Thanks for participating!

Idea: mesanu, MikeMirzayanov

**Tutorial**

### 1722A - Spell Check

Check if the string has length 5 and if it has the characters $$$\texttt{T}, \texttt{i}, \texttt{m}, \texttt{u}, \texttt{r}$$$. The complexity is $$$\mathcal{O}(n)$$$.

You can also sort the string, and check if it is $$$\texttt{Timur}$$$ when sorted (which is $$$\texttt{Timru}$$$).

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
#define forn(i, n) for (int i = 0; i < int(n); i++)
int main() {
string name = "Timur";
sort(name.begin(), name.end());
int n;
cin >> n;
forn(i, n) {
int m;
cin >> m;
string s;
cin >> s;
sort(s.begin(), s.end());
cout << (s == name ? "YES" : "NO") << endl;
}
}
```

Idea: flamestorm

**Tutorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
const int MAX = 200007;
const int MOD = 1000000007;
void solve() {
int n;
cin >> n;
string s, t;
cin >> s >> t;
for (int i = 0; i < n; i++) {
if (s[i] == 'R') {
if (t[i] != 'R') {cout << "NO\n"; return;}
}
else {
if (t[i] == 'R') {cout << "NO\n"; return;}
}
}
cout << "YES\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 = 200007;
const int MOD = 1000000007;
void solve() {
int n;
cin >> n;
map<string, int> mp;
string a[3][n];
for (int i = 0; i < 3; i++) {
for (int j = 0; j < n; j++) {
cin >> a[i][j];
mp[a[i][j]]++;
}
}
for (int i = 0; i < 3; i++) {
int tot = 0;
for (int j = 0; j < n; j++) {
if (mp[a[i][j]] == 1) {tot += 3;}
else if (mp[a[i][j]] == 2) {tot++;}
}
cout << tot << ' ';
}
cout << '\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 = 200007;
const int MOD = 1000000007;
void solve() {
int n;
cin >> n;
string s;
cin >> s;
long long tot = 0;
vector<long long> v;
for (int i = 0; i < n; i++) {
if (s[i] == 'L') {
v.push_back((n - 1 - i) - i);
tot += i;
}
else {
v.push_back(i - (n - 1 - i));
tot += n - 1 - i;
}
}
sort(v.begin(), v.end(), greater<int>());
for (int i = 0; i < n; i++) {
if (v[i] > 0) {tot += v[i];}
cout << tot << ' ';
}
cout << '\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...

You can read about 2D prefix sums if you haven't heard of them before.

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
long long a[1005][1005];
long long pref[1005][1005];
void solve()
{
long long n, q;
cin >> n >> q;
for(int i = 0; i <= 1001; i++)
{
for(int j = 0; j <= 1001; j++)
{
a[i][j] = pref[i][j] = 0;
}
}
for(int i = 0; i < n; i++)
{
long long h, w;
cin >> h >> w;
a[h][w]+=h*w;
}
for(int i = 1; i <= 1000; i++)
{
for(int j = 1; j <= 1000; j++)
{
pref[i][j] = pref[i-1][j]+pref[i][j-1]-pref[i-1][j-1]+a[i][j];
}
}
for(int i = 0; i < q; i++)
{
long long hs, ws, hb, wb;
cin >> hs >> ws >> hb >> wb;
cout << pref[hb-1][wb-1]-pref[hb-1][ws]-pref[hs][wb-1]+pref[hs][ws] << endl;
}
}
int main() {
int t = 1;
cin >> t;
while(t--)
{
solve();
}
}
```

Idea: MikeMirzayanov

**Tutorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
const int MAX = 200007;
const int MOD = 1000000007;
const int dx[3] = {-1, 0, 1}, dy[3] = {-1, 0, 1};
void solve() {
int n, m;
cin >> n >> m;
char g[n][m];
int id[n][m];
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
cin >> g[i][j];
id[i][j] = 0;
}
}
int curr = 1;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (g[i][j] == '*') {
vector<pair<int, int>> adjh, adjv;
if (i > 0 && g[i - 1][j] == '*') {
adjh.emplace_back(i - 1, j);
}
if (i < n - 1 && g[i + 1][j] == '*') {
adjh.emplace_back(i + 1, j);
}
if (j > 0 && g[i][j - 1] == '*') {
adjv.emplace_back(i, j - 1);
}
if (j < m - 1 && g[i][j + 1] == '*') {
adjv.emplace_back(i, j + 1);
}
if (adjh.size() == 1 && adjv.size() == 1) {
if (id[i][j] == 0) {id[i][j] = curr;}
else {cout << "NO\n"; return;}
if (id[adjh[0].first][adjh[0].second] == 0) {id[adjh[0].first][adjh[0].second] = curr;}
else {cout << "NO\n"; return;}
if (id[adjv[0].first][adjv[0].second] == 0) {id[adjv[0].first][adjv[0].second] = curr;}
else {cout << "NO\n"; return;}
curr++;
}
else if (adjh.size() > 1 || adjv.size() > 1) {
cout << "NO\n"; return;
}
}
}
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (g[i][j] == '*') {
if (id[i][j] == 0) {cout << "NO\n"; return;}
else {
for (int x = 0; x < 3; x++) {
for (int y = 0; y < 3; y++) {
if (0 <= i + dx[x] && i + dx[x] < n) {
if (0 <= j + dy[y] && j + dy[y] < m) {
if (id[i + dx[x]][j + dy[y]] != id[i][j] && id[i + dx[x]][j + dy[y]] != 0) {
cout << "NO\n"; return;
}
}
}
}
}
}
}
}
}
cout << "YES\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...

**Alternate Tutorial Sketch**

Output the integers from $$$1$$$ to $$$n-3$$$, then $$$2^{29}$$$, $$$2^{30}$$$, and the XOR of those $$$n-1$$$ numbers. Why does it work?

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
void solve()
{
int n;
cin >> n;
int case1 = 0;
int case2 = 0;
for(int i = 0; i < n-2; i++)
{
case1^=i;
case2^=(i+1);
}
long long addLast = ((long long)1<<31)-1;
if(case1 != 0)
{
for(int i = 0; i < n-2; i++)
{
cout << i << " ";
}
case1^=addLast;
cout << addLast << " " << case1 << endl;
}
else
{
for(int i = 1; i <= n-2; i++)
{
cout << i << " ";
}
case2^=addLast;
cout << addLast << " " << case2 << endl;
}
}
int main()
{
int t = 1;
cin >> t;
while(t--)
{
solve();
}
}
```

I didn't even think that Timur is lexicographically in order. What a brilliant solution!

Edit: It isn't, but it's still an interesting solution to think of.

Its not.

It should be Timru

i... totally knew that, good reading comprehension, well done, you passed the test

Problem D can also be solved in O(n) using two pointers

exactly

Can you give your solution here it would help

here if you need explanation feel free to ask

Hello, I saw your solution but I didn't understand this line . What is it for > for (k;k<=n;k++) ans[k]=s;

let's suppose that it was already optimal like RRLL so in the line you mentioned it just sets the remaining indexes (which could not be covered in while loop cuz of being optimal) they will just get the previous value as we will not be changing them...

Let's say the number of characters I needes to change was 8 and n is 10. I need go give an answer for all k such that 1<=k<=n so I just output for the remaining k

I got it. Thank you so much.

You're welcome

I solved it in O(n) aswell, but using prefix array

170271299

Can u pls explain,how did u do it using Prefix arrays

Take a look at the comments in the submission, if that doesn't help let me know once again.

Still i didn't get .M new to prefix array

You can have a look at others two-pointer solution. It's almost similar.

ExplanationHere you can go to this link for prefix array.

Now, for this question in each move you make, you want to maximise the value you can get, so it makes sense to pick either the left-most or right-most direction that has to be changed.

Take this example

12

LRRRLLLRLLRL

Split it into two half's,

LRRRLL LRLLRL

Now, for the left part, you want to make everything R, while for the right part you want everything to be L. (why? i will leave that to you)

For the first part, we will work from left to right, for right part we will work from right to left. (This can be done with two pointer aswell)

Now, whenever you find a value that has to be changed, remove what it was contributing to the sum, and add what it will contribute when changed the direction.

Keep storing them in array. Then make a prefix array.

Now, for each move from 1 to n, you can add their respective moves to the sum.

it seems that we have the same idea

What is two pointer approach... Can you share some resources from where I should learn... It would be a great help

Check Codeforces edu section. You can also read about it on USACO Guide

i do it with queue :)

It can be done too, because there indices that must be changed before others right?

Why is problem A $$$O(n)$$$? You check if $$$n = 5$$$ in $$$O(1)$$$; If it is, then sort the string which in this case will always have length $$$n = 5$$$ so sorting and checking is done in $$$O(1)$$$ (since the length is upper bounded). So overall $$$O(1)$$$, or am I missing something?

I think that's true. Change it, flamestorm.

You check if "T" is in the string, that can be done in $$$O(1)$$$ (since it only has 5 elements). Then you check if "i", also $$$O(1)$$$ And so forth. In total $$$5 \cdot O(1) = O(1)$$$

You have to read the string size of n, so it is O(n)

problem G:

## topic 1

Regardless of your correct answer, when you choose some elements and take their XOR, then the value will be equal to the XOR of the remaining elements. Why?

Sample: $$$(4,2,1,5,0,6,7,3)$$$ is one of answer for $$$n=8$$$, and $$$4 \oplus 0 \oplus 3 = 2 \oplus 1 \oplus 5 \oplus 6 \oplus 7 = 7$$$

AnswerRemember $$$x \oplus x = 0$$$ and $$$a \oplus b = c \leftrightarrow a = b \oplus c$$$ (you can move an element from one side to the other side).

Let's define your answer $$$A$$$. Your answer must satisfy:

$$$A_1 \oplus A_3 \oplus \dots = A_2 \oplus A_4 \oplus \dots$$$

$$$\leftrightarrow A_1 \oplus A_2 \oplus A_3 \oplus A_4 \oplus \dots = 0$$$

$$$\leftrightarrow$$$ (XOR of elements you choose) $$$=$$$ (XOR of the remaining elements)

## topic 2

There exists a randomized solution.

Explanation[1] Take $$$n-1$$$

distinctintegers from $$$[0,2^{31})$$$ and name the set $$$S$$$.[2] Let's define the XOR of $$$n-1$$$ elements $$$x$$$. If $$$S$$$ doesn't contain $$$x$$$, add $$$x$$$ to $$$S$$$ and $$$S$$$ will be the answer (Q. why? A. see topic 1). Otherwise, back to [1].

## topic 3

For $$$3 \le n \le 6$$$, the answers are provided in the sample, so let's use them as a prefix.

After that, add $$$(100,101,102,103,\dots)$$$ until the length of the sequence becomes $$$n$$$. (note that the length of added sequence is a multiple of $$$4$$$ ) Then, you can get a correct answer.

Submission: 170321109

Why?For all integer $$$k \ge 0$$$, $$$(4k) \oplus (4k+1) \oplus (4k+2) \oplus (4k+3) = 0$$$. Let's try to proof yourself!

master piece. i tried to construct a solution similar to topic 3, but unfortunately fell into tons of case work.

however, instead we can take advantage of samples, XD

Thanks alot. Can you share the proof of topic 3 ?

proof with an exampleLet's see the bitwise expression for some examples. For $$$k=37$$$,

The prefix of bitwise expression except the last $$$2$$$ digits are same, so their XOR is $$$0$$$. And the last $$$2$$$ digits are $$$00,01,10,11$$$, then their XOR is also $$$0$$$. So, overall XOR is always $$$0$$$.

A1⊕A3⊕⋯=A2⊕A4⊕…↔A1⊕A2⊕A3⊕A4⊕⋯=0can you swap randomly and still have the equality?

if yes, can you do the same in

X ^ Y < A ^ B -> x ^ B < A ^ Y ?

The latex for $$$2^{30}$$$ is bugged in the tutorial of problem G.

Aside from that, problem G is a really cool problem!

What no {} does to a mf

For anyone who is finding the solution of problem F, hard to understand/code, can take a look at my solution using DFS 170297341

Weak pretests for A , so many solutions got hacked.

Your solution is weaker

Agreed and that's a pity that such a solution passed.

How to write the code for Problem G with the alternate tutorial, I am unable to understand the approach

Now , I got it

Thanks

170326725 Here you are

Simple implementation for problem G

Thanks

Problem F can be done using

Surprise !DFS

:D

(￣︿￣)

I had school so I wasn't able to do the contest, however I've solved them all and I present my solutions. They may or may not be worse quality than the official editorial.

AI did a really really really really messy implementation. The much nicer way that it could have been done is to simply sort the string and check if it's equal to the string you get from sorting Timur. I'm not linking my submission b/c the code will make you lose braincells.

BJust change the 'G's to 'B's and check if the rows are the same. You can change the 'B's to 'G's if you want, but you need to pick one and change throughout the input. 170184897

CCreate a map. m[s] will store the number of strings equal to s in the input. Then for each player, get their score using the map. 170192041

DThe greedy solution of mine is to always prioritize changing people from the left and right rather than the middle as they simply affect more people. I first computed the answer you get with 0 changes and stored that. Then I kept 2 pointers i = 0 and j = n-1 to check the extreme left and right. If the element i was pointing at was an 'L', I would switch it to be an 'R' and updated my answer, vice versa for j. Finally I increased i by 1 and j by -1. What this does is it effectively does all necessary operations in the optimal order. Finally, print the answer. The answer for the ith case is the max of the answer for the ith case and the answer for the i-1th case. These values can easily be stored during the process. 170318148

EUse 2D prefix sums. ps[i][j] will store the total areas of all rectangles with a height <= i and a width <= j. You can easily build this when reading the input. This works b/c the maximum height and width values are 10^3 each. So you only need 10^6 space which is doable. Each query can be solved in O(1) time very easily using the prefix sums. 170329944

FUse DFS or BFS or whatever and find all the connected components of '*'s. If any of these components's size is not equal to 3, there is no answer. Now check if all the components form an L shape. After that, check all pairs of components and make sure no corners intersect. If all the conditions are met, then the answer is YES. 170329757

GNote that if you add ...01 ...10 ...00 and ...11 to a good array, the array will stay good. (Note that the prefix "..." has to be the same for all 4 numbers). Now there are just 4 cases for n: n = 3 + 4k, n = 4 + 4k, n = 5 + 4k, and n = 6 + 4k. The answers for 3, 4, 5, and 6, are given in the input. Then just add the extra elements needed by iterating over that prefix and adding the 00, 01, 10, and 11 suffix respectively. 170329345

Can you please tell me where my submission 170342106 for problem E is failing?

I think your code is quite overcomplicated

tgp07 Excellent editorial, better than official one! Could you please tell me if there is a way to solve E under the same time constraints for tighter dimensions of the rectangle say 1e5 or 1e9 or even higher?

That would be a lot tougher. I'm not immediately sure of any way to solve that.

your G one is really good Thanks for sharing.

I think an improvement for F would be to extend adjacency to include diagonals, i.e., each cell has up to 8 neighbors for the purposes of DFS. Then all we need to do is make sure each connected component has size 3 and forms an L shape.

This way, there is no need to check if different blocks touch by edge or corner, since the DFS would've merged such cases into larger components.

(also, checking whether three cells form an L-shape can be done by simply checking if, for each of the three pairs of cells, the difference between the x- and y-coordinates is at most 1)

This seems smart, have you submitted a solution with this approach?

I had a partially written code, but the contest ended and I didn't bother finishing it up. But your comment encouraged me to complete and submit it, so here it is: 170513958!

My first contest solving more than 1 problem :) I was excited to realize I could use a max heap to solve problem D.

Hey!, can someone help me up solving question F,

My logic for the solution was that for each * there can be only and only 2 '*' neighbors in all 8 directions(if a cell exists there) no more, no less than that.

But it fails on test no. 4, 69th ticket. can someone help me find a test case where this fails. WA link — https://codeforces.com/contest/1722/submission/170290140

Any help would be very appreciated.

`1`

`3 3`

`.*.`

`*.*`

`.*.`

Thanks a lot, how do you come up with such cases, like how do you think of cases where our code might fail, how to develop that? any tips on that would be very helpful.

If it's after the contest, you can simply view the test cases in your submission. Often the input would be truncated, but you can often still find a way to reveal it. Modify your code by adding conditions to specifically check for the problematic test case, and then use it to print the result.

In this case, for example, you can see that in Test #4, the first case has n = m = 3, unlike the earlier Test Suites. So you can write your program such that if the first test case begins with n = m = 3 (you can use scanf for this, if desynced with cin), then you call a different function that reads the first 68 cases while printing absolutely nothing (not even the answers), and then reads the 69th case to print the complete test case details. When submitting, this should pass Tests #1-3, and then print the problematic test case in #4 for you to view.

This can be cumbersome sometimes, but at least it's a generally definite approach to finding the test case that breaks your program. I haven't yet encountered a situation where I was unable to expose the problematic test case this way.

fst

Alternate solution for Problem G:Observe that $$$ a \oplus b$$$ $$$=$$$ $$$\sim a \oplus \sim b. $$$ Now you can construct a sequence $$$a0$$$, $$$\sim a_0,$$$ $$$a_1,$$$ $$$\sim a_1,$$$ $$$a_2,$$$ $$$\sim a_2, \dots$$$

(upto n terms)where $$$n \equiv 0$$$ ($$$mod$$$ $$$4$$$).For $$$n = 4k+1$$$ or $$$4k+2$$$ or $$$4k+3$$$, just take some prefixes of length $$$1$$$ (0), $$$6$$$ (4,1,2,12,3,8), and $$$3$$$ (2, 1,3) respectively from the given testcases itself and the remaining sequence will be of length $$$4k'$$$.

Note that $$$a_0$$$, $$$a_1$$$, $$$\dots$$$ can be taken to be $$$13,$$$ $$$14,$$$ $$$15,$$$ ... as none of the prefixes contain elements $$$>13$$$.

Also, $$$\sim a$$$ represents

1's complimentof $$$a$$$ inverting all 32 bits of $$$+a$$$, although (even the first 20 bits would work considering the constraints on $$$a_i$$$)what is the use of greater() in the sort function of the D problem?

Its a comparator which defines how you want it be sorted. You can read more about it here

In the problem E, trivial $$$O(n q)$$$ solution passes due $$$\mathrm{TL} = 6 \mathrm{s}$$$: 170297542

I wonder how your fastIO works...My BF solution 170407110 could pass only when adding your fastIO code. (And on test 4, it is 15 times faster than simply using untie cin/cout). It seems impossible because IO shouldn't be the bottleneck of this problem. I'm really confused about this.

We need optimize any constant in time complexity of solution. And it is fastest I/O (on Windows) that I have been seen.

In it, I completely abandoned small buffers (usually ~2KiB) that lead to a lot of file operations, and also decided to completely abandon non-system I/O functions due to overhead. Large buffers allow your program to use two file operations during work.

And when

`-O3`

is specified, gcc inlines my function`ReadInt32`

, but no`printf`

and`std::cin`

.I discuss this with some friends and get a weird conclusion. You can see the difference between 170435459 and 170435486. Seems that when we replace the last cin with a function (although we still use cin to read), the compiler will use SIMD to optimize the if statement. Replace function with inline function even macro gets the same result. But just using cin or scanf won't trigger optimization. I have no idea why this happens :(

I solved problem G quite differently from the editorial method utilising the property that XOR of consecutive numbers x and y is always 0 when x is an even number (if you didn't know this, it's easy to see why, by taking a few consecutive numbers and looking at their binary representation).

Let set a and b represent respectively the set of numbers at even positions and the set of numbers at odd positions. Then we break solution into 4 cases depending on the remainder of n on dividing by 4.

Case 1.) n%4 = 0Here both sets have an even number of elements. So we can just keep pairing two consecutive numbers and that will suffice.

Case 2.) n%4 = 1In this case, set a will have 1 more element than set b. In order for them to have the same end result (xor), we can just insert a 0 in set a as 0 is a neutral element. Now we just have to put pairs of consecutive numbers as in above case.

Case 3.) n%4 = 3This case is similar to Case 2 but set a will have an even number of elements and set b will have an odd number of elements. We can make up for the lack of a pairing element in set b by just inserting element 1 (which was to be the result of the XOR of two consecutive elements anyway).

Case 4.) n%4 = 2Here both sets will have odd size. That means they will both need a single extra element that can't be paired. That element can't be the same (that's why minimum n is not 2 in the problem).

We can break any such n into a combination of 3's and 2's, there will be a trio of 3 numbers and all the rest are pairs. Minimum n is 3, so this will fit. We can find 3 such numbers for both the sets s.t. they are all different and their XOR is same. I just looked at the samples which happen to have this test case for n = 6. So I took the numbers from the sample solution

(set a : 2, 3, 4 ; set b : 1, 8, 12), their XOR is 5. The remaining elements can all be pairs of consecutive numbers as described above.

I think some of the cases can probably be combined for a simpler solution. Please let me know in the comments.

Not so elegant and repetitive implementation : 170369765

I solved in the same way

Another way to solve F: L-shapes:

ref: https://codeforces.com/contest/1722/submission/170380611

Note that, a 2x2 grid will contain L shape if and only if there are exactly 3 '*' in it.

1)Firstly for all 2x2 subgrid that contain L-shape, check whether its surrounding has any '*' in it, if so, then the ans is NO. [except for the corner along void position(exactly one such position) of 2x2 grid ,

eg: '#' in below picture [and ^ is void position] ]

...#

.*^.

.**.

....

2)Now the only problem is, there can also be any other shapes than L-shapes. To find that,

In logic1, just whenever current 2x2 subgrid contain L,(ie:exactly 3 '*' in it) , mark those positions as #.

So that all L shapes will be marked, and if there exist any other shape than L shape then it will remain unmarked(ie:remains *).

So, atlast traverse the whole grid to find any such unmarked position,

if so, the ans is NO,

Otherwise YES.

My Solution for F

did the same thing as mentioned in the editorial but is easier to understand.

My solution : 170453262 for problem F

I got wa on F test 3 170393551 can somebody say the problem

Take a look at Ticket 16138 from

CF Stressfor a counter example.Tnx

Is there anywhere I can view solutions in java?

The status page (https://codeforces.com/contest/1722/status) lets you filter by language.

Here is a bit shorter code for F: 170427907

SpoilerObserve that for a valid construction:

1- Each * has exactly three neighboring *

2- Maximum connected chain of * is of length 3

second condition is neccesay to avoid patterns with shapes close to "circular" like these:

Is there any other way to solve #C without using a map?

Use sets, check out my submission here at https://codeforces.com/contest/1722/submission/170192052

Problem E can also be solved with offline queries and binary search on segment tree, which gives O(t*n*logn) that satisfied arbitrary large height and width of the rectangles.

My solution for G:

xor of 2x (even no.) and 2x+1 is always 1.

Case 1: n%4 == 0: A simple solution would be 20,20+2n,21,21+2n,22,22+2n,....

Case 2: n%4 == 1: Just add a 0 at the end of Case 1

Case 3: n%4 == 2: Just add 4 1 2 12 3 8 this sequence at the end of Case 1 (Provided in sample input for n=6)

Case 4: n%4 == 3: Just add 2 1 3 this sequence at the end of Case 1 (Provided in sample input for n=3)

Problem F can be solved easily using 8-directional floodfill

As a fun fact, you can solve G just by printing random numbers and equalizing Xor with the last two elements.

whats the logic behind it

For problem G: After noticing we need n numbers such that their xor is zero, we can use https://oeis.org/A003815 as follows:

My solution to G.

basically xor of x,x+1,x+2,x+3 is 0 and also xor of their alternate indexes. so every n could be writtern in form of 4x,4x+1,4x+2,4x+3. we will be taking avantage of this as an extension

Code ``` vector<vector> arr(4); arr[0]={0,1,2,3}; arr[1]={2,0,4,5,3}; arr[2]={4,1,2,12,3,8}; arr[3]={2,1,3};

```