Codeforces Round #590 (Div. 3) 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

Suddenly, all problems expect A and D were invented by me. The author of A and D is MikeMirzayanov.

Tutorial

Solution

#include <bits/stdc++.h>

using namespace std;

int main() {
#ifdef _DEBUG
	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);
#endif
	
	int q;
	cin >> q;
	for (int i = 0; i < q; ++i) {
		int n;
		cin >> n;
		int sum = 0;
		for (int j = 0; j < n; ++j) {
			int x;
			cin >> x;
			sum += x;
		}
		cout << (sum + n - 1) / n << endl;
	}
	
	return 0;
}

1234B1 - Social Network (easy version)

Tutorial

Solution

#include <bits/stdc++.h>

using namespace std;

int main() {
#ifdef _DEBUG
	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);
#endif
	
	int n, k;
	cin >> n >> k;
	vector<int> ids;
	for (int i = 0; i < n; ++i) {
		int id;
		cin >> id;
		if (find(ids.begin(), ids.end(), id) == ids.end()) {
			if (int(ids.size()) >= k) ids.pop_back();
			ids.insert(ids.begin(), id);
		}
	}
	
	cout << ids.size() << endl;
	for (auto it : ids) cout << it << " ";
	cout << endl;
	
	return 0;
}

1234B2 - Social Network (hard version)

Tutorial

Solution

#include <bits/stdc++.h>

using namespace std;

int main() {
#ifdef _DEBUG
	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);
#endif
	
	int n, k;
	cin >> n >> k;
	
	queue<int> q;
	set<int> vals;
	for (int i = 0; i < n; ++i) {
		int id;
		cin >> id;
		if (!vals.count(id)) {
			if (int(q.size()) >= k) {
				int cur = q.front();
				q.pop();
				vals.erase(cur);
			}
			vals.insert(id);
			q.push(id);
		}
	}
	
	vector<int> res;
	while (!q.empty()) {
		res.push_back(q.front());
		q.pop();
	}
	reverse(res.begin(), res.end());
	cout << res.size() << endl;
	for (auto it : res) cout << it << " ";
	cout << endl;
	
	return 0;
}

1234C - Pipes

Tutorial

Solution

#include <bits/stdc++.h>

using namespace std;

int main() {
#ifdef _DEBUG
	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);
#endif
	
	int q;
	cin >> q;
	for (int i = 0; i < q; ++i) {
		int n;
		string s[2];
		cin >> n >> s[0] >> s[1];
		int row = 0;
		int pos = 0;
		for (pos = 0; pos < n; ++pos) {
			if (s[row][pos] >= '3') {
				if (s[row ^ 1][pos] < '3') {
					break;
				} else {
					row ^= 1;
				}
			}
		}
		if (pos == n && row == 1) {
			cout << "YES" << endl;
		} else {
			cout << "NO" << endl;
		}
	}
	
	return 0;
}

1234D - Distinct Characters Queries

Tutorial

Solution

#include <bits/stdc++.h>

using namespace std;

int main() {
#ifdef _DEBUG
	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);
#endif
	
	string s;
	cin >> s;
	vector<set<int>> poss(26);
	for (int i = 0; i < int(s.size()); ++i) {
		poss[s[i] - 'a'].insert(i);
	}
	
	int q;
	cin >> q;
	for (int i = 0; i < q; ++i) {
		int tp;
		cin >> tp;
		if (tp == 1) {
			int pos;
			char c;
			cin >> pos >> c;
			--pos;
			poss[s[pos] - 'a'].erase(pos);
			s[pos] = c;
			poss[s[pos] - 'a'].insert(pos);
		} else {
			int l, r;
			cin >> l >> r;
			--l, --r;
			int cnt = 0;
			for (int c = 0; c < 26; ++c) {
				auto it = poss[c].lower_bound(l);
				if (it != poss[c].end() && *it <= r) ++cnt;
			}
			cout << cnt << endl;
		}
	}
	
	return 0;
}

1234E - Special Permutations

Tutorial

Solution

#include <bits/stdc++.h>

using namespace std;

int main() {
#ifdef _DEBUG
	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);
#endif
	
	int n, m;
	cin >> n >> m;
	vector<int> a(m);
	for (int i = 0; i < m; ++i) {
		cin >> a[i];
		--a[i];
	}
	
	vector<long long> res(n);
	for (int j = 0; j < m - 1; ++j) {
		res[0] += abs(a[j] - a[j + 1]);
	}
	
	vector<int> cnt(n);
	vector<vector<int>> adj(n);
	for (int i = 0; i < m - 1; ++i) {
		int l = a[i], r = a[i + 1];
		if (l != r) {
			adj[l].push_back(r);
			adj[r].push_back(l);
		}
		if (l > r) swap(l, r);
		if (r - l < 2) continue;
		++cnt[l + 1];
		--cnt[r];
	}
	for (int i = 0; i < n - 1; ++i) {
		cnt[i + 1] += cnt[i];
	}
	
	for (int i = 1; i < n; ++i) {
		res[i] = res[0] - cnt[i];
		for (auto j : adj[i]) {
			res[i] -= abs(i - j);
			res[i] += j + (j < i);
		}
	}
	
	for (int i = 0; i < n; ++i) {
		cout << res[i] << " ";
	}
	cout << endl;
	
	return 0;
}

1234F - Yet Another Substring Reverse

Tutorial

Solution

#include <bits/stdc++.h>

using namespace std;

int main() {
#ifdef _DEBUG
	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);
#endif
	
	string s;
	cin >> s;
	
	vector<int> dp(1 << 20);
	for (int i = 0; i < int(s.size()); ++i) {
		vector<bool> used(20);
		int mask = 0;
		for (int j = 0; i + j < int(s.size()); ++j) {
			if (used[s[i + j] - 'a']) break;
			used[s[i + j] - 'a'] = true;
			mask |= 1 << (s[i + j] - 'a');
			dp[mask] = __builtin_popcount(mask);
		}
	}
	
	for (int mask = 0; mask < (1 << 20); ++mask) {
		for (int pos = 0; pos < 20; ++pos) {
			if ((mask >> pos) & 1) {
				dp[mask] = max(dp[mask], dp[mask ^ (1 << pos)]);
			}
		}
	}
	
	int ans = 0;
	for (int mask = 0; mask < (1 << 20); ++mask) {
		if (dp[mask] == __builtin_popcount(mask)) {
			int comp = ~mask & ((1 << 20) - 1);
			ans = max(ans, dp[mask] + dp[comp]);
		}
	}
	
	cout << ans << endl;
	
	return 0;
}

WA?

#include <bits/stdc++.h>

using namespace std;

int main() {
#ifdef _DEBUG
	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);
#endif

	int tt = clock();
	
	string s;
	cin >> s;
	
	vector<int> dp(1 << 20);
	vector<int> masks;
	for (int i = 0; i < int(s.size()); ++i) {
		vector<bool> used(20);
		int mask = 0;
		for (int j = 0; i + j < int(s.size()); ++j) {
			if (used[s[i + j] - 'a']) break;
			used[s[i + j] - 'a'] = true;
			mask |= 1 << (s[i + j] - 'a');
			masks.push_back(mask);
		}
	}
	sort(masks.begin(), masks.end());
	masks.resize(unique(masks.begin(), masks.end()) - masks.begin());
	sort(masks.begin(), masks.end(), [](int x, int y){
		return __builtin_popcount(x) > __builtin_popcount(y);
	});
	
	int ans = __builtin_popcount(masks[0]);
	for (int i = 0; i < int(masks.size()); ++i) {
		if (clock() - tt > 1900) {
			break;
		}
		for (int j = i + 1; j < int(masks.size()); ++j) {
			if (!(masks[i] & masks[j])) {
				ans = max(ans, __builtin_popcount(masks[i]) + __builtin_popcount(masks[j]));
				break;
			}
		}
	}
	
	cout << ans << endl;
	
	return 0;
}

#include <vector> #include <string> #include <cstring> #include <map> #include <set> #include <algorithm> #include <cstdlib> #include <cmath> #include <unordered_set> using namespace std; const int N = 1e6; int main() { std::srand(std::time(nullptr)); int i = 0; for(; i < 19; ++i) { cout << (char)('a'+(i%19)); } for(; i < N-2; ++i) { cout << (char)('a'+(rand()%19)); } cout << "aat" << endl; }

Comments (109)

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Dsxv

5 years ago, # |

+21

Fenwick appraoch for 1234D : Solution

Implementation similar to: CodeMonk Problem

Awesome contest by the way! :D

→ Reply

Spheniscine

5 years ago, # ^ |

I used bitmasks and segment tree instead: 61684182

MrIo

I also used a segment tree but I used sets instead of bitmasks: 62130226

Apparently it is too slow... Can anyone explain why? by how much? or if it has a worse complexity time than the tutorial or other solutions?

← Rev. 4 →

Sets would be slower by a constant-ish factor (final complexity about $$$O(n \log n \cdot A \log A)$$$), since there are at most 26 distinct letters. This could be significant, as bitwise operations are one of the fastest hardware operations that a computer can do, compared to having to clear and rebuild trees every update.

pyduper

Can you please tell me how you came with the intuition of using Fenwick tree?

← Rev. 2 →

From what I can tell it's the same concept as explained in the editorial, except that he's using a Fenwick tree instead of an ordered set for "does the character exist in this range" queries

+12

Spheniscine Correct!

pyduper I am assuming you know basics of BIT :)
Notice that the question wants us to find if a character exists between the range l to r. ie. for all characters from a to z if their count(l to r) > 0 or not.

Which we can easily do by creating 26 BITs for every alphabet(and basically do what a BIT does)
So the ans for total distinct elements between l and r would be like :

 for(int i = 0 ; i < 26 ; i++){
       int count = query(BIT[i],r) - query(BIT[i],l-1) ;
       ans += count > 0 ;
 }

Updating BITs :
Assume we need to change arr[index] to char ch what we notice here is that we have to decrease frequency in BIT[arr[index]] at index by 1 and increase frequency of BIT[ch] at index by 1
which would be like :

 update(BIT[s[i]-'a'],i,-1) ;
 s[i] = ch ;
 update(BIT[s[i]-'a'],i,1) ;

Also the complexity of the code would be : NlogN + 26*QlogN which is quite feasible :)

Thats it!

Thanks.. I accidentally down voted instead of up voted. I cannot change it now. I am really sorry.

NBN_VN

funny thing that I saw a lots ppl like you, they knew a lots advanced things that I only heard about, like segment tree or fenwick tree or dp on tree, but the most important thing that you and they are the same, just only specialist. So the question I want to know that why you know a lots but you cant do anything gud in official time of a contest, and can I archive candidate master without knowing about segment tree or fenwick tree. I really need to know answer for this question, help me if you can, plz!

Well, it's just been a year since I knew about CP and if you see my rating graph I was not serious at all till april-may(mostly because I was into development at that time) so that makes about 7-8 months.

If you ask me the reason the obvious answer would be lack of practice(I have just solved about 400-500 problems (mostly easy-med) over coding websites all together).

looking at my graph I think I am kind of improving (maybe?)

important My personal thoughts: BIT should not be considered as hard as Segment Tree (according to me a person with 1400+ rating can easily understand and implement BIT).

BIT just requires 8-10 lines of code to implement. Also, I've never actually required to implement a seg tree in questions of rating < 1900 (mostly) but that's just my personal thoughts :D.

← Rev. 10 →

My opinion is that BIT is easier to code, but harder to intuit, while segment trees are the opposite. Segment trees are more powerful (it can solve any problem BIT can, and some problems BIT can't), but BIT tends to be faster constant-wise.

This problem is actually quite instructive about the difference, cause you need 26 sum-BITs, one for each character, but only one OR-segment-tree. Segment trees can work on any monoid (type/set with an identity value and an associative binary operation), but BIT has the additional requirements of the operation being reversible and commutative, hence working on + but not on OR (which is commutative and associative, but not reversible).

AbdAl-mateen

Hi Dsxv

1234D

I tried with an approach similar to the one mentioned in editorial. The only difference between the implementation is: I used array of sets while they used vector of sets. Could you please tell me what's wrong with my solution:

62051065

Ahmad7_7

I think the 'lower_bound' in your code is working in linear time. Just replace it with ( it = contain[i].lower_bound(l) )

Thanks Ahmad7_7, it was the culprit, now I got AC. But can you tell me why those two are not of similar complexity?

You will find the answer here

Saketd3769

3 years ago, # ^ |

Hey!!
What how will we solve the problem D if we are asked the count of distinct characters which occur more than two times in range??
Also can you suggest something where I can read about the BIT topic

Thallium54

← Rev. 3 →

My solution to E: https://codeforces.net/contest/1234/submission/61676673

Just consider how the distance of each pair in the array changes in permutations.

Let the smaller number in the pair be m and the other be n;

The distance increases by m-1 when i=m because m goes to the front and the distance increases 2m-n when i=n because n goes to the front and m shifts 1 backwards. For i from m+1 to n-1 the distance simply decreases 1 because one of the number between them goes to the front.

I hope my explanation makes sense as English is not my first language:D

arunikayadav42

can you please elaborate?

In the first to (m-1)th permutations the distance between m and n is constant. In the (m)th permutation, the number m goes to the front of the permutation so the distance increases by m-1. In the (m+1)th to the (n-1)th permutations, one of the number between m and n goes to the front so number m shift one unit back thus their distance decreases by 1. In the nth permutation the number n goes to the front while number m stays in the position of m+1 so now their distance is m+1-1=m. The distance used to be n-m so the change of the distance is m-(n-m)=2m-n In the rest of permutations, their relative position is back to normal so the distance doesn't change.

Does this make sense?

kelian

Weak Example in F makes me WA on 5........ I want strong example QAQ (sorry to my poor English :D

NoobCoder1998

How can we do C with dfs?

hahake1

-13

You will TLE

MZuenni

← Rev. 5 →

C can be solved by dfs by building a graph like this: example

and check if there is a path.

divyg07

Can you please explain how to do it with dfs with proper code and explanation?

ssvirk

look at this code https://codeforces.net/contest/1234/submission/61647685

what do the various directions specify in your code

direction in which water is entering.

Rockett

I got accepted with dfs, the ideia is basically what MZuenni said:

Firstly you make two types of edges, {1, 2} and {3, 4 ,5 ,6}, and you can see that with edges {1, 2} you can only continue on the line that you are, going to the next column, and with edges {3 ,4 ,5, 6} you go to the next column changing your line (see the picture above).

If there is a path, then there is only one of them, by the "nature" of the edges, so the complexity is O(n).

Here is my submission: 61677625, hope to be clear enough.

It is obvious that because of dynamic programming we need to remove at most one bit from our mask to cover all possible submasks that can update our answer. What is this means ？

Now I understand it , becasue some dp[mask] may not init

huzaifa242

I did the testing for F and found out few Test Cases. Surprisingly the WA code when ran on custom innovation, It gives same answer as that or correct code, But when it's executed locally or on some other online Compiler(other CP sites like codechef), the WA program fails on following Test Cases.

Test Cases Where WA should fail

Though If anyone finds the reason behind the behaviour of WA code in Custom Innovation, do let me know. I'm quite curious about this.

nishant403

+11

That's because CLOCKS_PER_SEC in codeforces is 1000, but in other compiler (I used Dcoder) is 1000000. So clock() gives microseconds. Instead you should set , (clock()-tt) > (1.90)*CLOCKS_PER_SEC).

My code for problem C. https://codeforces.net/contest/1234/submission/61656247 Can anyone help me with that? I considered that the number of 3,4,5,6 type pipes should be odd and if a pipe is of 3,4,5,6 type in one row,it should be of the same type in the corresponding index of the other row.

Leave it,got the error

gaurav_iiitian

I also applied segment tree approach 61686336. But it's giving TLE. Can anyone please tell me where I went wrong?

dhruv_garg

@gaurav_iiitian check my code I used segment tree and it got accepted. My code was easy to understand..

link — https://codeforces.net/contest/1234/submission/61689286

Thanks @dhruv_garg for replying, But I think our implementations are almost same. The only difference being I coded in python. I think codeforces hasn't set time multiplier for python since my code is giving TLE in just 2s which should be 10s for python. @admin look into this please. I'm really disheartened to see this in codeforces and it's my early days here.

sardina

Same story, but I did it after contest by adding some optimizations. Firstly I used bit representation for each char: 'a': 1, 'b': 2, 'c': 4. We have an alphabet with size of 26. So 'z' is 2^25. And if you want to know the sum, you just do bitwise OR (Node1 | Node2). Then I precalculated bit representation for every char and stored it in the dictionary. I added the popcount function (https://stackoverflow.com/questions/407587/python-set-bits-count-popcount) for [l, r] requests. Python got TL on 9th test, but PyPy got AC with 1980 ms. Here is the submission 61730030. Hope it will help.

chaudhary_19

I also got TLE when I implement segment Tree with Python...@admin please have a look at this..

rupav

I used approach of tutorial D. Can someone please explain why lower_bound(s[i].begin(), s[i].end(), l) is getting TLE but same solution with s[i].lower_bound(l) instead, not?

TLE solution: https://codeforces.net/contest/1234/submission/61687000
AC solution: https://codeforces.net/contest/1234/submission/61686913

Only difference is way of using lower_bound.

van_persie9

see this

Thanks a lot. Understood the reason.

ishaanshah

Why does ceil(double(sum)/n) give a wrong answer on but (sum+n-1)/n doesn't.

refer this thread, https://codeforces.net/blog/entry/70185?#comment-547110

darkhorse7

can someone help in finding issue in this code https://codeforces.net/contest/1234/submission/61661254. it's failing on 1 test case on codeforces compiler. but locally it passes

ILoveBitches

-8

Well. I have a deterministic solution for F. vovuh I don't know where your solution will break. But intuition is not coming. My approach:

dp[mask][len] will store maximum starting index for all the submask of mask with length len. then we can simply find the answer for it's complement for all length. link to my solution:

61694665

vovuh

+13

As you can see, my main solution is deterministic too (and it solution is described in the editorial). Moreover, my second (possibly WA) solution is also fully determined (except the order of masks with the same number of ones).

vovuh I don't understand how you are making sure that the mask will not override...

UPD: got it. silly me. Thanks

wol

Damn slow python... Got TL with Segment Tree

I also got TLE when I implement Segment tree in D problem ...I guess they did not give enough time for python

thecodearrow

Use Fast I/O in Python input=sys.stdin.readline

and Submit it using PyPy3

Refer this to get an AC-- https://codeforces.net/contest/1234/submission/61749285

I already use Fast I/O and I tried it in PyPy3 ... but actually I got my mistake there was no problem of time given

For extra masochism, try solving heavy bitmask problems with JavaScript :P

ChinmaySama

Very Nice Explanations and Clear Editorials.

CHHC

I wonder why I got TLE in pronlem D when using Segment Tree......(61656622)

(sorry for my poor English x(

Prateek

My solution got tle in D in system testing and the same solution is accepted on submitting after System testing

https://codeforces.net/contest/1234/submission/61710080

https://codeforces.net/contest/1234/submission/61641496

yougaindra

Your accepted solution took 1996ms which is too close to limit of 2s. difference of few ms in runtime can happen due to various factors.

Thanks

happypeople

I think this is not as simple as before

827852427

After the contest I tried to write B2 with STL deque,but when I use unordered_map it got TLE but map got AC.In my previous conciousness map should be slower than unordered_map,what's wrong with unordered_map in B2?
The unoredered_map version TLE:61711501
The map version AC:61711675

RainAir

You can read this

If you use unordered_map, your time complexity may up to O(n).

(and, sorry to my poor English

akifpathan

SquareRoot Decomposition approach for D: 61667541

prabalsingh24

Need help in this. https://ide.codingblocks.com/s/137727 for problem D this solution is getting WA for test case 2 https://ide.codingblocks.com/s/137726 I just switched line 91 and 92 got AC for all the test cases. Why am i getting this. I created 26 sum query segment trees and for every alphabet if the sum of segment is greater than 0 then there is that distinct character in it and I increased the answer for that query. Why does swapping two line get me WA?

kumaraditya1999

For B2:

Changing from unordered map to map gave AC. The unordered map is faster than the map then why did I get TLE?

Thanks.

https://codeforces.net/blog/entry/62393

I dont think prob E is a 2000-difficulty problem, just like a 1700 or 1800 problems in a div2 contest. I also saw a quite easy task in another div3 before that had 1900-difficulty. I feel weird when there are a lots overrating problems in div3, so I wonder how ppl rate difficulty for problems

Yeah that's true. It's probably because many high rated coders just solve the last problem and leave the contest. And the algorithm thinks they couldn't solve that problem. I am not sure about this. I am just guessing

nilesh8757

Problem rating are decided by relativey how many official contestants mangage to solve it during live contest. Majority of people do not get time to even read last 2 questions hence they could not solve. Ultimately, that leads to over-rating.

Rikikudo

Anyone can explain why using std::set gets AC but gp_hash_table (policy based data structures)(https://codeforces.net/blog/entry/60737) gets TLE? I'm very confused.

AC: 61716953

TLE on test 20: 61712878

dont use hashtable, its overkill in this case, and also hashtable sometime is not a gud choice because of hacks. Consider using a deque instead, this task is all about how to work with a double end queue and nothing else

Yeah, i just used hash table to check if elements existed in queue or not.

But I can not understand why set + deque = AC but gp_hash_table (as I know it's 4x faster than normal std::map) + deque = TLE. It's weird.

navneet.h

I am not sure but what he mean is that it can be hacked. refer this blog

slal

I used a segment tree for problem D. I tried the segment tree that had unordered_map(61658230),set(61650912) and map(61656330) as elements of its nodes but all three showed TLE. Can anybody help me figure out where is the mistake in my code.

rsFalse

As C have only 2 input strings, I zipped them, and used regex with eval, analyzing pairs of symbols (columns). An alternative to if-else blocks. Perl code 61668008 (or shorter without zip: 61669016)

kartik8800

Alternate approach for E, any two consecutive elements of the array x and y contribute abs(x-y) to all f(pi(n)) where i is less than min(x,y) or when i is greater than max(x,y). x and y contribute abs(x-y) -1 to all f(pi(n)) when i lies between x and y.

Using a Segment tree with lazy propagation we can do the above contributions/range updates for all pairs of consecutive elements. Answers will be the ranges i,i for i: 1 to n

Here is the code : https://codeforces.net/contest/1234/submission/61666909

HynDuf

In problem E, I have this stupid question which I cannot figure it out on my own -_- .

In the code provided above:

for (int i = 1; i < n; ++i) {
		res[i] = res[0] - cnt[i];
		for (auto j : adj[i]) {
			res[i] -= abs(i - j);
			res[i] += j + (j < i);
		}
	}

So why res[i] += j + (j < i) here. I thought when we move the element i to position 1 then the cost should be res[i] = j-1 + (j < i), shouldn't it?

Thanks in advance <3

I don't sure my answer what exactly do you mean but if I understood you correctly, that's because $$$i$$$ and $$$j$$$ are $$$0$$$-indexed in the code.

t1war1

He has decremented values of a[i] initially that's why res[i]=j+(j<i).

MubtasimShahriar

Problem D: I implemented editorial's solution in JAVA. For that I had to use TreeSet for lowerbound() function. Here's my JAVA code: 61727406

Mostafijur11

Can somebody explain me problem F's soln? Sorry but i am unable to understand the editorial?

CodeKoreKhai

Need DP idea of Problem C.

John_Cena

check out my submission

Roach00

can anybody explain, what does this mean ?

"Let dp[mask] be the maximum number of ones in some mask that is presented in the given string and it is the submask of mask" , here what does the submask of mask mean?

The submask has fewer or equal number 1s than the mask and the position of all the 1s in submask coincide with the 1s in mask (the mask includes the submask)

For example : 010011 is the submask of 011011

This is my opinion so correct me if I'm wrong <3

can you please explain how the submask is helping in dp , I have understood till the point in editorial that we have to find two non-intersecting mask in the string and the one of the pair with max sum from both mask will be answer, but I can't understand beyond because of not able to understand the role of submask in dp further from here.

if (dp[mask] == __builtin_popcount(mask)) {
			int comp = ~mask & ((1 << 20) - 1);
			ans = max(ans, dp[mask] + dp[comp]);
		}

So the point the submask is important is :

$$$dp[comp]$$$ will be the maximum number of 1s from all the submasks of $$$comp$$$ (include the mask $$$comp$$$ as well)
We don't know if the " $$$comp$$$ " mask is really present in the string or not, so we need to take the max of the submasks so that it is added to the answer.

For example:

$$$mask$$$ is ...100110 ( $$$dp[mask] == __builtin_popcount(mask)$$$ ) --> this ensures the mask is present somewhere in the string) (...bcf...)

--> $$$comp$$$ is ...011001 ($$$comp$$$ may or may not present in the string). So if we don't take the max of the submasks then $$$dp[comp]$$$ will be 0 then. But let suppose there is a substring (which is a submask of $$$comp$$$) ...010001 (...ae...) then $$$dp[comp]$$$ will be >0 (say, 2) and the answer is larger.

P/s: if $$$comp$$$ doesn't intersect with $$$mask$$$ so the submasks of $$$comp$$$ doesn't also

deadguy

Didn't understood the tutorial of E could someone pls elaborate on making prefix arrays.

My approach is somewhat different:

Let $$$F_j$$$ represent the $$$j$$$-th answer value, i.e. $$$f(p_j(n))$$$.

Now consider each consecutive pair in array $$$x$$$, i.e. $$$x_i$$$ and $$$x_{i+1}$$$. We want to consider this pair's "contribution" ($$$|pos(p_j, x_i) - pos(p_j, x_{i+1})|$$$) to the final values in $$$F$$$. Let the smaller of the pair be $$$l$$$ and the greater be $$$r$$$ (if they are equal, their contribution to all values of $$$F$$$ are zero, so we can skip it)

It turns out that we can split these contributions to five regions:

for indices $$$j < l$$$, we want to add $$$r-l$$$ to $$$F_j$$$,
for index $$$j = l$$$, we want to add $$$r-1$$$,
for indices $$$l<j<r$$$, we want to add $$$r-l-1$$$,
for index $$$j = r$$$, we want to add $$$l$$$,
for indices $$$j > r$$$, we want to add $$$r-l$$$.

Now the problem reduces to the "Array Manipulation" problem from HackerRank, i.e. we need to find a way to efficiently add values to entire ranges of indices. It turns out that there's a simple solution: instead of working on the final array $$$F$$$, we work on another array $$$D$$$ instead, where we want the eventual value of $$$D_i$$$ to equal $$$F_i - F_{i-1}$$$. We can then extract $$$F$$$ from $$$D$$$'s prefix sum.

So if we want to add e.g. value $$$x$$$ to $$$F_{a..b}$$$, simply add $$$x$$$ to $$$D_a$$$, and subtract $$$x$$$ from $$$D_{b+1}$$$.

Sample code (Kotlin): 61688222

How did you come up with those five contributions ?

Edit: Nevermind, that's simple observaton. Got it.

Suraj_IOI

Problem A can also be solved using binary search :: Solution : https://codeforces.net/contest/1234/submission/61633785

Ebiarat

but its not need

applemelon

【Problem E】 I did not understand what tutorial says. please tell me what cnt[min(x[j],x[j+1])+1] and cnt[max(x[j],x[j+1])] stands for.

Because all the number from $$$min(x[j],x[j+1])+1$$$ to $$$max(x[j],x[j+1])-1$$$ will have their presence increase by 1 so $$$cnt[min(x[j],x[j+1])+1]++$$$ and $$$cnt[max(x[j],x[j+1])]--$$$. From the number $$$max(x[j],x[j+1])$$$ till on doesn't increase so we decrease 1 at $$$cnt[max(x[j],x[j+1])]$$$

majorro

Has someone solved D with a segment tree with bitmasks, or am I alone dumb?

If you mean problem D, yes I did it that way too (after contest time though, cause the contest time wasn't convenient for me, but I solved it blind)

Lol, I forgot write problem which I meant, anyway thx for answer

mtewathia99

F. Yet Another Substring Reverse ( with explanation )

Check out my solution for more detailed explanation : 61824886

Cyber_Wizard

Thanks a lot :) Great Explanation!

I_Love_LeNguyenTrucQuynh

what is __builtin_popcount??? i don't know it

pulkit1411

it counts the set bit in a number

ohh thanks very much you are so nice

F is beautiful problem

tiany7

I have a question about pipe. In the context we could reasonably assume that the we are searching in 0th pipe and 1st pipe. If we encounter 3,4,5,6, we will make cur = !cur or cur ^= 1 The logic is simple, but when I tried to use an array of dimension 2 x n, the test 2 is always failed. Then I switched to string array dimension of 2, by the same algorithm, my answer was accepted. What is the difference?

EulerLee

I guess I find a counter test case you mentioned in Tutorial-1234F.

The answer should be 20, but may got 19 when using WA?-solution. counter test case like this

Shofiqur

Problem A should had to be explained more clearly and easily with notes

sultan_01

4 years ago, # |

Did Problem E, With segment tree lazy propagation Solution 88938957. For detailed explanation comment below.

Satish_S

I loved the question D..

nonrice

2 years ago, # |

D can be solved in a very straightforward way with Segment Tree and bitmasks. Each node should contain a bitmask detailing which letters are present in it's subtree. During calculation, nodes can be combined using bitwise OR, and the final result can be read with __builtin_popcount().

Mustafiz

21 month(s) ago, # |

i solved problem F using SOS dp. it is my first problem where i used SOS dp when solveing problem randomly.

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