Thank you for participating!

Idea: flamestorm

**Tutorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
const int MAX = 200007;
const int MOD = 1000000007;
string alph = "abc";
void solve() {
string s;
cin >> s;
int cnt = 0;
for (int i = 0; i < 3; i++) {
cnt += (s[i] != alph[i]);
}
cout << (cnt <= 2 ? "YES\n" : "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>
using namespace std;
void solve()
{
int n;
cin >> n;
vector<int> a(n);
int ans = 1;
for(int i = 0; i < n; i++)
{
cin >> a[i];
}
sort(a.begin(), a.end());
a[0]++;
for(int i = 0; i < n; i++)
{
ans*=a[i];
}
cout << ans << endl;
}
int32_t main(){
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;
int score[10][10] = {
{1,1,1,1,1,1,1,1,1,1},
{1,2,2,2,2,2,2,2,2,1},
{1,2,3,3,3,3,3,3,2,1},
{1,2,3,4,4,4,4,3,2,1},
{1,2,3,4,5,5,4,3,2,1},
{1,2,3,4,5,5,4,3,2,1},
{1,2,3,4,4,4,4,3,2,1},
{1,2,3,3,3,3,3,3,2,1},
{1,2,2,2,2,2,2,2,2,1},
{1,1,1,1,1,1,1,1,1,1}
};
void solve() {
int ans = 0;
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 10; j++) {
char c;
cin >> c;
if (c == 'X') {ans += score[i][j];}
}
}
cout << ans << '\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;
const int MAX = 200'007;
const int MOD = 1'000'000'007;
void solve() {
int n, k;
cin >> n >> k;
string s;
cin >> s;
int res = 0;
for (int i = 0; i < n; i++) {
if (s[i] == 'B') {
res++; i += k - 1;
}
}
cout << res << '\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;
const int MAX = 200'007;
const int MOD = 1'000'000'007;
void solve() {
int n;
long long x;
cin >> n >> x;
long long a[n];
for (int i = 0; i < n; i++) {
cin >> a[i];
}
long long lo = 0, hi = 2'000'000'007;
while (lo < hi) {
long long mid = lo + (hi - lo + 1) / 2;
long long tot = 0;
for (int i = 0; i < n; i++) {
tot += max(mid - a[i], 0LL);
}
if (tot <= x) {lo = mid;}
else {hi = mid - 1;}
}
cout << lo << endl;
}
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>
using namespace std;
const int N = 200'000;
int n, k;
int a[N+5], h[N+5], pref[N+5], length[N+5];
bool get(int dist)
{
bool found = false;
for(int i = 0; i < n-dist+1; i++)
{
if(length[i] < dist){continue;}
int sum = pref[i+dist]-pref[i];
if(sum <= k)
{
found = true;
break;
}
}
return found;
}
void solve()
{
pref[0] = 0;
cin >> n >> k;
for(int i = 0; i < n; i++)
{
cin >> a[i];
pref[i+1] = pref[i]+a[i];
}
for(int i = 0; i < n; i++)
{
cin >> h[i];
}
length[n-1] = 1;
for(int i = n-2; i >= 0; i--)
{
if(h[i]%h[i+1] == 0)
{
length[i] = length[i+1]+1;
}
else
{
length[i] = 1;
}
}
int l = 1, r = N;
while(l <= r)
{
int mid = (l+r)/2;
if(get(mid))
{
l = mid+1;
}
else
{
r = mid-1;
}
}
cout << r << endl;
}
int main() {
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 = 200'007;
const int MOD = 1'000'000'007;
void solve() {
string s;
cin >> s;
int n = s.length(), cnt = 0;
bool all = (s[0] == 'B' || s[n - 1] == 'B');
for (int i = 0; i < n - 1; i++) {
if (s[i] == s[i + 1] && s[i] == 'B') {all = true;}
}
vector<int> lens;
int curr = 0;
for (int i = 0; i < n; i++) {
if (s[i] == 'A') {curr++;}
else {
if (curr != 0) {lens.push_back(curr);}
curr = 0;
}
}
if (curr != 0) {lens.push_back(curr);}
sort(lens.begin(), lens.end());
if (lens.empty()) {cout << 0 << '\n'; return;}
int tot = 0;
if (all) {tot += lens[0];}
for (int i = 1; i < lens.size(); i++) {
tot += lens[i];
}
cout << tot << '\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>
using namespace std;
const int N = 200005;
vector<int> adj[N];
vector<bool> vis(N);
int entry_node = -1;
vector<int> path;
bool dfs1(int u, int p)
{
vis[u] = true;
for(auto v : adj[u])
{
if(v != p && vis[v])
{
entry_node = v;
return true;
}
else if(v != p && !vis[v])
{
if(dfs1(v, u))
{
return true;
}
}
}
return false;
}
int dfs2(int u)
{
vis[u] = true;
int distbruh = N;
for(auto v : adj[u])
{
if(v == entry_node)
{
return 1;
}
if(!vis[v])
{
int dist = dfs2(v)+1;
distbruh = min(dist, distbruh);
}
}
return distbruh;
}
void solve()
{
int n, a, b;
cin >> n >> a >> b;
for(int i = 0; i < n; i++)
{
int u, v;
cin >> u >> v;
adj[u].push_back(v);
adj[v].push_back(u);
}
dfs1(b, -1);
vis.assign(n+1, false);
int distMarcel = N, distValeriu = 0;
if(entry_node == a)
{
distMarcel = 0;
}
else
{
distMarcel = dfs2(a);
}
vis.assign(n+1, false);
if(entry_node == b)
{
distValeriu = 0;
}
else
{
distValeriu = dfs2(b);
}
if(distValeriu < distMarcel)
{
cout << "YES" << endl;
}
else
{
cout << "NO" << endl;
}
for(int i = 1; i <= n; i++)
{
adj[i].clear();
vis[i] = false;
}
}
int32_t main(){
int t = 1;
cin >> t;
while (t--) {
solve();
}
}
```

Problem H was awesome!

its nice but pretty easy for an H

When both Valeriu and Marcel would be in a cycle what would be needed to check the reaching time to the entry node? if Valeriu were in a cycle answer should be "NO". but here we are checking the reaching time to the entry node.

if they were both in a cycle i think the answer would be yes since V would always have 2 options to choose from , its the node out of the cycle where he could be caught.

This Case is considered in the first test case itself. Previously I just read the question, so I thought they might have missed it.

You can avoid hard coding in problem C.

Value of cell $$$(i,j)$$$ is $$$\min(i,j,11-i,11-j)$$$.

dislikeme>_<are you serious ?

XD

Bro ☠️

Another approach is to take min Chebyshev distance from (i, j) to the 4 middle squares.

Did you mean to invert the Chebyshev distance? I don't see how $$$min(x_2-x_1, y_2-y_1)$$$ would lead to the same pattern.

5 — min( max(abs(x — 4), abs(y — 4)), max(abs(x — 4), abs(y — 5)), max(abs(x — 5), abs(y — 4)), max(abs(x — 5), abs(y — 5)), )

Yes, that is to invert the Chebyshev distance. Also you don't actually need to take the smallest distance from all 4 middle points since the smallest distance it's the nearest corner of the center square.

idk what you are talking about, there is no inverting, maybe look at the definition: https://en.wikipedia.org/wiki/Chebyshev_distance

i just took chebyshev distance from all 4 middle points and took the distance to the closest one — if you know how to identify the closest corner without computing all 4 distances I would like to see that.

Chebyshev distance is defined as

$$$D_{Chebyshev} := max(|x_1-x_2|, |y_1-y_2|)$$$

When I talk about inverting $$$D_{Chebyshev}$$$ over $$$k$$$ I mean to do $$$k - D_{Chebyshev}$$$. Since every distance in one end gets mapped to the other end. More formally $$$D_{Chebyshev}: \{0, 1, 2, ... , k-1, k\} \mapsto \{k, k-1, ... , 2, 1, 0\}$$$

In our case $$$k=5$$$

I see, it is inverted in that sense.

Thank you for the formula. Could you please provide a more general formula that works for concentric matrices where the values decrease from the outermost layer to the inner layers? ("Do you think this formula consistently works for matrices where the values increase from the outermost layer to the inner layer

min(i,j,n+1−i,n+1−j)) and thanks.yes

I initially wanted to use the Chebyshev distance, but I became confused due to the uneven grid size. I then considered using the midpoint (4.5, 4.5), but that approach became quite messy. Do you have any suggestions ?thanks.

take smallest distance from all 4 middle points

I also did this.

it can be also measured by using this formula

if (mat[i][j] == 'X') total_sum += s[i * 11 + j] — '0';where predefined template string with values string s = "1111111111\n" "1222222221\n" "1233333321\n" "1234444321\n" "1234554321\n" "1234554321\n" "1234444321\n" "1233333321\n" "1222222221\n" "1111111111\n";

I want to know how do you come up with solutions like this??

problem E and F gives me some real stress but overall the contest was fun!!

Similar problem like E https://mirror.codeforces.com/contest/1840/problem/D.

"For more stress"

Problem G's walk-through was so clear I could understand the concept in a split second!

totally agreed Want more this kind of tutorial in future #flamestrom

For C, I asked chatGPT to provide me with a program to generate the matrix and then I used it. Happy to get skipped as the rating doesn't matter to me.

using chatGPT is legal tho

That is smart, wish I'd have thought of that

hello can you explain why direction of approach doesn't matter in B question like any proof if possible?

What do you mean?

B mei jaise greedy approach laga rhe hai na left to right jaate hue, so I'm asking right to left jaane pe same answer aayega and Better answer nahi aayega iski kya guarentee hai. I would love a proof of why Direction of approach does not matter.

Are you talking about this contest’s B? My solution was just brute force where I tried everything. I reset the variable after each iteration.

oh i'm really sorry i meant D (1d eraser)

Well if you’re convinced that it works backward you can think of reversing the string and doing it backwards as the same thing as doing it forwards

no i am not convinced that it works forward

according to me the ans should be min(forward approach, back approach)

greedy gives me the idea to approach from one single direction but how can you say that both directions will give same answer

here your reversing logic doesn't apply as say i claim forward approach takes 3 operation and backward approach takes 2 operations then on reversing also answer will be same i agree(2) but this time it will be from forward approach

how can you claim that only by checking from one direction you will get a answer same as backward approach

Oh, I see.

Well, let's try to 'prove' that the forward approach always gives an optimal answer.

What my code is doing is it's going from 0 to n and checking if each character needs to be converted or not. If it does, start a window, otherwise move on to the next index. Notice that when we move on to the next index, we'll never come back to the previous one.

So why is this important? That means that when we see an element that needs to be converted, we have to start a window. You might say that there are multiple ways to start a window (for example if we want to convert i+1, we could start a window of size k>=2 at index i) so that an index is converted, but the most optimal way will always be when it's started at the index that needs to be converted, since that way more indices ahead can be affected.

If you're convinced that the forward approach is correct, you can see why the backward approach will also be correct from my above reply, thus showing that both approaches will give the same answer.

Can someone explain why in problem B, it's optimal to increase the smallest digit ?

yes because if we notice that the ratio change of x+1/x (x is smallest digit) is large as compare to bigger digits so thats why it always gives ans..

if you add 1 to smaller number then you gain original product + (original — small). But if you add 1 to any other than smaller you gain -> Original product + (original — some other which is not smallest) . So first case will always be bigger

Say you have (3,3,2,3):

Let's pretend you need to find the largest possible product with all these elements except one.

This is obviously (3*3*3), as 2 is smaller than 3, and (3*3*2)<(3*3*3). Now we know that the excluded value is 2.

Now that we've got (3*3*3) as the largest product excluding the smallest value, we can add 1 to the excluded value, and now we've increased the product by (3*3*3).

We went from (3*3*3)*2 to (3*3*3)*3.

Based on AM-GM Inequality, a set of numbers with specific sum will have largest product if all the numbers are equal. Increasing the smallest digit will make these digits be closers to their average. It is not a solid proof but rather an intuitive way to quickly realize it.

hello can you explain why direction of approach doesn't matter in B question like any proof if possible?

Small number repeats more in product than big number

To avoid making product =0 ,since all numbers between 0 and 9 then if the number is 0 then it will be 1,of course if there is more than zero , the max product is 0 , otherwise we can proof by math that the (min(a)+1)×a(*) the rest of numbers is the maximum product

assume x=a*b*c*d

then (a+1)*b*c*d-x = b*c*d

basically, increasing a single number by one increases the answer by the product of all other numbers, so it's best if b,c,d are the largest possible, which means a is the smallest possible

$$$(a+1)*bcde\ldots < (b+1)*acde\ldots$$$

$$$\Leftrightarrow abcde\ldots + bcde\ldots < abcde\ldots + acde\ldots$$$

$$$\Leftrightarrow b < a$$$

If b is less than a, increasing b by one will result in a larger product.

You can also think of it as multiples. Which is greater, the next multiple of the smallest digit (i.e. increasing the largest digit) or the next multiple of the largest digit (i.e. increasing the smallest digit)

For example, let $$$a < b$$$.

You can think of $$$a*b$$$ as a multiple of $$$a$$$ or as a multiple of $$$b$$$. It can be shown that the next multiple of $$$b$$$ (i.e. $$$(a+1)*b$$$) is always larger than the next multiple of $$$a$$$ (i.e. $$$(b+1)*a$$$) since $$$a*b + b > a*b + a$$$

$$$E$$$ can be solved in $$$O(n \log(n))$$$ and $$$F$$$ can be solved in $$$O(n)$$$.

For $$$E$$$, first sort the array $$$a$$$. The optimal value of $$$h$$$ is $$$\max_i(v_i)$$$ where $$$v_i=\min(a_{i+1}-1, \left \lfloor (x+PrefixSum([a_1:a_i]))/i \rfloor \right)$$$ if $$$v_i \geq a_i$$$ and $$$v_i=0$$$ otherwise (set $$$a_{n+1}$$$ as $$$\infty$$$ or handle the case $$$i=n$$$ separately).

$$$F$$$ can be done by sliding windows method.

don't you use sort in E?, which is already O(n log(n))

You can use binary search for finding solution. l = 0, r = 1e9

but r = 1e9 get wrong answer. must use r = 2^31 -1 or r = 2e9 + 7. i used 1e14, 1e15 and 1e18 then i got wrong answer. it was so boring :>

we can also use r = max of all the array elements + 1e9.

Why ?

Its work fine for me , No need for this values .

My submission 224465049

And I think Binary Search is more simple than the formula and easier to work with for beginners.

I don't know. maybe the problem was from another thing. however this is my code who got wrong answer : 224488235

Mid is Overflow

1e18 / 2 = 5e17 which cann't stored in an int

oh really? you are kidding! I knew that. i used define for contest who i don't forget using long long

#define int long longF Can be done in O(N)

using two pointer one on the current node and other at the first element of contigous part (which satisfy the condition).

254795451

Can someone explain why this code doesn't work for Problem E. I understood immediately that Binary Search would be used but my code fails on TestCase-7. TIA

https://mirror.codeforces.com/contest/1873/submission/224447874

accepted solution by modifying your code. Your upper limit will overflow long long. worst case is 1 tile with 1e9 height and water also 1e9 , upper bound is 2*1e9 + 1 , use a bit more for edge cases.

https://mirror.codeforces.com/contest/1873/submission/224515771

Thanks

Try updating the method

`possible`

such that, you exit the loop (or return false) just after`sum`

exceeds`x`

, so that it doesn't overflow.calculating mid can overflow too

Calculating

`mid`

can overflow if we use`ll mid=(lo+hi)/2;`

.But, if we use

`ll mid=lo+(hi-lo)/2;`

, it shouldn't overflow.yes that won't overflow.

yes i agree, but why ???

as the constraints are 10 power 9, in the worst case i am adding n 10power9's, so in my case w = 1e9 * 1e5 which is 1e14. this should not get overflowed right.

https://mirror.codeforces.com/contest/1873/submission/224396755

You have set the upper bound as

`1e+17`

.So, the value

`mid`

can have a large number (say`5e+16`

). (Here, it doesn't have to be less than`10 power 9`

, I believe.)Inside the function

`func`

, when computing`h-a[i]`

, if`a[i]`

is small, then,`h-a[i]`

can be large.Because of this,

`w += max(h-a[i],0LL);`

will keep increasing, and it might overflow eventually. I think when`n`

is somewhere around`10 power 3`

or more, it will overflow.(I haven't used c++ for a while) Just curious. How does an

`int`

datatype in c++ store such a large number ?i am defining my int as long long which can store numbers till 1e18. so i wont think any overflow will occur

i got the same problem.i even doubt that this problem cant be solved by binary search

For problem F: Given, (l≤i<r). I don't understand why l and r can be same.

Author's right answer is 1 in this test:

`2 15 9 6 1 3`

.However, for 1<=i<r we can't find h[i] is divisible h[i + 1]. So I think this answer is 0.

Please read the blog https://mirror.codeforces.com/blog/entry/120638.

Didn't get the idea quite right, still cant figure it out how l and r can be equal, had to go through such a pain for this :)

Read the problem carefully, it says h[i] must be divisible by h[i+1] for each l<=i<r. Note how if l=r, no i's are checked.

I didn't check the case where l1 and r1, because of this I had wa2:(

most right one will always be picked if its apples value is less than or equal to k, so in this case minimum answer can be 1.

For E first i solved in Python and got AC but time was around 1400s.So for the fear of hacking i converted that code to C++ and submitted it but still the first AC time is being taken in account.so is the first ac soln final in div 4?

Video Editorial For Problem A to H

please make vedio editorials for atcoder regular contest.There are no english vedio editorial for that.

In f, we can use two pointers instead of binary search, so it can be reduced to O(2*n)

Great Editorial. Especially liked Problem G's illustration.

two liner solution for B.

Could someone help me, why this code is getting wrong answer in problem H https://mirror.codeforces.com/contest/1873/submission/224501501

It fails for the below input

For Problem F,

Can someone explain if I take l = 2 and r = 2 as well, then how is 2<=2<2 ?

pick l=2 and r=2 , you will find that no such i satistied l<=i< r . That means you don't need to check any h[i]. So you can collect fruit from a[l] to a[r] (that's a[2] actually) . The same thing works when you want to collect fruit from only one tree.

I found an exactly same problem as problem H (Mad City)

https://github.com/all1m-algorithm-study/uospc2021/blob/main/problems/Police_Thief/statements.md

This is from an intra-college contest held by University of Seoul in 2021.

There is no English statement, but if you use a translator you'll notice there are few differences.

Even the examples given in the descriptions are the same.

Thanks for a good contest, although i joined late the experience was enjoable

For E , i used the value of high as x + sum of array.

Problem E can be solved in $$$\mathcal{O}(n\log n + \log n\log (a_i+x))$$$.

First, we notice that the permutation of $$$a[i]$$$ does not matter at all.

So we can

`sort(a,a+n)`

first, and then when we want to calculate how much water we use when the height is $$$h$$$, we can use`idx=lower_bound(a,a+n,h)-a`

, and then the water we use`nowuse=idx*mid-s[idx-1]`

.$$$s[i]$$$ is the prefix of sorted $$$a[i]$$$, i from 0 to n-1. We can use this because for $$$idx\le i\le n-1$$$, we have $$$a[i]>h$$$, so the use of $$$i$$$ is $$$0$$$.

By using this way, we can avoid calculating

`max(0,...)`

with n times.Since sort takes $$$\mathcal{O}(n\log n)$$$ and in every check in binary search we use

`lower_bound`

which is $$$\mathcal{O}(\log n)$$$, we have $$$\mathcal{O}(n\log n + \log n\log (a_i+x))$$$ in all.In problem c:

I use this equation : point = min(min(i,9-i),min(j,9-j)) + 1 it's the minimum distance to the edge + 1.

It was my best CF contest ever through my CP journey. I overcame my limit at last..

for problem G, as mentioned in the tutorial, we can see the problem as each B choose one direction(left or right) to eat the A's, so we can use dp.

dp[i][0] means if the i-th B choose to go left, the maximum A's that the 1-th, 2-th, ..., i-th B can eat. dp[i][1] is similar, it means the i-th B choose to go right.

therefore, we can get the transition equation: dp[i][0] = dp[i-1][0] + i-th B's position — (i-1)-th B's position — 1 (because if the (i-1)-th B go right and the i-th B wants to left, it will have conflict, so (i-1)-th B can only choose to go left) dp[i][1] = max(dp[i-1][0], dp[i-1][1]) + (i+1)-th B's position — i-th B's position (if i-th B wants to go right, we don't need to care about the (i-1)-th B's choice)

and here is the implementation https://mirror.codeforces.com/contest/1873/submission/224561489

Overkilled

No.

One of those balanced contest that we wished for thank you

Can someone explain solution for problem G? Tutorial is unavailable((

can someone please explain why my code does't work for problem E, thanks!

https://mirror.codeforces.com/contest/1873/submission/224479403

upper bound is too small, maybe you can use 1e12(long long).

please can someone suggests similar problems to F in which you do binary search on the answer plus some other computations?

In problem F i have first searched for a segment of divisible no.s and then i have applied sliding window technique . The time complexicity was 2.n which is actually O(n). Try this approach.

Can anyone point out error in my code for Problem F. I am trying to do the

O(n)approach. My code will return max_count as the answer.https://mirror.codeforces.com/blog/entry/120634?#comment-1070270

solved F question in O(N) using 2 pointers[submission:224462891][Your text to link here...](https://mirror.codeforces.com/contest/1873/submission/224462891)

Problem C can be easily done by taking the nearest x distance and y distance (these are independent) from the 5-score zone [(4,4), (5,4), (4,5), (5,5)] (_0 based-index_). The answer is min(5-x, 5-y) per 'X' instead of hard-coding it:

Submission: 224566541

Tutorial for G was so helpful!!!

Video Editorial (A-H) LINK TO YOUTUBE EDITORIAL

Not getting how to solve F problem, can anyone explain the intuition?

https://mirror.codeforces.com/blog/entry/120634?#comment-1070270

You can refer to : YT LINK

I request Div4's authors to make future rounds a little more challenging. This was on the easier side.

Btw, I liked all problem statements as they were crystal clear! I hope to see such clear statements in all future rounds as well.

That's why it's DIV4. To get more challenge you can always participate in other divisions.

great editorial for problem 1873G - ABBC or BACB thank you :)

Can anyone tell me why my code is failing?

For the last problem, I didn't notice the number of edges is n, I skipped the first line immediately assuming it just says there are n nodes. Then I thought maybe around 20 mins on this and was amazed how people can solve it while I was trying to prove it is a hard problem. It was like finding a bunch of induced cycles, which isn't an easy problem in general graphs. I can't blame that it wasn't explicit enough, since if I was reading the input format or the first line, it was clear there are n edges.

I've also read the problem like this.

In such version, we can solve it at least as follows (or maybe easier):

anycycleI think you are right. I thought too deep about it: I thought the winning strategy is to walk along an induced cycle, otherwise, marcel can at some point go along a chord of a cycle and catch Valeriu. Finding a winning strategy is difficult I believe, the question didn't ask for the winning strategy, though. I think the reason I've thought too much about it is that, I overlooked that every vertex that lies on a cycle, also lies on an induced cycle, so I didn't need to check if a vertex is on induced cycle, if it is on a cycle, certainly there is an induced cycle the vertex is on it. So, yes it was easier to decide if he wins but not easy to find a winning strategy.

Also one last and less important point: the approach you mentioned in general graphs could be quadratic to #vertices, so if the number of the vertices is big (like in this case), it could fail if it has too many edges (well this case didn't have that).

The first item (for each vertex, determine whether it lies on any cycle) can be performed with a single depth-first search, similar to finding articulation points or biconnected components.

Finding distances is two breadth-first searches, one from Marcel and the other from Valeriu.

So the solution is O(E) in total.

That’s correct, and O(|E|) in dense graphs could lead to O(|V|^2) and if |V| is big, like 1e5, then it will be TL. But, of course for this particular question it doesn’t happen.

When the graph is given explicitly, which is true in the majority of the problems, $$$O(|E|)$$$ is quite distinct from $$$O(|V|^2)$$$.

If we can read it, which takes $$$O(|E|)$$$ already, we can certainly work with it in $$$O(|E|)$$$ as well.

Fair point and applies for most of the cases. But notice that we can read a quadratic sized graph in sub quadratic or even linear time, if it is given by the edges which don’t belong to the graph. Basically by its complement definition. Can we still run it linear to size of V? Or modify it to be linear to V?

Here's my simple

O(n)approach using sliding window onF — Money Trees@mesanu :224470856

I think it calls "two point".

How to understand "Because we have a tree with an additional edge, our graph has exactly one cycle."? I think we are only gived a simple graph and there may be more cycle.

It follows from the statement "The roads are given that it is possible to get from any building to any other building going along the roads.". This means the graph is a connected graph. A connected graph with n nodes and n-1 edges can be proven to be a tree. There is edge left over after making the tree and adding it must create a single cycle (as every node is already connected).

It dawned on me。Thanks

you killed it , thanks

In hindsight my bridge-finding algorithm seems like an overkill :) 224589798

Could you explain the idea?

If you don't understand anything just read the offical solution :))) It's much faster, simpler and more elegant.

With the given conditions, every bridge in the graph does not belong to the cycle, and thus every non-bridge belongs to the cycle. After we traverse on the graph we will find the full set of vertices that belong to the cycle. Start traversing from Valeriu's positon until you reach one that is in the cycle (possibly Valeriu's original position itself) and calculate the distance between the vertices. As the tutorial above indicates, this is Valeriu's entry to the cycle. We then calculate the distance between that entry vertice and Marcel's. The answer is No if the former distance is not smaller that the latter, or both Valeriu and Marcel share the same starting position.

For an implementation of the bridge-finding algorithm, see https://cp-algorithms.com/graph/bridge-searching.html#implementation.

What i have done is very similar to yours, I used a bfs to find nodes in a cycle and then used another bfs to find the shortest distance both of them need to get in the cycle

H was beauty I was played in B ;-;, although i observed quickly that answer is always incrementing the smallest element but i wasnt able to prove it at the moment. What is the proof tho?

Can somebody explain Problem F using BS Please.

My video solution for F uses Binary Search and Prefix Sums.

Can some please explain why I got wrong answer when the value assigned to 'hi' was 1e14 or 1e18?

https://mirror.codeforces.com/contest/1873/submission/224393419

This resulted in wrong answer in test case 4 but was accepted when I updated the value of 'hi' to 1e10.

Thanks.

Because when 1e14 or 1e18 is multiplied by 2*10^5 (maximum size of array) an overflow will occur (long long maximum value is approximately 9*1e18)

A is a pulchritudinous problem. I solved it with Bogosort!

Why in problem H, in test 2, in test case 3, the answer is NO? Test case: 1 18 7 15 14 6 14 8 6 18 4 13 3 12 11 9 14 2 14 17 14 11 11 5 1 7 16 11 15 11 14 18 1 13 10 8 13 14 12 6 UPD: Sorry I missed one test case, the answer for test case above was YES

⠀

System testing, nothing much you can do about it

thanks for editorial,thanks for doing div 4.

I applied sliding window in F but getting wrong answer. Can anyone help

https://mirror.codeforces.com/blog/entry/120634?#comment-1070270

224598998 224600540 May I ask why the first submission cannot solve the problem while the second one can? The only difference between them is:

You can also watch the video editorials I made on the problems E , F, G and H

Enjoy watching and let me know what you think about them!

for problem A, you can check if any of the letter is in the right position then you can return "Yes" else "No".

Finally I became Pupil after a one year of trying. Since i acheived my goal. I am done with these stupid contests. since anyway I feel there is no much difference anyway

why are they not rating the questions in the recent contests?

H problem Clearing the array vis[N] is the most important !

Could someone can take a look at my F: 224638271, thx. I look for all the continuous intervals that meet the conditions, then use binary search on the length of intervals, but got wa2.

1873F — Money Trees Sliding window approach

if current tree high is not divisible then we re-init it, also update ans, else try to reduce the window to match the condition.

My solution for G.

The logic was intuitive. Going left to right, it doesn't make sense to use operation 1 the operation 2 and then operation 1 again(try simulating it through the test cases given). Our best possible score is achieved through a series of operation 1s followed by a series of operation 2s.

how is the time complexity in the 1873C — Target Practice is O(1) when you have 2 nested for loops iterating the input?

Problem H Idea is simple but implementation bit complex.

can anyone tell me where my code fails

https://mirror.codeforces.com/contest/1873/submission/224537788

Here I simply find the nodes which are part of cycle and run a disktra to find whether b can reach any part of cycle before a if true then return 1 else return 0.

Solved Actually my code for finding cycles nodes is not correct.

Problem F can also be solved using Binary Search ... My code

Can someone explain the second last sample test case for problem H?

8 5 3

8 3 5 1 2 6 6 8 1 2 4 8 5 7 6 7

I think Valeriu can always escape Marcel by choosing node 3 or 4, whichever one Marcel doesn't choose.

G is the hardest problem I think, but solving it was gratifying

for H,what if there are 2 entrances of the circle?

I absolutely loved the editorial for problem G! Wow! It was so fun to read, and so easy to understand! It would be amazing if all editorials were like this: being longer is not necessarily bad if a longer editorial is better at explaining the solution than a shorter one.

What exactly I am doing wrong here in Problem E. I solved this on a pen and paper, and seems to me that the calculations are correct. I am first sorting the heights and then start filling up from the leftmost side of the tank. My submission My solution

At the end, you're adding x/n.. instead, you need to add x/(i+1) where i is the index of the largest coral that gets submerged. Also, you need to use a larger data type.. I modified your code.

Thanks a lot brother. Gotta love the Codeforces community for helping learners grow.

Awesome round. Enjoyed how the editorial was written.

F can be done in O(n) using Sliding Window. Much more intuitive too imo.

https://mirror.codeforces.com/contest/1873/submission/224490319

why is my F failing i just updated len and the sum in one execution rather than using a binary search approach??

I tried solving "ABBC or BACB" problem with this code but I can't find the reason why my code is not working for 2 test cases ???(given below is my code) signed main() {

ios::sync_with_stdio(false); cin.tie(NULL); int size; cin>>size; while(size--) { string s; cin>>s; int n = s.length(); bool check=false; if(s[0]=='B' || s[n-1]=='B') check=true;

}

why the answer of test case "ABAAB" in problem G is 2

ignored this comment

Awesome contest with awesome editorial !! Really had a great time solving this amazing problems

G can also be done using dp. for every 'B' we store the count of A's on its left and right and then we try every possible way of using those A's.. 225181389

Can't we do problem F in

O(n)?simplest solution of G ~~~~~

~~~~~

I have implemented Problem H a little bit differently than the editorial if you are interested then check out my code.

Codefor F this solution is somewhat dp if you prefer. My submission: 225711679 The idea is to first find each maximal segment. then check maximum size we can get starting from any i in that seg. in order to pass TL remember sum and where it ended for previous starting point and start from there and not from i.

Can someone help me understand why my solution for F gives WA judgement: https://mirror.codeforces.com/contest/1873/submission/225741996

For money trees problem my solution is giving TLE only if I use sort else without sort it gets accepted can anyone explain why sort here is giving TLE FOR INFO: if i remove sort and optimization part the solution gets accepted include <bits/stdc++.h> using namespace std; int mm=0; define ll long long void solving(void); int main() {

} void solving(void) {

vector<pair<ll, pair<ll, ll>>> v; // this will store <length of subarray satisfying the divisibility rule<e means starting,l means ending of subarray >> elements

if (maxi >= abs((i.second.second — i.second.first) + 1)) // optimization only if vector is sorted in decreasing order of subarray length break;

}

Hey, Can anyone help me with this runtime error? I have tried debugging and wasted a lot of time. Any help is appreciated.

https://mirror.codeforces.com/contest/1873/submission/226134316

In function dfs() last else block you are popping from the set without checking if it is possible to pop from the set Here is your AC code

You can always you assert to figure out the mistake in your code.

EDIT -> The word set in the first para is actually the stack you are using. I called it set as I was thinking about memory and pointers. In the last statement, "use assert"

226968842

Can someone help me debug this code for problem F . I can't seem to find on which test case it fails.

Thanks in Advance :)

227178473 my dp with memoisation solution,store every preceding and proceeding B's and then use dp to choose the best possible combinations(note that:once u eat all the A's from right,u cant eat any left A's for all the B's proceeding that B).

In Problem E

why the concept mid=(low+high)/2; is not working?

problem/F

i have being trying to understand the editorial of this question. I can't understand how at the end when we cout<<r<<endl; how is r the maximun subarray. like the logic behind how this is happening. Can anyone explain y this is happening?

Could someone tell me for problem e why do we have to consider high = 2e + 7?

Problem F can also be solved in O(n) using sliding window approach. It is the modified version of the problem of Longest subarray having sum of elements atmost K. My submission — https://mirror.codeforces.com/contest/1873/submission/233766852

Can anyone please tell me the error in this code 237621263 for the 1873F - Money Trees.

I am first finding whether $$$h[i]$$$ is divisible by $$$h[i+1]$$$ for all $$$0 \leq i < n$$$. Then, for all the contiguous subsequences $$$[h_l,h_{l+1},…,h_r]$$$ and storing $$$l,r$$$, and the $$$sum$$$ in a map as a single entry for a sequence.

Now, for a sequence using two pointers $$$start$$$ and $$$end$$$, I am traversing the subsequence; if $$$a[start] > a[end]$$$, then increase the $$$start$$$ pointer and decrease the $$$sum$$$ by $$$a[start]$$$ otherwise decrease the $$$end$$$ pointer and decrease the $$$sum$$$ by $$$a[end]$$$ until the $$$sum \leq k$$$ condition is satisfied.

The above process is done for all the sequences, and the maximum length will be the answer.

In Problem F. Money Trees can anyone tell me why my this code is not passing test case 46 of test case 2 . https://mirror.codeforces.com/contest/1873/submission/238364394

What does the phrase "Notice the order of operations doesn't matter." precisely means for the problem D

The tutorial of problem G is beautiful, it was very informative for me to learn how to approach a problem :)

243211598 i dont understand why counting sum is giving overflow on problem E.

Overall Good contest, No offense but I don't think problem G was kind of an Ad hoc problem. I solved it using DP. Later came to see your approach and the explanation is amazing.

Problem C can be solved using a triple for loop: since every "square" has a delta with the outer one of 1, that is constant, we can iterate trough the whole sub-square adding 1 to the final answer each time we find a 'X'. In this way, we will find an 'X' as many time as it worths.

My sumbission as reference: 247543890

ik I'm late but in E, instead of computing value of 'tot' for every height(mid), why not just check if the height is greater than the height of longest coral, if it is, then instead of looping through all coral heights to calculate 'tot' we can calculate it by subtracting total volume of coral from total volume of aquarium. Total volume can be precomputed while taking input of coral heights.

1873C — Target Practice (alternative way to it may be a bit convenient if we don't want to write that number matrix)

// By: TESFAMICHAEL TAFERE

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

using namespace std;

int main() { ios::sync_with_stdio(false); cin.tie(nullptr);

int t; cin >> t; while (t--) { char s[10][10];

}

}

Problem G becomes way easy you just have to find smallest consecutive A substring . The ans = Number of A(s) — Length of smallest consecutive A(s) substring . Submission = https://mirror.codeforces.com/contest/1873/submission/264596469

Problem G was very good for the beginners and thanks for the lovely Editorial .

But the given implementation part (code) is bit complex and not easy to understand. I think I have written a code which is very easy to understand code out of all the solutions I've watched till now

Here's my submission link for those who are struggling in the implementation part : https://mirror.codeforces.com/contest/1873/submission/265911111

Approach is simple :

If the string start with 'B' or ends with 'B' or if it has any two consecutive 'B' in it , then the answer is simply the sum of all A's in the string.

Otherwise we have to eat the A's where it occurs more in a chunk , hence the chunk with minimum no. of A's would be ignore , so in this case we can simply print, Sum of all A's — (minimum number of A's in any chunk/group);