I'm so thankful to all testers, especially to Gassa and Rox for their invaluable help!

1409A - Yet Another Two Integers Problem

Idea: vovuh

**Tutorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int t;
cin >> t;
while (t--) {
int a, b;
cin >> a >> b;
cout << (abs(a - b) + 9) / 10 << endl;
}
return 0;
}
```

Idea: vovuh

**Tutorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int t;
cin >> t;
while (t--) {
int a, b, x, y, n;
cin >> a >> b >> x >> y >> n;
long long ans = 1e18;
for (int i = 0; i < 2; ++i) {
int da = min(n, a - x);
int db = min(n - da, b - y);
ans = min(ans, (a - da) * 1ll * (b - db));
swap(a, b);
swap(x, y);
}
cout << ans << endl;
}
return 0;
}
```

1409C - Yet Another Array Restoration

Idea: vovuh

**Tutorial**

Tutorial is loading...

**Solution (Gassa)**

```
// Author: Ivan Kazmenko ([email protected])
module solution;
import std.algorithm;
import std.conv;
import std.range;
import std.stdio;
import std.string;
void main ()
{
auto tests = readln.strip.to !(int);
foreach (test; 0..tests)
{
int n, x, y;
readf !(" %s %s %s") (n, x, y);
auto answer = int.max.repeat (n).array;
foreach (start; 1..51)
{
foreach (d; 1..51)
{
auto a = iota (start, start + d * n, d).array;
if (a.canFind (x) && a.canFind (y))
{
if (answer.back > a.back)
{
answer = a;
}
}
}
}
writefln !("%(%s %)") (answer);
}
}
```

**Solution (vovuh)**

```
#include <bits/stdc++.h>
using namespace std;
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int t;
cin >> t;
while (t--) {
int n, x, y;
cin >> n >> x >> y;
vector<int> ans;
for (int d = 1; d <= y - x; ++d) {
if ((y - x) % d != 0) continue;
vector<int> res;
bool foundx = false;
int cur = y;
int need = n;
while (cur >= 1 && need > 0) {
res.push_back(cur);
foundx |= cur == x;
--need;
cur -= d;
}
cur = y;
while (need > 0) {
cur += d;
res.push_back(cur);
--need;
}
sort(res.begin(), res.end());
if (need == 0 && foundx) {
if (ans.empty() || ans.back() > res.back()) {
ans = res;
}
}
}
assert(!ans.empty());
for (auto it : ans) cout << it << " ";
cout << endl;
}
return 0;
}
```

**Solution (Rox)**

```
#include <bits/stdc++.h>
using namespace std;
int main() {
int tcs;
cin >> tcs;
while (tcs--) {
int n, x, y;
cin >> n >> x >> y;
int diff = y - x;
for (int delta = 1; delta <= diff; ++delta) {
if (diff % delta) continue;
if (diff / delta + 1 > n) continue;
int k = min((y - 1) / delta, n - 1);
int a0 = y - k * delta;
for (int i = 0; i < n; ++i) {
cout << (a0 + i * delta) << ' ';
}
cout << endl;
break;
}
}
}
```

1409D - Decrease the Sum of Digits

Idea: MikeMirzayanov

**Tutorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
int sum(long long n) {
int res = 0;
while (n > 0) {
res += n % 10;
n /= 10;
}
return res;
}
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int t;
cin >> t;
while (t--) {
long long n;
int s;
cin >> n >> s;
long long ans = 0;
if (sum(n) <= s) {
cout << 0 << endl;
continue;
}
long long pw = 1;
for (int i = 0; i < 18; ++i) {
int digit = (n / pw) % 10;
long long add = pw * ((10 - digit) % 10);
n += add;
ans += add;
if (sum(n) <= s) {
break;
}
pw *= 10;
}
cout << ans << endl;
}
return 0;
}
```

Idea: vovuh

**Tutorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int t;
cin >> t;
while (t--) {
int n, k;
cin >> n >> k;
vector<int> x(n), y(n);
for (auto &it : x) cin >> it;
for (auto &it : y) cin >> it;
sort(x.begin(), x.end());
int j = n - 1;
vector<int> l(n), r(n);
for (int i = n - 1; i >= 0; --i) {
while (x[j] - x[i] > k) --j;
r[i] = j - i + 1;
if (i + 1 < n) r[i] = max(r[i], r[i + 1]);
}
j = 0;
for (int i = 0; i < n; ++i) {
while (x[i] - x[j] > k) ++j;
l[i] = i - j + 1;
if (i > 0) l[i] = max(l[i], l[i - 1]);
}
int ans = 1;
for (int i = 0; i < n - 1; ++i) {
ans = max(ans, r[i + 1] + l[i]);
}
cout << ans << endl;
}
return 0;
}
```

1409F - Subsequences of Length Two

Idea: vovuh

**Tutorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9;
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int n, k;
string s, t;
cin >> n >> k >> s >> t;
vector<vector<vector<int>>> dp(n + 1, vector<vector<int>>(n + 1, vector<int>(n + 1, -INF)));
dp[0][0][0] = 0;
for (int i = 0; i < n; ++i) {
for (int ck = 0; ck <= k; ++ck) {
for (int cnt0 = 0; cnt0 <= n; ++cnt0) {
if (dp[i][ck][cnt0] == -INF) continue;
int e0 = s[i] == t[0];
int e1 = s[i] == t[1];
int e01 = t[0] == t[1];
dp[i + 1][ck][cnt0 + e0] = max(dp[i + 1][ck][cnt0 + e0], dp[i][ck][cnt0] + (e1 ? cnt0 : 0));
if (ck < k) {
dp[i + 1][ck + 1][cnt0 + 1] = max(dp[i + 1][ck + 1][cnt0 + 1], dp[i][ck][cnt0] + (e01 ? cnt0 : 0));
dp[i + 1][ck + 1][cnt0 + e01] = max(dp[i + 1][ck + 1][cnt0 + e01], dp[i][ck][cnt0] + cnt0);
}
}
}
}
int ans = 0;
for (int ck = 0; ck <= k; ++ck) {
for (int cnt0 = 0; cnt0 <= n; ++cnt0) {
ans = max(ans, dp[n][ck][cnt0]);
}
}
cout << ans << endl;
return 0;
}
```

**Solution (Gassa, greedy, O(n^4))**

```
// Author: Ivan Kazmenko ([email protected])
module solution;
import std.algorithm;
import std.conv;
import std.range;
import std.stdio;
import std.string;
void main ()
{
int n, k;
while (readf !(" %s %s") (n, k) > 0)
{
readln;
auto s0 = readln.strip;
auto t = readln.strip;
auto s = s0.dup;
int res = 0;
foreach (b0; 0..k + 1)
{
auto e0 = k - b0;
loop_x0:
foreach (x0; 0..b0 + 1)
{
loop_y0:
foreach (y0; 0..e0 + 1)
{
int b = b0;
int e = e0;
int x = x0;
int y = y0;
s[] = s0[];
for (int i = 0; i < n && b; i++)
{
if (s[i] == t[0])
{
}
else if (s[i] != t[1])
{
s[i] = t[0];
b -= 1;
}
else if (x > 0)
{
s[i] = t[0];
x -= 1;
b -= 1;
}
}
for (int j = n - 1; j >= 0 && e; j--)
{
if (s[j] == t[1])
{
}
else if (s[j] != t[0])
{
s[j] = t[1];
e -= 1;
}
else if (y > 0)
{
s[j] = t[1];
y -= 1;
e -= 1;
}
}
int cur = 0;
int add = 0;
for (int i = 0; i < n; i++)
{
if (s[i] == t[1])
{
cur += add;
}
if (s[i] == t[0])
{
add += 1;
}
}
res = max (res, cur);
if (x > 0)
{
break loop_x0;
}
if (y > 0)
{
break loop_y0;
}
}
}
}
writeln (res);
}
}
```

fastest tutorial <3

Runtime Error C++ Exit code is -1073741571Submission : https://codeforces.com/contest/1409/submission/91879266

Can anyone explain what is going wrong in my code that is giving me runtime error?

Any help would be appreciated

Thanks :)

r[r.length() — 1]++; I guess this line is what causing error when r="" that is length = 0, r[-1] will give segmentation fault

nice problemset

My screencast + live commentary, enjoy watching :)

https://youtu.be/nloGFTpdTJo

+1 thanks

https://codeforces.com/contest/1409/submission/91894057

could you please help me why it is giving wrong answer?

re_start

Test Case2

828374536691952768 75

348970515787375312 29

Wonderful problems, tutorials and solutions!

Thanks to writers!

.

Don't spam the editorial blogs with these dead memes dude.

Why does pow(10,18) return 10^18-1

pow() doesn't always give the right answer, because it was intended to work with doubles, not integers

if u want to use pow for high powers u can typecast with long long int , whichs works quite effectively.

I do not know what that is could you show me an example?

I think he means if you want to do pow(10,x) and not have it mess up, do (long long)pow(10,x) and it will cast it to a long long and will give the right answer.

Why B solution is optimal? I did it, but didn't understand why it worked

Suppose you have 2 numbers a and b and you can decrement them by 1, 2 times.

Then you have 3 options

- a-1,b-1

- a-2,b

- a,b-2

In first case product is ab+1-a-b.

In second case product is ab-2b.

In third case product is ab-2a.

Now let b>a

Then ab+1-a-b > ab-a-b > ab-2b (because a<b)

So you now saw first case gives bigger product than 2nd case.

Similarly you can prove for other pair of cases too.

Thanks I arrived at the same conclusion but I missed one case and concluded decreasing smaller number will be more optimal. Got the correction thanks.

If we decrease a 2 times, then you decreased the product by 2*b. If we decrease b 2 times, then we decreased the product by 2*a. If we decrease a once and b once, then we decreased the product by a+b.

If a>b, 2*a>=a+b>=2*b, whereas, if b>a, 2*b>=a+b>=2*a

I guess you can see why decreasing one as much as possible is optimal.

thanks_wicardobeth_now i understand why it is optimal;`Can someone tell me why my submission for E doesn't work? It's coordinate compression + greedy. https://codeforces.com/contest/1409/submission/91881485

Dont know why you used coordinate compression but greedy wont work for this. Greedily choosing best for one plank might incur loss in overall max answer.

Could you give me an example please? I still don't get why greedy won't work. I used coordinate compression beacause the values of the xs and ys could range from 1 to 10^9.

I got why. Thanks anyways!

Can you explain why it wouldn't work?? I too had a similar approach.

Take this case:

Answer should be 6, but greedily it is 5.

Well, with a little bit of maths, C can even be solved in O(n)

E can also be solved with a simple DP. First, sort the $$$x$$$ array. Then for each $$$i$$$ from $$$0$$$ up to $$$n - 1$$$:

$$$dp^0_i$$$ contains best answer up to $$$x = x_i$$$ with one platform, and $$$dp^1_i$$$ with two platforms. If you are lazy (like me), just take the maximum of all values. Note that $$$j$$$ can be $$$n$$$ in the above loop, so the DP array will have $$$n + 1$$$ items.

Very nice problems!

Thanks vovuh! Nice contest to learn the basics for noobs like me

How to solve E if we have K platforms? $$$O(KNlogN)$$$ is fine but better than this??

$$$dp_{i, j}$$$ — we at the $$$x=i$$$ and placed $$$j$$$ segments. $$$dp_{i, j} = max(dp_{i - 1, j}, dp_{i - k - 1, j - 1} + cnt(i - k, i)$$$ where $$$cnt(l, r)$$$ is the number of points between $$$x=l$$$ and $$$x=r$$$. Can be improved by coordinates compression and two pointers (if the outer loop iterates over $$$k$$$ and inner loop iterates over the positions). Total complexity is $$$O(nk)$$$.

Nice solution, so this works even if we have $$$q$$$ platforms, and for each of them, we are given $$$cost_i$$$ and $$$length_i$$$, and you have a total budget $$$B$$$. It's some knapsack then?

Yes, it's a knapsack, but you need an additional state in your dynamic programming to solve such a problem.

You can binary search to find the maximum index can reach from each index i and store data in two arrays X, Y

first of all you need to sort x points "y points are useless"

by using binary search you can search for ind "the maximum index you can reach from index i" and store in X and Y the number of nodes "which is ind-i+1"

then you can maximize the elements of X from the back so you have the maximum number of nodes you can take starting from node i to the end

the answer will be max(Y[i]+X[i+Y[i]]) "make sure (i+Y[i]) doesn't go out of the array"

O(nLog(n))

https://codeforces.com/contest/1409/submission/91933328

You should re-read the problem

It was misunderstand.

Sorry.

I'm using similar logic but not sure why the 2nd test case is giving wrong answer. 101451461

First I'm compressing the array of Xs, and then for each X[i], trying to find farthest X[j] such that it's at least K units away. Then I'm basically iterating over each X[i] again and evaluating answer for i as max_number_of_points_for_X_i+ max_answer for farthest (eligible_index+1).

This test cases drop your code

1

3 10

10 100 1000

1 1 1

your code print 3 and the answer here is 2

Lower_bound function return you points which is larger than c[i].second+k which makes you count them

to solve this problem i searched in lower_bound for c[i].second+(k+1) to find the smallest number out of my range then decrement ind by 1

an accepted solution for your code after this small edit https://codeforces.com/contest/1409/submission/101465446

it doesn't work for me with upper_bound on c[i].second+k instead of lower_bound on c[i].second+(k+1), i really don't know why.

I believe you can solve this version in $$$O(NlogN)$$$ using Alien's trick. Here's a good tutorial on it: http://serbanology.com/show_article.php?art=The%20Trick%20From%20Aliens

Yeah, F can be solved in O(n^3).

Hey could someone please take a look. I did the same thing as the editorial for

Problem Ebut still getting WA.[Submission here](https://codeforces.com/contest/1409/submission/91884363)any help is appreciated, thanks!

I think pts2 = upper_bound(all(xc),xc[next_ind]+k) — (xc.begin()+next_ind); should be replaced with a suffix MAX array

I think both of them should yield the same result, is that incorrect? Anyway will try that!

Thanks for your help!

can someone help me the tutorial of Problem B ? .. I didnt get it ... or any better soln will be appreciated !

Well the tutorial says that if we decide decrease a by 1 in the step then it's better to continue to decrease a until we can and then start with b. Similarly if we start with b then keep on decreasing it until we can then start with a.

consider both of the above cases and print the smallest product. Hope this helps:)

Problem D — Can be solved in

constant time(O(18))ApproachFirst, find the sum of the digits as

summMove from the first place digit (ones) to the higher one, check if

summ — (sum from ones digit to current position) + 1 <= s(needed sum)if

true: the answern — (number if you make digit after ith 9 and add one to the result)...Exampleif

n = 500ands = 4all

3digits should be 0 to make them zero the number should be increamented until 999 and adding +1 so500 --> 1000so the answer will be1000-500 = 500...Python ImplementationThis is the same time complexity as the intended solution.

I believe the editorial solution is actually O(18*18) as it goes through the array once to calculate the prefix/suffix and it also goes through the array to calculate the sum. To reduce the extra log(N) factor, an int could be used to keep track of the sum and just update it accordingly in between iterations. However, in the grand scheme of things, it doesn't really matter.

It can also be solved with binary search.

TimeTraveler Can you please explain how your Binary search solution works here?

For a given number x, we need to check if we can make the sum less than s in moves <= x, i.e is there any number between n and n + x which has sum of digits <= s ? Then we can binary search over the value of x.

and how do we do this check (fast): "is there any number between n and n + x which has sum of digits <= s "

By finding the least sum possible between n and n + x. Here's the function (log n) I wrote in my submission.

Check function1409E - Two Platforms I do not get the point why the y[] are in the statement. If it is obvious that we do not need them, then they are obviously unnecessary. If it is not so obvious then it is a misleading statement. So, why?

spookywooky I do not get the point why this comment is under the editorial. If it is obvious that we do not need it, then it is obviously unnecessary. If it is not so obvious then it is a misleading comment. So, why?

I mean, I don't think this makes sense. Clearly spookywooky thought we

doneed the comment, so your clever response isn't actually valid.I thought this legend needs $$$y$$$-coordinates as well as he thought we need this comment.

Partially, your answer below is right, this part teaches to notice some probably useless things in the problem and to make some transitions from one problem to another. And yeah, this is essentially not the worst way to allow repeated $$$x$$$-coordinates.

The y coordinates are of no use I agree but they are necessary as a part the author to chose to describe the original problem, sometimes some variables are just to help understand the actual statement clearly , but here these y coordinates are not misleading!

I do not think they are necessary in any way. The statement would be much simpler if formulated in 1 dimensional space. "Here are some x[], find two segments of size k covering as most possible of those points on x-axis, how much?"

Unfortunately problems are not all supposed to be simple. You have to decipher for yourself what is important and what is not. Just because it was obvious to some does not mean that others don't make that connection, at least immediately.

This is not atcoder.

I don't know whether they are misleading or not, but they are definitely unnecessary.

Unnecessary from readers point of view, think as a setter and you have this question in your mind but you have to plot a frame to fit it into an understandable statement. Now since everyone have different ways to describe the problem thus a statement in order to describe the original point of you may go in various directions. If you were a setter may be you would describe it in an easier fashion. But everyone cannot have same thoughts. Making a perfect problem statement that match thought of every participant is a tough kind of job. When we will jump into it we will realize the importance of clearity!

One benefit of $$$y$$$-coordinates is that it makes it clear and unambiguous that we can have repeated $$$x$$$-coordinates, as well as overlapping platforms. However, the statement said platforms could overlap anyway, so yeah I'm not sure.

Also, maybe it was intentional that they're unnecessary because the 2D to 1D transformation is something worth testing/teaching, especially in Division 3. In general, part of the difficulty of Codeforces problems does stem from transforming the problem into an equivalent but simpler one (in a way, this is all problem-solving).

For Problem D: Can anyone tell me what's wrong in the following approach: For example 217871987498122 10, traverse from starting digits, 2,1,7,8 and add them, when sum become greater than or equal to 10, that is on 7. increment one digit before 7 from 1 to 2 and rest all digits after it to 0, the number will become 220000000000000. if it violates on first digit then make 1000... this number. please help?

you need to stop when the sum becomes greater than or equal to S. In this test case, you are actually already doing that.

Otherwise, the approach looks good to me. I've also implemented an almost similar approach.

i had stopped, can you tell me, what's wrong in the code? 91871968

I tried stress testing your submission against mine on random inputs and found that your code fails on some inputs.

ExampleExample input: 1785120922 34

Your Output: -1783335802

Answer: 78

Upon observing, it fails on all such inputs where the digit at which you stop is 9, so your code tries to increment it, but it fails, and ends up dropping the remaining digits altogether.

My solution to C was as follows. I "stuffed" as many elements as I could betweeen x and y. Then I calculated this minimum difference and added some elements less than x. I would add as many as I could until I was done or these values became negative. In the case they became negative, I would add the remaining values above y. I believe this should be O(N)

My solution: https://codeforces.com/contest/1409/submission/91831887

https://codeforces.com/contest/1409/submission/91894057

can somebody please help me with this problem why it is giving the wrong answer? any help would be highly appreciated

Thanks

In the loop where you work with Sum and carry stuff, you have put the check and increment condition w.r.t j where it should be in respect to i.

As a result i is not changing therefore affecting your answer.

Brother, You are too good. thanks for showing your generosity towards me. Got AC

Always happy to help a fellow coder!!

Can somebody explain me the greedy approach in problem F please....

Let the string $$$t$$$ be

`be`

(begin and end). Sure, there is also a case where the letters of $$$t$$$ are equal, but it can be solved separately, or the solution could be carefully implemented to account for that.Now, suppose that we are going to make exactly $$$k$$$ moves. We have two kinds of useful moves: put

`b`

somewhere and put`e`

somewhere. Let the number of`b`

-moves be $$$b$$$, and the number of`e`

-moves be $$$e = k - b$$$.Of all the ways to put exactly $$$b$$$ letters

`b`

, the most impactful way is to put them into $$$b$$$ leftmost places in the string... unless we change some`e`

into`b`

in the process. Let the number of`e -> b`

moves we are going to make be $$$x$$$ ($$$0 \le x \le b$$$). So, with our`b`

-moves, we pick $$$x$$$ leftmost letters`e`

and change them into`b`

, then pick $$$b - x$$$ leftmost letters which are not`b`

and not`e`

and change them into`b`

.Similarly, when we are going to put exactly $$$e$$$ letters

`e`

, the most impactful way to do it is to put them into $$$e$$$ rightmost places in the string... unless we change some`b`

into`e`

. Let the number of`b -> e`

moves we are going to make be $$$y$$$ ($$$0 \le y \le e$$$). So, with our`e`

-moves, we change $$$y$$$ leftmost letters`b`

into`e`

, and also change $$$e - y$$$ leftmost letters which are not`b`

and not`e`

into`e`

.What remains is to look over all possible tuples of $$$(b, e, x, y)$$$. Remember that $$$e = k - b$$$, so there are only $$$O (n^3)$$$ such tuples. For each of them, simulate the above greedy process (in linear time) and then find the answer (in linear time too). The total complexity is $$$O (n^4)$$$. A possible speedup is to precompute the required stuff for prefixes and suffixes, bringing it down to $$$O (n^3)$$$ or $$$O (n^2)$$$ (didn't actually try).

Why we can assume $$$e = k - b$$$? If we actually do less than $$$k$$$ moves in the optimal solution, it means all the letters become

`b`

or`e`

. So we can cheat: for the optimal $$$e$$$, we make unnecessary moves from the left, and then overwrite their results by exactly $$$e$$$ moves from the right.All in all, this greedy solution requires more observations than the dynamic programming solution, but requires less proficiency to come up with it.

1409C - Yet Another Array Restoration can be solved greedily in $$$O(n)$$$ instead of $$$O(n+\sqrt{y})$$$.

HintCode (Python 3)In 1409D - Decrease the Sum of Digits, if we maintain the digit sum, time complexity will be reduced to $$$O(\log n)$$$.

Code (Python 3)Hi guys, I'm a python coder and I was wondering if someone could explain to me what's slowing my code down?

Submission: https://codeforces.com/contest/1409/submission/91900467

It should work quite fast since each loop can only go 18 times, and it seems to work fast when I tested it. However, I keep running out of time on test 3.

Number of operations is $$$2 \cdot 10^4 \cdot 18 \cdot 18 \approx 6.5 \cdot 10^6$$$ operations, and when you include the powers

`10**(i+1)`

and`10**i`

, as well as the fact that pypy bigint's are pretty slow, it's too much for python to handle I guess.I tried submitting it in Python3 instead, to solve the bigint problem, but that didn't work.

It could've been easily fixed by having $$$100$$$ or $$$1000$$$ test cases maximum, but ¯\_(ツ)_/¯

Chinese tutorials updated.

Can someone tell me why my code is giving the wrong answer, please... Submission: https://codeforces.com/contest/1409/submission/91873431 Thank you.. big help

For $$$10, 1$$$, your code

`if(sum == s && i!=v.size()-1)`

is triggered at $$$i=0$$$, so your shortcut to output "0"`if(i == v.size())`

fails.Similar problem to D.

https://codeforces.com/problemset/problem/1143/B

My doubt is pretty naive. Still I want to know why wrapper class Integer/ Long array sorting is faster than primitive data types like int / long ? I know that primitive data type use quicksort internally in Arrays.sort().

Also, when to use primitive for sorting and when to use wrapper for the same.Also look at both of my submissions for E in this contest.

In a nutshell, you can be hacked with an adversarial case that makes quicksort degrade to quadratic. Object sort uses merge sort instead which doesn't have this issue, but wrapper objects are slow and inconvenient.

I think the best solution is to randomly shuffle before sorting (I have a prewritten

`safeSort`

method, you can see it here: 91830757).Thanks for your reply. I got what the actual problem is.

By the way, shuffling the array and then sorting it may (very rare) also cause TLE if the array after shuffling comes out to be the worst case for quicksort.

Then let's build suffix maximum array on r and prefix maximum array on l. For l, just iterate over all i from 2 to n and do li:=max(li,li−1). For r, just iterate over all i from n−1 to 1 and do ri:=max(ri,ri+1).In problem E — the editorialist tells to build suffix maximum array on r and prefix maximum array on l. But Why?? I understand the use of the prefix array and suffix array. Why to find the maximum prefix array and suffix array?

Let's assume the answer is segL and segR (which do not overlap of course). Assume that there is a point midp which lies between two segments but do not overlap. We know segL stands on the left side of the midp and segR stands on the right side of the midp. Note that segR is the best platform possible on the right side of the midp otherwise the answer would contain the better platform (which means segR is not a valid solution). Greedly you can see the score of the best platform that is on the right side of the midp can be found using maximum suffix array. Also the the score of the longest platform that is on the left side of the midp can be found using maximum prefix array. This operation takes O(1) time if you have the prefix and suffix arrays. Now, because we don't know the midp, our goal is to iterate over all possible midps in O(n) time and store the best possible score.

Please don't hesitate asking any questions I'm trying my best to make things clearer.

Thanks for your reply. I understood it.

Can someone share some thoughts about how $$$F$$$ can be solved when length of $$$t$$$ is arbitrary. During contest i didn't read the part that length of $$$t$$$ is fixed and is 2. I kept trying to solve it for t of arbitrary length and when after the contest i read that length of t is fixed i realised it was trivial (: for length t = 2.

A slightly different and a bit easier to understand (maybe?) implementation of E. 91876338.

:)Regarding problem

E. Two Platforms, I used TreeMap to solve. Them time complexity is also O(N log N), but it is TLE. I got stuck on investigating the cause. Could anyone help me to have a look? thanks.https://codeforces.com/contest/1409/submission/91911507

Why this greedy approache for F is wrong?Can you help me?

Detail:

Suppose we change i positions into t[0],and i2 into t[1].Let's greedy:The i positions which are changed into t[0] is as small as possible,and the i2 positions which are changed into t[1] is as big as possible. (If s[i]=t[0]/t[1] ,it doesn't need to be changed).

Code:

Spoileri did a greedy approach too and it was like you said the t[0] position should be the least and the t[1] position should be the last but there is a problem maybe we have some character t[0] in the end of the string not exactly the last character but lets say the second half of the string lets say t[0] = a and t[1] = b

there can be a situation like this : aaa .. b .... (a) .. bb in this case its better to change (a) to (b) to get more occurrences.

in my solution i count if we change the i position to t[0] or t[1] how much we can get and how much we loss as we can see in the sample if we change (a) to (b) then we get 3 more occurrences and loss 2 occurrences and its good but some times its not however my solution got wrong answer in testcase 41 :))))

Thank you very much.I have completely understand.

For problem D, why do we need to have

`(10−dig)%10`

moves instead of`(10−dig)`

moves? As`dig=n%10`

, is that`%10`

really necessary?dig can be 0 as well so (10-dig) will give you pw*10 instead it should be pw*0

Can someone help me with F.

My

dp[ind][k][left]states that we are at index'ind', we can do atmostkchanges, and we haveleftoccurances of t[0] from [0:ind-1]. Then my answer will be dp[0][k][0]. My transition are, either change current character to t[0], or t[1] or don't change at all.I am getting wrong answer at testcase 20. My submission 91926014

Binary Search Solution for D

For some reason spoiler isn't working properly.

Thanks for sharing your code. I did pretty much the same thing but get TLE https://codeforces.com/problemset/submission/1409/99688938

Try using lambda function.

But why should it make a difference?

Lambda functions are a bit more efficient.

This submission with your code(with less arguments) got accepted. Also this got accepted. The former with lambda functions ran slightly faster.

Never saw this happen before...Thanks though

Dear, MikeMirzayanov Please Check Why all my Correct Solutions got Skipped. I had submitted 3 Solutions in Yesterday Div 3. Contest and they were correct but saw today all were skipped which dipped my rating a lot. Please Check why Did it happen.

pranshukas Because you sumbit your code again in your alt account anonymouse_13 to hack them afterword.(Remember hacks in div3 give no points).

I know that hacking any solution don't give any points. I was new to it so I thought to learn it. That's Why I just submitted my own solutions form some other ID and give it a try. I thought it was in no way harming any policy and it wasn't any mischief either because I haven't cheated, nor Plagrised. Then Why did my Submissions got Skipped?

Did you read the rules of the contest before participating ? If not then the last rule says:

"I will not use multiple accounts and will take part in the contest using your personal and the single account".In the solution of Two platforms problem,why are we considering l[i]+r[i+1] rather than l[i]+r[i]?

I don't understand where the author says in problem E that li:=max(li,li−1) will help. Why? How? Can somebody explain it more clearly?

Don't forget that you have calculated l and r beforehand.

`where li is the number of points to the left from the point i (including i) that are not further than k from the i-th point`

After that you convert the same array to a prefix maximum array (ith element contains best possible value of elements in range(0, i)). By induction you can iterate from left to right and you can say that ith element of maximum array is either i-1th element or the new added element's value. Because the author didn't separate the two arrays it might get a little confusing. They just converted the array to aa prefix maximum array without assigning it to a new variable.

oh okay I got it thanks!!

basically maximum contribution upto i from left side + maximum contribution from i+1 to end in right side for each i, while i is the fixed barrier between the two segments(worst case they will share a common end).

Exactly. But I don't think worst case is sharing a common end. The algorithm will always run n times on an array.

yeah, poor choice of words there actually! I meant extreme case if they share common end not overlap

I think the explanation of Problem E is a little hard to understand. I think the key point of this question was "Dividing the number line into two segments.". The answer is the best platform that lies in the left segment + the best platform that lies in the right segment. Simply iterate over all possible divisions and the best result is your answer. I couldn't understand the solution when I read this explanation but when I watched SecondThread's video I really understood the problem.

I also really liked the difficulty levels of the first 4 questions.

What is the proof for the fact in problem B solution ?

Assume you are going step by step (decrease 1 at a time) and a > b (you can switch them very easily in code).

if you decrease a, the final product will be (a-1)b = ab — b if you decrease b, the final product will be a(b-1) = ab — a

because a > b, decreasing a from ab will result in a smaller product then decreasing b from ab.

The problem starts when a=x. After a=x, you start decreasing from b. I don't think you can have a solid proof of which is optimal without starting from a or starting from b after this point. Therefore try both possibilities.

What is 1ll in editorialist solution? ans = min(ans, (a — da) * 1ll * (b — db));

LL stands for

`long long`

data type. It basically returns 1 as a long long instead of integer default.Thanks but why it is used here?

Because both a and b are integers that may be up to 1e9, so the multiplication would be up to 1e18, meaning it would overflow, that's the reason we want to convert it to long long.

In the solution code of problem C (Rox); the condition

diff/delta + 1 > nCan someone explain it for me?i.e x=4, y=16, n=5

when diff=1 you can get 16 — 4 = 12 so you need to have 12 — 1 = 11 integers to fill that gap. Thats why you check if you have enough integers to fill that gap.

How to solve F with bigger constraints? I was trying to solve for O(n^2) solution but couldn't make significant progress.

I felt soo dumb on myself when i did the correct thing but wasted 45 min and 5 WA submissions for erasing the vector for graph only upto n-1 and not n,on question D.

del

I thought to solve this Problem F with all my best but someone decided to irritate me as usual No matter how many time I tell them they irritate me so I started getting irritated but I will only get irritated I will ensure that they wont find any postive thing in irritating me The greater the irritation the dumber I will become even if not I will ensure I will become