Codeforces Round #842 (Div. 2) Editorial

№	Пользователь	Рейтинг
1	tourist	4009
2	jiangly	3823
3	Benq	3738
4	Radewoosh	3633
5	jqdai0815	3620
6	orzdevinwang	3529
7	ecnerwala	3446
8	Um_nik	3396
9	ksun48	3390
10	gamegame	3386

№	Пользователь	Вклад
1	cry	167
2	Um_nik	163
3	maomao90	162
3	atcoder_official	162
5	adamant	159
6	-is-this-fft-	158
7	awoo	157
8	TheScrasse	154
9	Dominater069	153
9	nor	153

Hope you liked the problems!

(from thanhchauns2) Before the round starts

Tester	A	B	C	D	E	F
xiaoziya	900	1000	1400	1800	2100	3000
feecle6418	900	1200	1400	1700	2200	2900
valeriu	800	1100	1550	1900	2200	-
LastDance	800	1200	-	2000	2200	-
neko_nyaaaaaaaaaaaaaaaaa	800	-	-	2100	2400	3000
tibinyte	800	1100	1550	1800	2100	2550

Hints

Tutorial

Solution

answer = [print(int(input()) - 1) for testcase in range(int(input()))]

Feedback

1768B - Быстрая сортировка

Author: Vladithur Preparation: Vladithur and Alexdat2000

Hints

Tutorial

Solution

#include <bits/stdc++.h>

#define all(x) (x).begin(), (x).end()
#define allr(x) (x).rbegin(), (x).rend()
#define gsize(x) (int)((x).size())

const char nl = '\n';
typedef long long ll;
typedef long double ld;

using namespace std;

void solve() {
    int n, k;
    cin >> n >> k;
    vector<int> p(n);
    for (int i = 0; i < n; i++) cin >> p[i];
    
    int c_v = 1;
    for (int i = 0; i < n; i++) {
        if (p[i] == c_v) c_v++;
    }
    
    cout << (n  - c_v + k) / k  << nl;
}

int main() {
	ios::sync_with_stdio(0); cin.tie(0);
	
	int T;
	cin >> T;
	while (T--) solve();
}

Feedback

1768C - Поэлементное ограничение

Author: thanhchauns2

Hints

Tutorial

Solution

#include <bits/stdc++.h>

using namespace std;

const int N = 200005;
int n;
int a[N], b[N], c[N], ra[N], rb[N];

void out()
{
	for (int i = 0; i < n; i++)
	{
		cout << a[i] << ' ';
	}
	cout << '\n';
	
	for (int i = 0; i < n; i++)
	{
		cout << b[i] << ' ';
	}
	cout << '\n';
}

void solve()
{
	cin >> n;
	vector<pair<int, int> > V;
	for (int i = 0; i < n; i++)
	{
		cin >> c[i];
		a[i] = b[i] = 0;
		ra[i + 1] = rb[i + 1] = 1;
		V.push_back(make_pair(c[i], i));
	}
	sort(V.rbegin(), V.rend());
	
	for (int i = 0; i < n; i++)
	{
		int k = V[i].second;
		if (ra[c[k]] == 1) a[k] = c[k], ra[c[k]]--;
		else b[k] = c[k], rb[c[k]]--;
	}
	
	int r1 = n, r2 = n;
	for (int i = 0; i < n; i++)
	{
		int k = V[i].second;
		if (a[k] == 0)
		{
			while (ra[r1] == 0) r1--;
			ra[r1]--;
			if (r1 > b[k])
			{
			    cout << "NO" << '\n';
			    return;
			}
			a[k] = r1--;
		}
		else
		{
			while (rb[r2] == 0) r2--;
			rb[r2]--;
			if (r2 > a[k])
			{
			    cout << "NO" << '\n';
			    return;
			}
			b[k] = r2--;
		}
	}
	for (int i = 1; i <= n; i++)
	{
		if (ra[i] != 0 || rb[i] != 0)
		{
			cout << "NO" << '\n';
			return;
		}
	}
	cout << "YES" << '\n';
	out();
}

int main(int argc, char* argv[])
{
	ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
	int t;
	cin >> t;
	while(t--)
		solve();
}

Yet another better solution

There is actually a $$$O(n)$$$ solution to this problem.

Fresh meme from SPyofgame to the intended solution of the author

Step 1

Step 2

Step 3

Step 4

Bonus

Full Code

#include <iostream>
#include <vector>

using namespace std;

int query()
{
    /// Input number of element
    int n;
    cin >> n;

    /// Input the array a[1..n]
    vector<int> a(n + 1);
    for (int i = 1; i <= n; ++i)
        cin >> a[i];

    /// Storing position of each a[i]
    vector<vector<int> > b(n + 1);
    for (int i = 1; i <= n; ++i) {
        b[a[i]].push_back(i);

        /// max(p[i], q[i]) = a[i] so either p[i] = a[i] or/and q[i] = a[i]
        /// so if the number appear the third time or more, then "NO"
        if (b[a[i]].size() >= 3) {
            cout << "NO\n";
            return 0;
        }
    }

    /// Initialize permutation p[1..n], q[1..n]
    vector<int> p(n + 1, -1), q(n + 1, -1);

    /// Initialize permutation position, fp[p[i]] = i, fq[q[i]] = i
    vector<int> fp(n + 1, -1), fq(n + 1, -1);
    for (int x = n; x >= 1; --x) {
        for (int i : b[x]) {
            /// Because of max(), we must save up the larger value
            /// So we assign p[i] or q[i] by x, one by one from x large -> small
                 if (fp[x] == -1) p[fp[x] = i] = x;
            else if (fq[x] == -1) q[fq[x] = i] = x;
        }
    }

    for (int x = n, vp = n, vq = n; x >= 1; --x) {
        for (int i : b[x]) {
            /// Assign the remaining integers
            while (fp[vp] != -1) --vp;
            while (fq[vq] != -1) --vq;
            if (p[i] == -1 && vp > 0) p[fp[vp] = i] = vp;
            if (q[i] == -1 && vq > 0) q[fq[vq] = i] = vq;
        }
    }

    for (int i = 1; i <= n; ++i) {
        if (max(p[i], q[i]) != a[i]) {
            /// Statement condition is not satisfied
            cout << "NO\n";
            return 0;
        }
    }

    /// Output the answer
    cout << "YES\n";
    for (int i = 1; i <= n; ++i) cout << p[i] << " "; cout << "\n";
    for (int i = 1; i <= n; ++i) cout << q[i] << " "; cout << "\n";
    return 0;
}



signed main()
{
    ios::sync_with_stdio(NULL);
    cin.tie(NULL);



    int q = 1; /// If there is no multiquery
    cin >> q;  /// then comment this

    while (q-->0)
    {
        /// For each query
        query();


    }



    return 0;
}

Feedback

1768D - Удачная перестановка

Author: Vladithur Preparation: Vladithur and Alexdat2000

Hints

Tutorial

Solution

#include <bits/stdc++.h>

#define all(x) (x).begin(), (x).end()
#define allr(x) (x).rbegin(), (x).rend()
#define gsize(x) (int)((x).size())

const char nl = '\n';
typedef long long ll;
typedef long double ld;

using namespace std;

void solve() {
	int n;
	cin >> n;
	vector<int> p(n);
	for (int i = 0; i < n; i++) {
		cin >> p[i]; p[i]--;
	}
	
	int ind = 1, ans = 0;
	vector<int> comp(n, 0);
	for (int i = 0; i < n; i++) {
		if (comp[i]) continue;
		{
			int v = i;
			while (comp[v] == 0) {
				comp[v] = ind;
				v = p[v];
				ans++;
			}
			
			ind++; ans--;
		}
	}
	
	for (int i = 0; i < n - 1; i++) {
		if (comp[i] == comp[i + 1]) {
			cout << ans - 1 << nl;
			return;
		}
	}
	cout << ans + 1 << nl;
}

int main() {
	ios::sync_with_stdio(0); cin.tie(0);
	
	int T;
	cin >> T;
	while (T--) solve();
}

Feedback

1768E - Частичная сортировка

Author: thanhchauns2

Hints

Tutorial

1768E- Partial Sorting

We need at most $$$3$$$ operations to sort the permutation: $$$1 -> 2 -> 1$$$

For $$$f(p) = 0$$$, there is only one case: the initially sorted permutation.

Calculate

For $$$f(p) \leq 1$$$, this scenario appears when the first $$$n$$$ numbers or the last $$$n$$$ numbers are in the right places.

Calculate

For $$$f(p) \leq 2$$$, this scenario appears when the smallest $$$n$$$ elements' positions are in range $$$[1, 2n]$$$, or the largest $$$n$$$ numbers' positions are in range $$$[n + 1, 3n]$$$.

Calculate

If the smallest $$$n$$$ elements are all in position from $$$1$$$ to $$$2n$$$, then:

There are $$$C_{2n}^{n}$$$ ways to choose $$$n$$$ positions for these numbers.
For each way to choose these positions, there are $$$n!$$$ ways to choose the position for the smallest $$$n$$$ numbers, and $$$2n!$$$ ways for the rest.
The total number of valid permuations are: $$$C_{2n}^{n} \times n! \times 2n!$$$

If the largest $$$n$$$ elements are all in position from $$$n + 1$$$ to $$$3n$$$, we do the same calculation.

Intersection: intersection appears when the first $$$n$$$ numbers are all in range $$$[1, 2n]$$$ and the last $$$n$$$ numbers are all in range $$$[n + 1, 3n]$$$.

Let intersection = 0

Sketch and calculation in details

The number of intersections will be equal to: $$$ \sum_{i = 0}^{n} C_{n}^{n - i} \times C_{n}^{i} \times C_{2n - i}^{n} \times n! \times n! \times n! $$$

So, the answer will be $$$2 \times C_{2n}^{n} \times n! \times 2n! - \sum_{i = 0}^{n} C_{n}^{n - i} \times C_{n}^{i} \times C_{2n - i}^{n} \times n! \times n! \times n!$$$

For $$$f(p) \leq 3$$$, it will be the count of all valid permutations.

Calculate

Time complexity: $$$\mathcal{O}(n)$$$

Solution

#include <bits/stdc++.h>

using namespace std;

long long n, M;

long long frac[3000005], inv[3000005];

long long powermod(long long a, long long b, long long m)
{
	if (b == 0) return 1;
	unsigned long long k = powermod(a, b / 2, m);
	k = k * k;
	k %= m;
	if (b & 1) k = (k * a) % m;
	return k;
}

void Ready()
{
	frac[0] = 1;
	inv[0] = 1;
	for (int i = 1; i <= 3000000; i++)
	{
		frac[i] = (frac[i - 1] * i) % M;
	}
	
	inv[3000000] = powermod(frac[3000000], M - 2, M);
	
	for (int i = 3000000; i > 0; i--)
	{
		inv[i - 1] = (inv[i] * i) % M;
	}
}

long long C(long long n, long long k)
{
	return ((frac[n] * inv[k]) % M * inv[n - k]) % M;
}

int main()
{
	cin >> n >> M;
	Ready();
	long long ans[4]{};
	
	// X = 0
	
	ans[0] = 1;
	
	// X = 1
	
	ans[1] = 2 * frac[2 * n] - frac[n] - ans[0] + M + M;
	ans[1] %= M;
	
	// X = 2
	
	ans[2] = frac[2 * n];
	ans[2] = ans[2] * C(2 *n, n) % M;
	ans[2] = ans[2] * frac[n] % M;
	ans[2] = ans[2] * 2 % M;
	
	for (int i = 0; i <= n; i++)
	{
		int sub = C(n, i);
		sub = sub * C(n, n - i) % M;
		sub = sub * C(2 * n - i, n) % M;
		sub = sub * frac[n] % M;
		sub = sub * frac[n] % M;
		sub = sub * frac[n] % M;
		ans[2] = (ans[2] - sub + M) % M;
	}
	ans[2] = (ans[2] - ans[1] + M) % M;
	ans[2] = (ans[2] - ans[0] + M) % M;
	
	// X = 3
	
	ans[3] = frac[3 * n];
	ans[3] = (ans[3] - ans[2] + M) % M;
	ans[3] = (ans[3] - ans[1] + M) % M;
	ans[3] = (ans[3] - ans[0] + M) % M;
	
	long long answer = ans[1] + 2 * ans[2] + 3 * ans[3];
	answer %= M;
	
	cout << answer << endl;
}

Feedback

1768F - Великолепный прыжок

Author: Vladithur Preparation: Vladithur and Alexdat2000

Hints

Tutorial

Solution

#include <bits/stdc++.h>

#define all(x) (x).begin(), (x).end()
#define allr(x) (x).rbegin(), (x).rend()
#define gsize(x) (int)((x).size())

const char nl = '\n';
typedef long long ll;
typedef long double ld;

using namespace std;

const int K = 650;

int main() {
	ios::sync_with_stdio(0); cin.tie(0);
	
	int n;
	cin >> n;
	vector<ll> a(n);
	for (int i = 0; i < n; i++) cin >> a[i];
	
	vector<ll> dp(n, 1e18);
	dp[0] = 0;
	
	vector<int> st = {0};
	for (int i = 1; i < n; i++) {
		// Case 1
		{
			ll mi = a[i];
			for (int j = i - 1; j >= max(0, i - K); j--) {
				mi = min(mi, a[j]);
				dp[i] = min(dp[i], dp[j] + mi * (i - j) * (i - j));
			}
		}
		
		// Case 2
		if (a[i] <= K) {	
			for (int j = i - 1; j >= 0; j--) {
				dp[i] = min(dp[i], dp[j] + a[i] * (i - j) * (i - j));
				if (a[j] <= a[i]) break;
			}
		}
		
		// Case 3 
		{
			for (int j: st) {
				dp[i] = min(dp[i], dp[j] + a[j] * (i - j) * (i - j));
			}
			
			if (a[i] <= K) {
				while (!st.empty() && a[st.back()] >= a[i]) {
					st.pop_back();
				}
				st.push_back(i);
			}
		}
	}
	
	for (int i = 0; i < n; i++) cout << dp[i] << ' ';
	cout << endl;
}

Shorter solution (tfg)

#include <iostream>
#include <vector>
#include <chrono>
#include <random>
#include <cassert>
 
std::mt19937 rng((int) std::chrono::steady_clock::now().time_since_epoch().count());
 
int main() {
    std::ios_base::sync_with_stdio(false); std::cin.tie(NULL);
    int n;
    std::cin >> n;
    std::vector<int> a(n);
    for(int i = 0; i < n; i++) {
        std::cin >> a[i];
    }
    std::vector<long long> dp(n, 1e18);
    dp[0] = 0;
    for(int i = 0; i < n; i++) {
        int dist = n / a[i] + 1;
        // take from behind
        for(int j = i-1; j >= 0 && i-j <= dist; j--) {
            dp[i] = std::min(dp[i], dp[j] + (long long) a[i] * (i - j) * (i - j));
            if(a[j] <= a[i]) break;
        }
        // propagate forward
        for(int j = i+1; j < n && j-i <= dist; j++) {
            dp[j] = std::min(dp[j], dp[i] + (long long) a[i] * (i - j) * (i - j));
            if(a[j] <= a[i]) break;
        }
        std::cout << dp[i] << (i + 1 == n ? '\n' : ' ');
    }
}

Feedback

#include <bits/stdc++.h> using namespace std; typedef long long ll; void solve() { ll n; cin >> n; vector<ll>v(n); for (int i = 0; i < n; i++) cin >> v[i]; vector<ll>col(n); ll c = 1, comp = 0, pos = 0; for (int i = 0; i < n; i++) { if (col[i]) continue; pos = i; while (col[pos] == 0) { col[pos] = c; pos = v[pos] - 1; } comp++; c++; } ll ans = n - comp; for (int i = 0; i < n - 1; i++) { if (col[i] == col[i + 1]) { ans--; break; } } if (ans == n - comp) ans++; cout << ans << endl; } int main() { ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0); int t; cin >> t; while (t--) { solve(); } return 0; }

Комментарии (148)

Показать архивные | Написать комментарий?

thanhchauns2

23 месяца назад, # |

+100

From SPyofgame during the contest:

→ Ответить

__Tony_Sama__

23 месяца назад, # ^ |

← Rev. 2 →

yeah, bro

YocyCraft

First time ranked top100 in div2. Hope for no FST!

Update: It's frustrating that I always get downvoted and don't know the reason

div4only

Why so many downvotes? This is bully!

vasuganesha1

7 месяцев назад, # ^ |

Giving you Downvote to motivate your frustration to frustrate you more....

PepeTheBuilder_74

+88

Permutation Forces !

daifuku

Delivered so fast..

mangoDD

+33

2023 should not be a permutation

kriepiekrollie

love the fast editorial

kaifkhan.

I got the idea for C but not able to implement it. :-)..

nikhilsaini546

mee too

utsav_upadhyay

← Rev. 3 →

-21

Can anybody help me to figure why this code is not accepted??? https://mirror.codeforces.com/contest/1768/submission/188117185

Neaj_Morshad_101

Do not paste the code in the comment.

ok thanks to indicate I have removed the code

soul_hunter0807

Write codes in spoiler or only provide the link as you have provided here, as it becomes quite difficult to read this way

ok I will see to it from next time

Although pasting code directly is awful, I've checked your code in the link. In fact, the most part of your code is right except one point: when doing

m1[*it1]=*it2;

you need to check if (*it1>*it2). If not, there's no solution because max(p[i],q[i]) will be *it2.

ok!!!!!!!! Right I will think harder next time to not miss something like this

HappyIvan

+43

You know what it feels like when you finish E when there're only 10 sec left but fail to submit the code before the last second?

Swishy123

dude same, i was literally about to finish it, rip 2250

SPyofgame

Not my first time actually :3

Sorry, never solved D.:(

d_sqaure

Amazing , was very curious for fast tutorial especially for Problem C.

i_pranavmehta

Hi guys! although the editorial they've provided is great! still if you want a video format editorial you can find it here

v0s7er

PermutationForces?)

chromate00

+12

"it took me 2 minutes to actually prove the solution"

and it took me probably a minute and a half ^{(may be inaccurate)} to find out that $$$k! + (k-1)! = (k-1)!(k+1)$$$ and the rest of the proof is just factoradic blah blah

rliu4936

i submitted 6 unsuccessful attempts so i lost 300 points in the contest? if there is a maximum amount u can lose, how much is it?

+11

No matter how many times you submit for a problem, you get at least 30% of points if you solved it finally.

vitosevskich

If you have wrong submission for problem $$$X$$$ you will get 50 points less then you usually would on that problem.

If you have few wrong submissions on a problem you didn't solve, you won't lose any points.

isheoran

Why so tight TL on E.

practiceCpp

You can calculate the inverses of factorials more efficiently by calculating them from n to 0.

OK! thanks

+13

Actually there's O(n) solution for C. We don't need to do any sorting. We can just count the occurence of 1-n and let occ[i]= the times i appeared in array a[]. if there's some occ[i]>2 there's no answer. Then we build two list more[] and less[]. We iterate for i from 1 to n, if occ[i]==0 we add it to less[], if occ[i]==2 we add it to more[]. since 0<=occ[i]<=2, the size of more[] and less[] will be equal. then we iterate for more[i] and less[i] reversely, if some more[i]<less[i] there's no answer. If there's no such i, we can construct p and q: for each pair of more[i] and less[i], let's call they M and L, and j = the smaller index where a[j]==M, k = the larger index where a[k]==M, we let p[j]=q[k]=M, q[j]=p[k]=L. For those j which a[j] appear only once, we just let p[j]=q[j]=a[j].

My submission:188076704

SPyofcode

Yah nice solution :3

__MOUGOUPAN_22

It's a well know fact that n — cycles is the minimum number of swaps to get 1, 2, ..., n.

Can someone explains me this ? and where is this well know ?

each cycle of length $$$l$$$ needs $$$l-1$$$ swaps to be sorted, you can try it yourself

lmn207

nice E...stuck in case f=2 too longgg...

Your_Bunny_Wrote

codeforces!

bulijiojiodibuliduo

+73

F is so fantastic! It's hard to solve, but not hard to understand the solution.

Kartikgs

Can we see pretest after the contest. This submission failed pretest 2: Link

22 месяца назад, # ^ |

Pretest 2 is every combination possible for $$$n \leq 5$$$, you can write a code then generate pretest 2 yourself.

I'm looking for problem like today problem D and C. Specially problems on arrays that can be transformed into problems on graph.

Example 1 : You want to obtains array b from array a with some specific operation

Example 2 : You want to obtains array a and b from array c with some specific operation

Example 3 : optimizing or computing something on an array after some operation (This turns out to be somewhat related to paths on a graph ...)

In almost all the solutions we need to build some graph with specific edges. I don't know how to tackle that kind of problems.

__c_j_n__

+25

F let me feel :F

D let me feel :D

Euclid73

I got o(n) for C lolol. 188093045

nvm its o(nlogn)

iiand

sort(putP.rbegin(), putP.rend()) It is minimum O(n logn)

nsqrtlog

Since I didn't implement, please point out if I did anything wrong on F, or if my approach would TLE.

After the monotonic stack observation, it should be possible to use a kinetic tournament on quadratics to handle cases with $$$min(a_i,\ldots, a_j)=a_i$$$ and li-chao tree or another kinetic tournament otherwise. The time complexity would be $$$O(n2^{\alpha(n)}\log^2(n))$$$.

rock_00

I cannot understand the editorial of Problem C, which says there are (i — 2)*2 + 2 numbers. (there can be a repetition of numbers in p and q) also, why is this a contradiction?

vgtcross

What do you mean by "there can be a repetition of numbers in p and q"? $$$p$$$ and $$$q$$$ are permutations, they can't have any repeated numbers (although the same numbers that appear in $$$p$$$ also appear in $$$q$$$ since they are both permutations from $$$1$$$ to $$$n$$$).

In the editorial of problem E, when calculating the intersection for f(p)<=2, the sum should be sum(i=0...n) instead of sum(i=1...n). Please fix it

Yes it is, I fixed the editorial.

18o3

I had a slightly different $$$O(n)$$$ 188071468 from the one given for Problem C.

BiouX

No CHT solution for F? That's very interesting.

Vladithur

It is possible to use it to solve the second case, but you'll have to squeeze it into the tight ML.

Math_is_Good

It is a well know fact that n−cycles is the minimum number of swaps needed to get the permutation 1,2,3,…,n from our initial one (in other words, to sort it).

Can anyone Please provide me the material where i can see this tutorial and learn . Thank You!!.

ChenZ01

Someone mentioned it. Each cycle of length $$$l$$$ needs $$$l - 1$$$ swaps to sort.

Thanks, i got it now .

vrintle

Can anyone explain E in a simpler way ? I am having a hard time understanding the editorial..

I modified it a little bit, maybe you'll find it easier to read.

canurag

Ugh my solution to C FST'd due to TLE can someone tell me why :(( https://mirror.codeforces.com/contest/1768/submission/188118022

modernbeast02

Here is the reason -> https://mirror.codeforces.com/blog/entry/48793

:((((( oh no I didn't know

Well, U know it now.

Mr_Sa3dola

Is it possible for you to make a report on the channels that download solutions on YouTube during the contest? Very thanks.<3

Dart-Xeyter

Maybe I missed something, but why there is O(n*sqrt(n)) with [n, n-1, n-2, ..., 1] array? Or any similar.

+10

In this subcase, $$$a_j < \sqrt{A}$$$.

zjcccc

I think this time aj is smaller than sqrtA, consider all the cases of such j, the total compexity of the j with same value will not exceed n, so the overall complexity will not exceed..nsqrtA sorry my poor english

abdur_rahman_41

⌊n−w+k−1⌋/k
could you please eplain me why we add K-1 with n-w ? .

so as to do the ceil operation otherwise it's by default floor operation as the integer division truncates anything after the decimal point

lbm364dl

I'm having a hard time trying to understand the intersection formula for $$$f(p) \le 2$$$ from E's editorial. Can you please proofread the 'sketch and calculation' spoiler and fix some mistakes/typos?

I mean I read for example this sentence:

In the last $$$n$$$ numbers there are $$$n$$$ numbers in range $$$[n+1,2n]$$$, we have used $$$1$$$ numbers for the first $$$n$$$ numbers. There are $$$C^n_{2n-1} \cdot n!$$$ cases.

I think you wanted to write:

In the last $$$2n$$$ numbers there are $$$n-1$$$ numbers in range $$$[n+1,2n]$$$, we have used $$$1$$$ numbers for the first $$$n$$$ numbers.

And if this is the case, then I don't really understand the $$$C^n_{2n-1} \cdot n!$$$ formula.

No, there are $$$n$$$ numbers, not $$$n - 1$$$. Since we used $$$1$$$ number to fill in the first $$$n$$$ numbers, there are only $$$2n-1$$$ options to choose for the last $$$n$$$ numbers, since they must be in range $$$[n + 1, 3n]$$$. I think the $$$n!$$$ part is easy to understand.

Edit: okay I think I spotted the wrong thing, it is fixed.

Ok, I see the interval in the third case was fixed. Now I get it. Thanks!

I still want to add something, maybe someone finds it helpful. I don't find the $$$n!$$$ parts easy to understand as they are right now in the editorial, I find them a bit unnatural if not just 'wrongly' explained. The third case actually counts the number of ways to fill the last $$$n$$$ positions. What are the other two cases counting exactly, including the $$$n!$$$ 's, explained with words?

This is how I would split it, each part counting one third of the positions:

Take $$$n-i$$$ elements from $$$[1, n]$$$ and $$$i$$$ from $$$[n+1, 2n]$$$. This makes $$$n$$$ elements for first $$$n$$$ positions, and we take all the rearrangements. This is $$$C^{n-i}_n \cdot C^i_n \cdot n!$$$
As you explained, we need $$$n$$$ numbers in the range $$$[n+1, 3n]$$$ for the last $$$n$$$ positions, but $$$i$$$ of them were already used in the first $$$n$$$ positions, so we only have $$$2n-i$$$ left to choose from. Also, we can rearrange them. This is $$$C^n_{2n-i} \cdot n!$$$
The first $$$n$$$ and last $$$n$$$ positions are filled. The rest are the center $$$n$$$ positions. Rearrange them freely. This is just $$$n!$$$

Nxxlt

This is much clearer, thanks a lot.

sonlh07

Kudo for lbm364dl, great explanation that help me a lot, take a long time to understand the editorial before found your comment.

Sorry for late reply, suppose we have $$$n$$$ fixed indexes for the numbers in range $$$[1, n]$$$, then there are $$$n!$$$ cases for these numbers switching places with each other, the second and third $$$n!$$$ are exactly the same.

In conclusion:

The combinatorics is how we choose indexes for number in a range.
The factorial is the number of cases after having fixed indexes.

The third one is $$$n!$$$ because we already chose indexes for $$$n$$$ numbers out of $$$2n$$$.

Anyway thank you for providing a much clearer explanation for this problem!

Hi guys I have made editorial videos on A,B,C,D and uploaded on the channel — https://www.youtube.com/@GrindCoding, you can have a look and let me know if you like the explanation or please provide your valuable feedbacks.

[D is currently being processed by youtube so might take a few mins to upload]

tfg

+14

There's no reason to make the mod not fixed as something like 1e9 + 7 in problem E. Any fft solution that passes that mod will pass for any other mod that you might ask for.

JoseSoto

Any idea why I got Wrong Answer instead of Runtime?

During the contest in problem C I was getting WA so I tried to modify my solutions but now that the we can see the test cases it says "wrong output format Unexpected end of file — int32 expected"

The thing is that I used an array of size $$$10^5$$$ instead of $$$2 \times 10^5$$$

WA code

AC code

So sad, with this modification I got AC with my first attempt :(

Vipin

In problem D,

Turns out that we can easily calculate cycles′ if we know cycles:

cycles′=cycles+1 if the vertices k and k+1 were in the same cycle in the initial graph, cycles′=cycles−1 otherwise.

Can someone explain this???

MeetAmin

Here is the explanation

itachi_fam

+16

thanks for the great round <3

finally become pupil after 82 contest !!

shivn

Somebody please help me with my submission of problem C . https://mirror.codeforces.com/contest/1768/submission/188118465 Can't able to detect error.

Can someone explain for problem Lucky Permutation:

It's not easy to explain. I noticed this by using brute force on some small cases, but I can't prove it until the end of contest.

dbtalaviya

← Rev. 4 →

Cycles are for sort the array, we want 1 inversion. Let's say $$$i$$$ is connected with $$$x$$$ and ($$$i$$$+1) is connected with $$$y$$$.for inversion, we break the link from $$$i$$$ to $$$x$$$ and ($$$i$$$+1) to $$$y$$$ and make a new link from $$$i$$$ to $$$y$$$ and ($$$i$$$+1) to $$$x$$$. That's it, now if we analyse breaking and making link for $$$i$$$ and $$$i$$$+1 ,

we can see if $$$i$$$ and $$$i$$$+1 are from the same cycle they will brake the cycle and we have 2 cycles from the 1 cycle.

If $$$i$$$ and $$$i$$$+1 in different cycle they will join together and make 1 cycle from joining 2 cycles.

i hope it is understandable.

graph

Nice explanation

Thanks

brainstorm56

"If we iterate from the smallest value to the top, there will be scenarios where all the remaining positions i will result in max(p[i],q[i])≥a[i] because you don't have enough smaller integers." This is written in the tutorial of problem C. Can anyone pls provide me a testcase where this scenario happens?

5 5 2 2 1

In this case, there are no enough place for 3,4,5 to put in.

But this test case anyways has no solution, provide a test case in which iterating from the smallest value give wrong answer.

bhatte2.0

Why does my submission for C give TLE although it looks nLog(n).

Spoiler

Abito

You are using the function upper_bound(something). Instead, use setname.upper_bound(int). The one you ar eusing can run O(n) sometimes and is bad. The one I mentioned uses binary search on sets and is much faster

smart_work

can anyone please help me in figuring out why is this code of mine giving runtime error on test 7? tried enough not able to figure out.188129954

Rubikun

F is cool!

Is "1. min(ai...aj)>=A and 2. min(ai...aj)<A" (in F's Tutorial) typo ?

I think they should be sqrt(A).

Thanks! It's a typo, should be fixed now.

tgp07

Got the idea for D after the contest lol

sharmaharisam

Video Solutions for A-E

TheHoodyGang

Nlog(n) code giving Tle for problem C please help! here is the code 188134958

In vector upper-bound is log(n) right? So why not in sets?

It is not. The normal upper_bound() function works differently and at worst cases might iterate over the entire vector/set. But there is a special upper_bound() function for sets. Use setname.upper_bound(x). This is actual logn

Dipankar_Kumar_Singh

A approach to Problem D using Group theory ,

Link To solution : 188139688

If Permutation is not sorted, then it will contain some cycles, using Cycle Decomposition, we can break the Cycle of length k into** k – 1 Transpose** (Two Length cycles). If Identity transpose ex (1 3) is multiplied to itself ( 1 3 ) => it will cancel the effect , i.e. Elements have been placed to their correct respective location . So to make the permutation sorted, if there are k transpose, we can multiple k identity transpose, thus canceling out the effect and make them sit on their place. Note : sorted permutation means zero Inversion. And at last, do multiple any transpose of adjacent elements ( 3 4 ) or (1 2 ) or ( 5 6 ) , etc. to get One Extra Inversion .

So, using DFS one can check for Cycles, find the length of cycles, cycle length – 1 is the transpose.

So answer = total no of transpose from all cycles + 1 [ General case ] (for last extra One Inversion that is been added on top of sorted permutation).

There is an edge case, what if one adjacent transpose is already present in one of the cycles, For example (1,2) is already present, and this contributes to only one Inversion, so we can take help of already present good transpose for this case note: we only need any One good transpose (adject transpose) . Let’s say good transpose is T , so first , in general case , we will multiply T with T to cancel out T and make permutation sorted, then again multiply T to add extra one inversion . So we can skip multiplication of T two(x2 ) times , because at last we want to use T as good transpose .

T x T x T = T

( extra multiplication of T two times ) so we can subtract 2 operation from the general case answer .

Ans = ( transpose + 1 ) – 2 [ when good transpose is already present in any of the cycles

valeriu

Fucking shit I have been associated with a psychopath

Ujjwalkool

Isn't a case possible in D where number of cycles will remain same?

yuan-shen

Why does my solution for C work when using stack but not unordered_map? Solution using unordered_map: Solution Solution using stack: Solution I am doing the same thing in both of them, except it works when I use stack instead of unordered_map. I see it gives wrong answer on test case 2, but I can't see the test case. Can someone give a test case that doesn't work for unordered_map? Thanks.

When you do it=extra.begin() when extra is an unordered_map, It's not guaranteed which element you will get, so probably it->second==true for both it you get for a certain i, which cause i appear 2 times in q.

But I'm taking the iterator, then erasing it before I get the next element. Shouldn't it delete the one I just used?

It's not "getting the one just used". It's "There might be multiple entries where it->second==true and you get 2 such entries for one x, then you put two x in q"

Can you give me an example of when this would happen? Thanks.

It depends on the implementation of the iterator of unordered_map, so there's no determinate example.

But why would it work when I use a stack? Couldn't this also happen with a stack???

Stack is an ordered data structure, you always get what you just put into it

So the two elements I get are guaranteed to be one in p, and one in q, right?

Yes

Thanks!!!

atharvd

The problemset should should have a tag for permutations, to practice for these kinds of problems.

Rivuozil

Can someone explain why there is a +k in the final result of problem B, and not just (n-w)/k??

The reason is because we want to using ceiling division. If we used (n — w) / k, we would round down.

Dinshey

Yet another $$$O(n)$$$ solution for C, but can be implemented more easily. Without the headers, it's only 700 B long with two for loops.

solution

vegetable_king

Is $$$O(n)$$$ solution for C too hard to think? Why not hack $$$O(n \log n)$$$ solutions?

071_Prashant

Can someone help me to find the reason for TLE in Problem C for the submission 188102874

But an AC for the submission 188105380

use set.lower_bound(value) instead of set.lower_bound(start,end,value). The latter costs much more time.

Thanks!! Got Accepted now

t_t_m

lower_bound on set is linear time, use set.lower_bound instead

amitkumar2003.edu

At Line 157 use auto it = s.lower_bound(-q[i]); O(n^2)->O(nlogn)

ok thanks

The Submission 188119572 is Successfully Hacked but the Participant got points for the problem. How is This possible?

It's hacked after contest ends

dogoeswoof

Concise solution for D:

~~~~~

YC_Taffy

I'm still curious about that "the minimum number of swaps" theory in problem D, is there any prove of that theory (blog or something)?

the minimum number of swaps to sort is the classic problem, here is link

Thx :)

HHY_zZhu

About D, How they know the way to solve this problem during the contest? I mean is this a classical question ? Or they just figure out?

i'm interesting of how could they just know whatever what is swaping in a cycle it will only minus 1,and whatever what is swaping between two cycle ,it must cost one movement to change it to the answer Do they prove it or just consider it should work

0.01

Though my rating falls drastically in this contest. I love both the contest and the editorial. Keep organizing such contests Vladithur ❤️.

codechaser

Loved the contest, amazing problems, amazing editorial!!

By the way,
for problem C: 1768C — Elemental Decompress

"There is also another way that you can skip testing if max(p[i],q[i])=a[i]) is correct."

Another Way Here

puspendra_09

Can anyone explain or tell about resources where i can learn about counting the number of cycles in a directed graph. as a can't find it online. thanks

shubham84

I am just really glad this contest happened and would like to thank the authors of this round, as finally I could reach expert after this round.

double_zero

I love F!I think it is a quite good problem!

FishAndCat

In F tutorial,Is the "Just maintain the leftmost occurences" for the 2.1 cases a typo?Seems that it needs to maintain the rightmost occurences<j from 1 to √A

Should be fixed now.

peltorator

22 месяца назад, # |

Problem F is awesome. Couldn't solve it during the contest. After the contest, I looked at the "$$$a_i \le n$$$" hint from the editorial and solved it from there :) Sometimes just stressing a part of the statement can help you solve a problem :)

wsyear

F is a really good problem, I thought $$$O(n\sqrt n)$$$ solution for a long time during the contest but still did not know how to solve it. But later I was amazed by the tutorial of problem F :)

The facts to solve this problem are easy to understand and easy to prove, but it's really hard to find them :)