do operations with intervals)

It is definitely a segment tree (if by "interval tree" you mean the Wikipedia definition). I don't think I've ever seen an interval tree used in a contest problem. But it's confusing because some people refer to segment trees as interval trees.

Thank you. My solution was to construct interval tree (from horizontal segments), then for each vertical segment find the count of horizontal segments which contain its x coordinate. But I haven't done that.

If it is a segment tree I will try harder to understand this solution:)

Something like that should work as well!

Can someone share his/her approach for solving problem E. The editorial is too breif for me.

Here's what I did:

First put all the horizontal lines down in the square. For every line that spans $$$x = 0$$$ to $$$x = 10^6$$$, the number of regions increases by 1 (originally 1).

Then, we can start putting vertical lines from left to right. We run a line sweep from left to right and maintain a data structure of "active" y-coordinates during every vertical line. I included $$$y = 0$$$ and $$$y = 10^6$$$ as always active lines. You can then notice that for every vertical line $$$V$$$ added, if the number of horizontal lines intersecting $$$V$$$ is $$$z$$$, then the number of regions increases by $$$z - 1$$$. For this part using a Fenwick tree would be a simple data structure that works.

Can you please provide the link to your code. It would be very helpful to understand .

Code: Submission

Your idea and implementation both are awesome. Just 1 small thing, I was checking your code and saw that there is sorting of lines done which is not required, I guess.

Yeah the sorting is not required, I just never removed them from my code :)

I used the same logic just switching roles of horizontal and vertical lines. I used ordered set for querying. But getting wrong answer on 5th test case. Can u please help?? 90619940

Found !!

:)

Very nice problems!

What's special in Pretest 5(Question number D). Can anybody provide a smaller test so that I can manually verify my code? My submission I had considered the case when m>n-1 too. Any help will be highly appreciated.

Pretty sure it's overflow. I had the same WA, and switching some of my ints to long longs fixed the problem.

You take mod before sorting frequencies.

Thanks, bro! One mode affect my delta to more than 50..really upset

it was common mistake for maximum.

Ratings are calculated already , this contest is fast.

what is the wrong with my solution for D

for(ll i=2;i<=n;i++) { vc.push_back((counts[i]*(n-counts[i]))%mod); }

remove mod

I done the same mistake. Take mode after sorting not before sorting!

yep. makes sense

B using recursion (when you don't want to use brain but want to do labour work) https://codeforces.com/contest/1401/submission/90623923

This is written in your code"//.size() return unsigned int so never use i<v.size()-1 in loop"

What is the risk of doing that ?

I never thought someone will ask about the commented part.

In some question, you have to loop till the third last character of string. So you will write i<s.size()-2 But when string size is 1, then s.size()-1 will give wrong result as .size() always give in unsigned int.

Really Enjoyed the contest man ! Keep up the great works tlsdydaud1 especially and all other co-ordinators too in general :)

Please somebody expalin me in more detail, how to prove that we should multiply numbers in both cases in exactly this way. Sorry for my English

During the contest I submitted solution for F and it passed pretests. But I quickly realized that it works in $$$O(2^n \cdot q)$$$, though it was easy to fix so it would work in $$$O(nq)$$$ and I resubmitted. But then I submitted initial slow solution after the end of the contest, and it passed systests! So I quickly hacked it 90619845 But I'm quite disappointed that systest didn't manage to break my initial slow solution(

The idea behind D was exactly same as 1280C - Jeremy Bearimy.

And thank you for the contest.

Actually I motivated by that problem and 1187E - Tree Painting. :)

Also this problem is a way much similar to the current, except the tree is rooted here and only paths from roots were considered. 1399E1 - Weights Division (easy version)

My poor puny brain still didn't understand problem A. Could anyone kindly explain a bit more as to why ans would be 0 or 1 in case of k<n ?

When k is less than n, consider B is x distance left of A, then k = n — 2*x. So x = (k — n)/2. If if the difference is odd, you need to move 1 unit, otherwise, with 0 steps you can do it. Visualize it with an example and you might understand better.

Can anyone explain in Problem D, why should we do the "product of the number of vertices belonging to each of the two components divided when the ith edge is removed from the tree"? I understood that the edge removed will be visited that many times, but am not getting the intuition or reasoning behind that.

all vertices to the left of edge x, right of edge y for each vertex in left side that edge is used when visiting y vertices on the right. so in total that edge is used x*y times

If you understand what is required in D, the implementation is not trivial, but still easy. So the tutorial should explain what is needed. Instead, the wording is even more bulky than the problem statement.

C can also be solved in segment tree 90626651

Bruh ! Thats next level of overkill xD

How to prove the equivalence of the two segments in Problem F?

In problem B: Ternary sequence, along with the pairs 1,0 0,2 2,1 , I am also maximising the pair 2,2 to decrease the pairing of 1,2. But I am getting wrong answer. Please explain.

you should increase 2 1 pair as this will add 2 to the answer , 2 2 pair will add 0, i hope you get the points.

wow fast editorial fast rating change.. thanks for fast

Can someone help me with a smaller test case which fails my submission 90615069. Or point the mistake?

nevermind, silly mistake

https://codeforces.com/contest/1401/submission/90627708

can anyone please help me with this code getting WA in test case 5

You're taking modulo before sorting, so the order is wrong.

Thanks a lot.. :) stuck for more than 1 hour

Good problem set !! :)

You can find elaborate video solutions to A-E (not F) here if you like videos, timestamps are in the description

Though the problems were good. But It would have been much better if at least one sample tests was considering the case m>n-1

In problem D, can someone tell me why my solution fails, please? I checked all the common mistakes (modulo before sort, more factors than edges), it doesn't seem that I fall into any of those categories.

Never mind, worked it out myself — I had a problem with dfs, for chrissake

How not to mess up implementation of problem D today if you have one hour to solve it? I got the solution in 5 minutes but couldn't finish it You see my code during contest and after a slight change in it.

And tell me what improvement should I do, to not repeat them in future.

code during contest ->90617959

code after slight change -> 90630780

hoping someone will help me.

WA on test5 problem D .checked all common erros overflow,not taking mod for storing frequencies. D wa on test5

sorry didnt check properly.

afterFor problem E, I used ordered_set, but the same logic as in the editorial but it doesn't seem to be working for large test cases. Code — here

Help would be appreciated.

Edit — Found ! Thanks !

Best codes written. Hats off to you brooooo. Other editorialists use dirty templates. Templates so dirty that no one is able to understand. Why don't they understand that it is difficultto read codes, when untidy templates are used. Thanks for giving such clean codes.

Thank you so much. Again thanks.

Can some one elaborate on the proof for the second case in problem D ? Why is it guaranteed that thethat are obtained fromare the maximum group of primes that can be obtained? I can see why if there is no

n-1 primes[p_{0}, p_{1},p_{2},p_{3}, ... , p_{n-1}* p_{n}*....*p_{m-1 }]MODtaken, but shouldn't the mod make things different?You're not trying to find the answer with the highest mod value, you're trying to find the exact answer first, and then mod it because it's too big

Do you mean that because of the wording of the problem, this is the required answer but not necessarily the maximum answer that can be achieved ?

The problem exists independent of the mod. You only take mod for the answer becuase handling such big numbers is hard(probably impossible).

Any idea why does this code not work for D? Code

I think there is another interesting way to solve the problem F using a lazy segment tree. The idea is to maintain the "lazy array" as a "segment tree" and then, the querie 2 and 3 are equivalent to updates in a range in the "segment tree" and the others queries are normal ones in the lazy segment tree. You have to keep in mind that you obtain the values of the "lazy array" with the "segment tree" in every moment. This way you can even do queries of type 2 or 3 in a certain range of blocks. What I mean is that not only works for the whole blocks of subarrays (update in the whole block of subarrays), but for certain range of blocks of subarrays. The complexity is O(n^2*q). Anyway, nice problemset bro!.

My solution for E:

By Euler's polyhedron formula, we have $$$V-E+F=2$$$ in a planar graph. We want to compute the number of faces of our planar graph, excluding the outside of the square, which is $$$F-1 = E-V+1$$$. For each segment (including the edges of the square), let's suppose that it has $$$v$$$ vertices lying on it. Then it contributes $$$v-1$$$ edges. Summing over all segments, $$$E = \sum v-1 = 2V-n-m-4$$$, since each vertex lies on exactly two segments and there are $$$n+m+4$$$ segments. Thus, $$$F-1 = V-n-m-3$$$. Computing $$$V$$$ is easy using sweepline.

*In planar

connectedgraph, which is why we need the assumption that all segments touch one side of the square. Without it the formula becomes: $$$V - E + F = 1 + C$$$Does all vertices lies on exactly two segments? What about cases like this:

One vertex is exclusive to one edge only

I define a vertex as an intersection between a vertical and horizontal edge, so the vertices and edges would be counted like this:

Humble Request. I am facing issue with understanding D problems how have we calculated no. of nodes in the separated component(d) of which we are calculating d*(N-d). Actually I did it using Dfs N-1 time on each node. I knew that it will give TLE. Whereas here it is done in one Dfs only. Can someone explain to me whether this is a general algorithm or anything like that or where can I learn it from more intuitively?

Root arbitrarily the tree. Now let's compute $$$sz[u]:$$$ the amount of nodes on u's subtree.

Now, for an edge $u, v$ (with $$$v \in G[u]$$$), the amount you want is $$$sz[v]\cdot(n-sz[v])$$$.

Thanks i got it !!

Hey why is sz[v].(n-sz[v]) coming? Like i didnt got the reason why it came.

Thanks in advance

edge parent->cur(or p->v) will be used sz[v]*(n-sz[v]) times in given function.

Which is no nodes connected to edge p->v.(sz[v]=no nodes connected to v and n-sz[v]=nodes connected to node p

Let's see, $$$G[*]$$$ donotes adjacency list of in the rooted tree. The directed edge $$$u \rightarrow v$$$ (with $$$v \in G[u]$$$), is counted on any path between each node in $$$v$$$'s subtree and each node not in $$$v$$$'s subtree. This amount is $$$sz[v]\cdot (n-sz[v])$$$.

My solution to D is running well on pc but giving wrong answer when submitted. Can anybody help me please https://codeforces.com/contest/1401/submission/90616879

can you explain me the dfs part ?

.-. does anyone think that problem C is easier than B?

yes, implementation wise B was hardest among A, B, C.

yeh, I did B with a lot of if-else statements, but it still accepted XD

Hindi Video Solution for Problem D 1401D — Maximum Distributed Tree Youtube Video Solution Link for Prolbem D (hindi)

please have a look

Nice problems

In Problem B, I am using the same logic, then why I am getting the wrong answer.. https://codeforces.com/contest/1401/submission/90655781

I think you should first subtract from a[2] then set b[1] to zero

Can anyone help me in finding why my code is giving WA on testcase 5. I used dfs to find the subtree size of node i, for all i, 1<=i<=n. Then used that subtree size to find the number of times some edge contributes to the answer. Then used greedy to find the maximum possible answer, and it matches the approach of the editorial.

Here is my submission.

90656681

refer to this.

It solved my problem for testcase 5

I also had similar mistake try changing

to`sort(all(e),greater<int>())`

`sort(all(e),greater<long long>())`

Can some body point out what's problem with my solution to Problem D.

upd- problem was taking mod before sorting resulted in wrong results. Refer to this

I got the thinking process of D but I am not able to prove it for $$$m \gt n-1$$$, someone please help in making the proof more simple. Thanks !

In particular I am not able to see why all the big primes will go to first weight and other all weights get only single primes...

Let us consider $$$n$$$ weights and $$$m$$$ primes.

I'm using 1-indexed arrays.

We must take into account 2 facts.

Let $$$W_i$$$ be the $$$i$$$th smallest weight, i.e. sorted in ascending order. Let $$$c_i$$$ be the coefficient of $$$W_i$$$. Let us prove $$$c_i<=c_j$$$ if $$$i<j$$$. We can swap any two coefficients and retain the total product. Let us assume $$$c_i>c_j$$$ and $$$i<j$$$. Then we can show $$$c_iW_i + c_jW_j <= c_jW_i + c_iW_j$$$. So it is optimal for $$$c_i$$$ to be non-decreasing.

We can show that $$$c_iW_i + c_nW_n <= c_iW_i/k + kc_nW_n$$$ for some $$$k>1$$$. So optimal structure has all primes in $$$c_n$$$, as each prime shift is essentially just substituting $$$k$$$ as $$$p$$$.

But since we need to minimize the number of ones, we need to give some primes to the previous values. So it also obvious that if there is more than 1 prime on some $$$c_i$$$ for $$$i<n$$$, it is optimal to shift it to $$$c_n$$$.

Let us say that we gave a prime $$$p_j$$$ such that there exists $$$p_k$$$ such that $$$p_j>p_k$$$ and $$$p_k$$$ is in $$$c_n$$$. It is optimal to swap $$$p_j$$$ and $$$p_k$$$ as shown in the previous equation, $$$k$$$ for this swap is $$$p_j/p_k$$$ which is greater than $$$1$$$. Therefore the mentioned structure is optimal.

makes sense thanks!!!

Can someone help me with a more formal proof for E?

Hi Aditya,

Here is a semi-formal proof. I mixed in graph and non-graph words, and realised midway that we do not really need the terminology for the problem.

For each of the intersection points of the line segments, and each of the "dangling" endpoints of a line segment, we put a vertex there. The edges between these vertices are naturally formed. This gives us a planar graph which is also connected, as each line segment has one endpoint on the outer square. This gives, by Euler's formula,

V-E+F = 2. This 2 includes the outer, infinite face, so F_finite = 1+E-V.

Now, consider a "dangling" vertex (that is, a leaf vertex). We can delete this vertex along with the edge joined to it. This does not change the answer, as removing it decreases E and V both by 1.

Now, consider two "internal" line segments (that is, one of the line segments other than the segments of the outer square) that intersect. This intersection increases E by 2, but increases V only by 1.

Also, consider a line that completely crosses the outer square. Such a line segment increases E by 3 and increases V by 2.

Therefore, our answer is

1 + (number of intersections amidst the internal lines) + (number of internal lines that completely cut through the outer square).

why it is wrong on test 5? http://codeforces.com/contest/1401/submission/90669604

check this

If anyone need detail explanation & implementation of D Here

In problem D can someone explain why we are multiplying the largest prime numbers together(the case when m > n-1)? How do we prove that this is the optimal way?

Video for D Maximum Distribution Tree

https://codeforces.com/contest/1401/submission/90678039 My E submission is failing on test 5 :( Could use some help :P

Could anyone help me with debugging my code submission for problem D https://codeforces.com/contest/1401/submission/90679884 failing on test case 5 ? It seems ok to me.

In your vector coef, don't take modulo. else { (10^7 + 4 ) < (5) } after computing modulo. Take it while computing the final answer.

Got it. Thanks a lot !!

problem C(Mere Array) can be done in one for-loop instead doing much stuff as given in editorial my submission link - https://codeforces.com/contest/1401/submission/90681298

still nlogn time, you've used sort :)

My bad I didn't noticed

Can someone tell me why this solution is giving TLE test 5{question D}. https://codeforces.com/contest/1401/submission/90657948

E is easier than I thought. What a pity, I had silly trouble with problem D and did not have enough time for E :((

Here's a more straightforward approach for F:

Let's build a standard range sum Segment Tree. Aditionally, we will store which powers of 2 have been swapped (in a simple boolean array).

Note that a reverse operation of power $$$k$$$ is simply swapping of all powers $$$i$$$ such that $$$i < k$$$. We have now reduced the problem to just range-sum with updates and swaps.

Whenever you swap a certain size, just flip its value in the boolean array we created. Now, because a segment tree is based on dividing the array into segments of powers of two, it's pretty easy to perform the swaps. Let's say you have assigned children of node $$$p$$$ to be $$$2*p+1$$$ (left child) and $$$2*p+2$$$ (right child).

While traversing the segment tree for anything (updates or sums), you just need to flip the children if needed. That is, if the power of two that this node is responsible for is swapped, your left child will now be $$$2*p+2$$$ and right child will be the node $$$2*p+1$$$.

This is pretty straightforward, and only adds a couple of lines to your standard segment tree implementation. Handles swap queries in constant time, and the other two in $$$O(n)$$$.

Nice approach!

About: "Note that a reverse operation of power $$$k$$$ is simply swapping of all powers $$$i$$$ such that $$$i<k$$$". How did you notice this?

Mmm I see, like it says in editorial reverse is doing xor with $$$2^k - 1$$$ and swap is doing xor with $$$2^k$$$. So $$$2^k - 1 = 2^0 \oplus 2^1 \oplus \ldots \oplus 2^{k-1}$$$.

Maybe one of those things that just click :)

I didn't think of the XOR approach explicitly as mentioned in the editorial (although it is pretty intuitive if you think about it). The equation you mentioned also makes sense.

I initially tried to simplify the problem by only considering queries of one kind. Swapping seemed pretty easy to do with a segment tree, and naturally I realised that reverses are also just swaps!

This is awesome. I was thinking on the same lines but just couldn't realize that reverse operation can be seen as swap.

There is a mistake in the editorial for F, $$$[l,r]$$$ should map to $$$[l\text{^}(x\&\text{~}(2^k-1)), r\text{^}(x\&\text{~}(2^k-1))]$$$ instead.

Thank you for noticing me. I'll edit it.

A very subtle mistake which can be done in the D question is that, while calculating the frequency of the edges, you might use modular multiplication, which will give the wrong answer, as very high edge frequency, under modulo might become somewhat lesser, and in the end when sorting the frequency array, it might be more towards the smaller side and hence answer obtained will be lesser than the actual answer.

Auto comment: topic has been updated by tlsdydaud1 (previous revision, new revision, compare).What is wrong with my submission?

Once again,I'm very much willing to recommand my blog for the Chinese editorial

In que C, it is given that we can rearrange only if the gcd(a1,a2) is the minimum element but the tutorial says only being divisible by the minimum is enough, can someone explain??

Let a be the minimum element, and b,c two multiples of a. Then you can swap (b,c) by doing the three swaps (a,b),(a,c),(a,b).

Can anyone check my code of problem D 90800991 ?.Could you suggest how to overcome these type of errors while taking mod.

You should use 64 bit type instead of 32 bit. Ie long long instead of int.

We are taking modulus right I thought it would fit in int.Anyway tried but still not working 90801804.Is there anything wrong in the logic?

You sort the edges by frequency, which is correct. But you sort after taking them mod, which is not correct.

Gee.. thanks for looking into it.It worked.

For Question E wouldn't the runtime complexity for a sweep algorithm using a segment tree set be O((m+n)*log(m+n)) instead of O(nlogn + mlogm)?

Suppose we are sweeping from left to right. Then we would need to stop at every x value that corresponds to {beginning of horizontal segment, end of horizontal segment, or a vertical segment}. This adds up to (2n+m).

And for the segment tree, if we were to use index compression, we would still have to include leaf nodes for both the y-values corresponding to horizontal segments and the top/bottom y-values for vertical segments to be able to efficiently obtain a range query. So this would add to (n+2m).

Another possible solution is to use an ordered set instead of a segment tree, and we would only store the y-values of active horizontal segments. And for every vertical segment we can perform the range query using lower and upper bound, which would be O(logm). But this would still lead to a runtime of O((n+m)*logm).

I'm mainly just thinking out loud here, but if anyone can confirm/refute what I have here your help would be greatly appreciated!

Edit: So I just realized that using lower and upper bound with an ordered set would require O(m) instead of O(logm) since finding the distance between two iterators in a data structure without random access is linear. So this would mean that the segment tree is the fastest option. But my point still stands that the time complexity is O((n+m)*log(m+n)).

tlsdydaud1

I wonder why you use

`for(;t--;)`

unusual isn't it? Is there any special reason? I have seen most of the people using`while(t--)`

.I know that. But I use

`for(;t--;)`

, so also used it.Alternate O(nq) solution for F:

Make a standard RSQ segment tree. Define the 'level' of a node in the tree as log2(size of range covered by node). The level of the root is obviously n.

An operation "swap k" can be simulated by swapping the left child and right child of all nodes with level = (k+1).

An operation "reverse k" can be simulated by swapping the left and right child of all nodes with level <= k.

The swap and reverse operations can be done fast with a lazy propagation type optimisation. After this, it's just basic query and updates on a segment tree.

My code for this: https://codeforces.com/contest/1401/submission/92201994

EDIT: found the mistake

Help needed !

Can someone please help me find my mistake in problem E's solution? I have been stuck on it for many days now, so help would be much appreciated!

My WA submission: 92319896

the solution code of problem D can't pass the test case?????

There was one

`j++`

missed, and I fixed it. So the solution code passes the testcases now. Thank you for noticing!Could someone tell me the error in 102290234 this code for problem d? I'm getting wrong answer on test 5

A is not explained well... like for the testcase:

`n = 5, k = 8`

.If we increase A by 1 then

`A = 6, k = 8`

. If we take`B = 7`

then absolute of`((B-0)-(A-B)) = 8`

which is equal to`K`

.Hence the minimum move is 1 and not 3. But the testcase has 3 as output.... ¯_(ツ)_/¯

Hello, I am getting wrong answer for 1401D — Maximum Distributed Tree in TC#5. Please help to understand what is wrong with my solution. Solution link :- https://codeforces.com/contest/1401/submission/199095063