Codeforces Round #817 (Div. 4) Editorial

#	User	Rating
1	tourist	4009
2	jiangly	3831
3	Radewoosh	3646
4	jqdai0815	3620
4	Benq	3620
6	orzdevinwang	3529
7	ecnerwala	3446
8	Um_nik	3396
9	gamegame	3386
10	ksun48	3373

#	User	Contrib.
1	cry	164
1	maomao90	164
3	Um_nik	163
4	atcoder_official	160
5	-is-this-fft-	158
6	awoo	157
7	adamant	156
8	TheScrasse	154
8	nor	154
10	Dominater069	153

Thanks for participating!

Tutorial

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;
    }
}

1722B - Colourblindness

Idea: flamestorm

Tutorial

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();
}

1722C - Word Game

Idea: flamestorm

Tutorial

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();
}

1722D - Line

Idea: flamestorm

Tutorial

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();
}

1722E - Counting Rectangles

Idea: mesanu

Tutorial

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();
    }
}

1722F - L-shapes

Idea: MikeMirzayanov

Tutorial

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();
}

1722G - Even-Odd XOR

Idea: mesanu

Tutorial

Alternate Tutorial Sketch

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();
    }
}

void solve() { int n; cin >> n; int xorr = 0; vector<int> ans; for (int i = 1; i <= n - 2; ++i) { ans.push_back(i); xorr ^= i; } if (xorr != 0) { ans.push_back(1 << 30); ans.push_back((xorr) ^ (1 << 30)); } else { /* The maximum xor of all the elements from 1 to n-2 or n-3 will be less than 2**21. (So, you can choose any power of 2 from [21, 30) and replace n-2 with it) If we have XOR([1...n-2]) as zero, we can remove (n-2) and can push (1<<21) And now the XOR([1....n-3, (1<<21)]) will be some number having the 21st as set Now, we just need two more nos, that can be (1<<30) and XOR([1...n-3, (1<<21)])) ^ (1<<30) */ xorr ^= (n - 2); ans.pop_back(); ans.push_back(1 << 21); xorr ^= (1 << 21); ans.push_back(1 << 30); ans.push_back(xorr ^ (1 << 30)); } for (auto &i : ans) { cout << i << " "; } cout << nline; }

if n = 4k+1: print(0, 2,3,...,n); # that is print zero instead of 1 :) if n = 4k+3: print(1,2,3,...,n); # already 0 if n = 4k+4: print(0,1,...,n-1); # that is converted to the 4k+3 case if n = 4k+2: print(*[0,3,5],*[7,8,...,n+3]) # consider xor of (0,1,...,n+3). It will be 1. Now remove 1,2,4,6. We get n numbers with zero-xor.

Comments (97)

Show archived | Write comment?

CaveVenom1

2 years ago, # |

← Rev. 2 →

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.

→ Reply

Naomi_Russell

2 years ago, # ^ |

Its not.

FrozenKandy

It should be Timru

+29

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

Abito

+15

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

sandry24

exactly

imMortalSuv

Can you give your solution here it would help

here if you need explanation feel free to ask

iMarwan

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

Ghulam_Junaid

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

SilverWing05

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

170271299

hemanth.508136

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.

Explanation

QDJenny

12 months ago, # ^ |

it seems that we have the same idea

ibtesam_ansari

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

Snapper_001

i do it with queue :)

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

ealjkl

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?

vitosevskich

← Rev. 3 →

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 $$$5 \cdot O(1) = O(1)$$$

kizaru

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

physics0523

+41

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 $$$4 \oplus 0 \oplus 3 = 2 \oplus 1 \oplus 5 \oplus 6 \oplus 7 = 7$$$

Answer

topic 2

There exists a randomized solution.

topic 3

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

$$$n \equiv 0 \mod 4$$$ then prefix = $$$(2,1,3,0)$$$
$$$n \equiv 1 \mod 4$$$ then prefix = $$$(2,0,4,5,3)$$$
$$$n \equiv 2 \mod 4$$$ then prefix = $$$(4,1,2,12,3,8)$$$
$$$n \equiv 3 \mod 4$$$ then prefix = $$$(2,1,3)$$$

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?

UUUnmei

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

islam231

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

proof with an example

ISeeDS-V2.0

A1⊕A3⊕⋯=A2⊕A4⊕… ↔A1⊕A2⊕A3⊕A4⊕⋯=0

can you swap randomly and still have the equality?

if yes, can you do the same in

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

willy108

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

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

IAmTheHazard

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

Numinous

+18

Weak pretests for A , so many solutions got hacked.

MoHafez11

Your solution is weaker

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

yadavDipesh

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

Now , I got it

Thanks

woonder

170326725 Here you are

hotsonhonet

-9

Simple implementation for problem G

-6

Problem F can be done using

Surprise !

(￣︿￣)

tgp07

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.

practice_52

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

I think your code is quite overcomplicated

adsolve

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.

aditya_01

your G one is really good Thanks for sharing.

Andalus

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!

tn757

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

aprilharshit

← Rev. 6 →

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.net/contest/1722/submission/170290140

Any help would be very appreciated.

shadow9236

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.

OMT_TaiLung

3 weeks ago, # ^ |

Can u share how to write the fuction how to out the input of the test

Let's consider your latest attempt at Problem 1722F: L-Shapes. Your latest submission is 286022069, where you failed Test Suite #2, and the feedback says that the 50th token does not match. Since the output is just one line of YES/NO per test case, this means that the 50th test case is incorrect. Now, the objective is to find out what is the 50th test case in Test Suite #2.

To do this, we can first detect Test Suite #2 by checking if the number of test cases is exactly 100. If not, then the code should run as normal and pass Test Suite #1. But if it is exactly 100, then we are in Test Suite #2. From there, you can simply read the first 49 test cases without printing anything, and then read the 50th test case and print only the input for the 50th test case. I have done this (using your submission) to construct the following submission: 286061011; it's the same code as yours, except with two new functions, burn and reveal, and a check in the main function where if the number of test cases is 100, then burn the first 49 cases and reveal the 50th case.

Of course, the submission verdict is Wrong Answer, but you can view the output that the submission produced, which is the 50th test case:

2 3
.**
*..

You can test and see that your code does, indeed, output YES, when the answer should be NO.

Oh tks u so much

Can u send me your code :(((, I don't understand :(

Oh sorryy I see it

Ghieath

fst

faizanhussain2310

Is it only Me or anybody else also noticed that F was comparatively easy E for those who till now havent encountered problem based on 2-D PREFIX Sum . I am feeling very frustrated as I spent more than one hour on E and also during the contest F have least successful submission so I just dont focused on F

KFPanda

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 compliment of $$$a$$$ inverting all 32 bits of $$$+a$$$, although (even the first 20 bits would work considering the constraints on $$$a_i$$$)

KumarPrashant

what is the use of greater() in the sort function of the D problem?

adityagamer

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

Cuellius

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

LinkinPony

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 :(

vantaablackk

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 = 0

Case 2.) n%4 = 1

Case 3.) n%4 = 3

Case 4.) n%4 = 2

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

beyondpluto

Another way to solve F: L-shapes:

ref: https://codeforces.net/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.

Sharan_Swaroop

My Solution for F

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

← Rev. 4 →

My solution : 170453262 for problem F

Mreza5760

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

variety-jones

Take a look at Ticket 16138 from CF Stress for a counter example.

Tnx

Anish111

Is there anywhere I can view solutions in java?

robostac

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

Here is a bit shorter code for F: 170427907

Spoiler

Observe 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:

.*.
*.*
.*.

.**.
*..*
*..*
.**.

pirate_hunter_zoro

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

secret-void

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

ImplementingForces

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.

priyesh_coder

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)

AksLolCoding

14 months ago, # |

Problem F can be solved easily using 8-directional floodfill

Cogegallinas

11 months ago, # |

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

Ylandolsi

8 months ago, # ^ |

whats the logic behind it

sachin_kumar1

3 months ago, # ^ |

because if xor of odd position is same as even position then xor of whole array will be zero so we need to create an array such that xor of all element is zero

iedfa

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

KingRish_30

9 months ago, # |

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};

ll n,x=16;
cin>>n;

ll val=n%4,extra=n-arr[val].size();

for(ll i=0;i<(ll)arr[val].size();i++){
    cout<<arr[val][i]<<" ";
}
while(extra--){
    cout<<x<<' ';
    x++;
}
cout<<nl;

```

5 months ago, # |

E :solution with 1d prefix sum https://codeforces.net/contest/1722/submission/265651414

thatchemguy

In fact in G , we can choose any number from [18,30] as the lastbutone'th number as because the XOR of the (n-2) elements would atmost be (n+1) {by the consecutive XOR property of elements} and as elements range from 2*10^5 which is approximately < 2^18, but n+1 is less than 2^18 so we can use anything from 18 & beyond. Hope it helps

a_gt3_rs

4 months ago, # |

E can be solved in $$$O((n+q)\log(max(h_i))\log(max(w_i)))$$$ with the same amount of memory (if you choose the container as nested hash tables) or $$$O(max(h_i)max(w_i))$$$ (if you choose arrays which are faster) if you used 2D BITs. The solution is 2x faster than the normal solution: https://codeforces.net/contest/1722/submission/267849082, https://codeforces.net/contest/1722/submission/267849782.

flamestorm's blog

1722A - Spell Check

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