Codeforces Round #552 (Div. 3) Editorial

#	User	Rating
1	Benq	3792
2	VivaciousAubergine	3647
3	jiangly	3631
4	Kevin114514	3574
5	maroonrk	3521
6	strapple	3515
7	Radewoosh	3461
8	tourist	3428
9	turmax	3378
10	Um_nik	3376

#	User	Contrib.
1	Qingyu	162
2	adamant	148
3	Um_nik	146
4	Dominater069	143
5	errorgorn	141
6	cry	138
7	Proof_by_QED	136
8	YuukiS	135
9	chromate00	134
10	soullless	133

Tutorial

Solution

#include <bits/stdc++.h>

using namespace std;

int main() {
#ifdef _DEBUG
	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);
#endif
	
	vector<int> a(4);
	for (int i = 0; i < 4; ++i) {
		cin >> a[i];
	}
	sort(a.begin(), a.end());
	cout << a[3] - a[0] << " " << a[3] - a[1] << " " << a[3] - a[2] << endl;
	
	return 0;
}

1154B - Make Them Equal

Idea: 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 n;
	cin >> n;
	vector<int> a(n);
	for (int i = 0; i < n; ++i) {
		cin >> a[i];
	}
	sort(a.begin(), a.end());
	a.resize(unique(a.begin(), a.end()) - a.begin());
	if (a.size() > 3) {
		cout << -1 << endl;
	} else {
		if (a.size() == 1) {
			cout << 0 << endl;
		} else if (a.size() == 2) {
			if ((a[1] - a[0]) % 2 == 0) {
				cout << (a[1] - a[0]) / 2 << endl;
			} else {
				cout << a[1] - a[0] << endl;
			}
		} else {
			if (a[1] - a[0] != a[2] - a[1]) {
				cout << -1 << endl;
			} else {
				cout << a[1] - a[0] << endl;
			}
		}
	}
	
	return 0;
}

1154C - Gourmet Cat

Idea: le.mur

Tutorial

Solution

#include <bits/stdc++.h>

using namespace std;

int main() {
#ifdef _DEBUG
	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);
#endif
	
	vector<int> a(3);
	cin >> a[0] >> a[1] >> a[2];
	
	vector<int> idx({0, 1, 2, 0, 2, 1, 0});
	
	int full = min({a[0] / 3, a[1] / 2, a[2] / 2});
	a[0] -= full * 3;
	a[1] -= full * 2;
	a[2] -= full * 2;
	
	int ans = 0;
	for (int start = 0; start < 7; ++start) {
		int day = start;
		vector<int> b = a;
		int cur = 0;
		while (b[idx[day]] > 0) {
			--b[idx[day]];
			day = (day + 1) % 7;
			++cur;
		}
		ans = max(ans, full * 7 + cur);
	}
	
	cout << ans << endl;
	
	return 0;
}

1154D - Walking Robot

Idea: MikeMirzayanov

Tutorial

Solution

#include<bits/stdc++.h>

using namespace std;

int a, b, maxa;

void use_battery(int s)
{
	if(s == 1) a = min(a + 1, maxa);
	--b;	
}

void use_accum()
{
	--a;
}

int main()
{
	int ans = 0;
	int n;
	cin >> n >> b >> a;
	maxa = a;
	vector<int> s(n);
	for(int i = 0; i < n; i++) cin >> s[i];
	for(int i = 0; i < n; i++)
	{
		if(a == 0 && b == 0)
			break;
		else if(a == 0)
			use_battery(s[i]);
		else if(b == 0)
			use_accum();
		else if(s[i] == 1 && a < maxa)
			use_battery(s[i]);
		else use_accum();
		ans++;
	}
	cout << ans << endl;
}

1154E - Two Teams

Idea: 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 n, k;
	cin >> n >> k;
	vector<pair<int, int>> a(n);
	for (int i = 0; i < n; ++i) {
		cin >> a[i].first;
		a[i].second = i;
	}
	
	sort(a.rbegin(), a.rend());
	queue<int> q;
	for (int i = 0; i < n; ++i) {
		q.push(a[i].second);
	}
	
	set<int> idx;
	for (int i = 0; i < n; ++i) {
		idx.insert(i);
	}
	string ans(n, '0');
	int who = 0;
	while (!idx.empty()) {
		while (!idx.count(q.front())) {
			q.pop();
		}
		int pos = q.front();
		q.pop();
		
		vector<int> add;
		auto it = idx.find(pos);
		for (int i = 0; i <= k; ++i) {
			add.push_back(*it);
			if (it == idx.begin()) break;
			--it;
		}
		it = next(idx.find(pos));
		for (int i = 0; i < k; ++i) {
			if (it == idx.end()) break;
			add.push_back(*it);
			++it;
		}
		for (auto it : add) {
			idx.erase(it);
			ans[it] = '1' + who;
		}
		who ^= 1;
	}
	
	cout << ans << endl;
	
	return 0;
}

1154F - Shovels Shop

Idea: MikeMirzayanov

Tutorial

Solution

#include <bits/stdc++.h>

using namespace std;

const int INF = 1e9;

int main() {
#ifdef _DEBUG
	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);
#endif
	
	int n, m, k;
	cin >> n >> m >> k;
	vector<int> a(n);
	for (int i = 0; i < n; ++i) {
		cin >> a[i];
	}
	
	sort(a.begin(), a.end());
	a.resize(k);
	reverse(a.begin(), a.end());
	
	vector<int> offers(k + 1);
	for (int i = 0; i < m; ++i) {
		int x, y;
		cin >> x >> y;
		if (x <= k) {
			offers[x] = max(offers[x], y);
		}
	}
	
	vector<int> pref(k + 1);
	for (int i = 0; i < k; ++i) {
		pref[i + 1] = pref[i] + a[i];
	}
	
	vector<int> dp(k + 1, INF);
	dp[0] = 0;
	for (int i = 0; i < k; ++i) {
		dp[i + 1] = min(dp[i + 1], dp[i] + a[i]);
		for (int j = 1; j <= k; ++j) {
			if (offers[j] == 0) continue;
			if (i + j > k) break;
			dp[i + j] = min(dp[i + j], dp[i] + pref[i + j - offers[j]] - pref[i]);
		}
	}
	
	cout << dp[k] << endl;
	
	return 0;
}

1154G - Minimum Possible LCM

Idea: MikeMirzayanov

Tutorial

Solution

#include <bits/stdc++.h>

using namespace std;

const int INF = 1e9;
const int N = 10 * 1000 * 1000 + 11;

int n;
vector<int> a;

int mind[N];

pair<int, int> mins[N];
vector<pair<int, int>> divs;

void build_sieve() {
	vector<int> pr;
	mind[0] = mind[1] = 1;
	for (int i = 2; i < N; ++i) {
		if (mind[i] == 0) {
			pr.push_back(i);
			mind[i] = i;
		}
		for (int j = 0; j < int(pr.size()) && pr[j] <= mind[i] && i * pr[j] < N; ++j) {
			mind[i * pr[j]] = pr[j];
		}
	}
}


void add_to_mins(int curd, int idx) {
    if(mins[curd].first == -1)
        mins[curd].first = idx;
    else if(mins[curd].second == -1)
        mins[curd].second = idx;
}

void rec(int pos, int curd, int idx) {
	if (pos == int(divs.size())) {
		add_to_mins(curd, idx);
		return;
	}
	int curm = 1;
	for (int i = 0; i <= divs[pos].second; ++i) {
		rec(pos + 1, curd * curm, idx);
		curm *= divs[pos].first;
	}
}

void add(int idx) {
	int value = a[idx];
	divs.clear();
	while (value > 1) {
		int d = mind[value];
		if (!divs.empty() && divs.back().first == d) {
			++divs.back().second;
		} else {
			divs.push_back(make_pair(d, 1));
		}
		value /= d;
	}
	rec(0, 1, idx);
}

int main() {
#ifdef _DEBUG
	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);
#endif
	
	cin >> n;
	a.resize(n);
	for (int i = 0; i < n; ++i) {
		cin >> a[i];
	}
	for(int i = 0; i < N; i++)
	    mins[i] = make_pair(-1, -1);
	build_sieve();
	
	vector<pair<int, int> > vals;
	for(int i = 0; i < n; i++)
	    vals.push_back(make_pair(a[i], i));
	sort(vals.begin(), vals.end());
	for (int i = 0; i < n; ++i) {
	    if(i > 1 && vals[i].first == vals[i - 2].first) continue;
		add(vals[i].second);
	}
	
	long long l = INF * 1ll * INF;
	int ansi = -1, ansj = -1;
	for (int i = 1; i < N; ++i) {
		pair<int, int> idxs = mins[i];
		if (idxs.second == -1) continue;
		long long curl = a[idxs.first] * 1ll * a[idxs.second] / i;
		if (l > curl) {
			l = curl;
			ansi = min(idxs.first, idxs.second);
			ansj = max(idxs.first, idxs.second);
		}
	}
	
	cout << ansi + 1 << " " << ansj + 1 << endl;
	
	return 0;
}

Comments (63)

Show archived | Write comment?

babu1998

7 years ago, hide # |

← Rev. 2 →

Could somebody explain the LCM question clearly ? I could not understand the editorial

→ Reply

yoddhaa

**O(n)** solution for problem **E**:

For every index i, we can maintain two links, one for its left, l[i], and another for its right, r[i], pointing to the first yet undecided student on either sides.
When we assign a sub-array to a particular team, m, update only the right link of left of leftmost cell, and the left link of right of rightmost cell, provided these cells are not out of bound.
In this manner, we iterate over every index at most two times.

Link of Code

quackdratic

7 years ago, hide # ^ |

+16

In your code, you're sorting the input array, so complexity is still O(nlogn).

Oh sorry, I missed that:)

Sturdy

You don't need a nlogn sort algorithm in your code. You can do linearly too

AliShahali1382

E can be solved using Linked List if O(n). you just need to implement the operations. since every student will be chosen exactly once so the time complexity is O(n). code: 52920413

architrungta120

sorting?

INMAK

If you use a non-comparing sort algorithm. It could be done in linear time.

CriticalPundit

I don't know of any non-comparison sort algorithm that works in O(n)time. The best I know of is works in O($$$n\sqrt {log(log(n)}$$$). Can you tell one that works in linear time.

Counting sort

OK, that would work in this case, as it also depends on the range of input, won't work for larger ranges.

if you read my submission you can see that I didn't sort anything!!!

sanket

Can you explain your O(n) solution.

build a Linked List and every time you are removing a student remove him and his k left and his k right students from the Linked List.

-6

But finding the max in remaining list will take O(n). so it will be O(n*n) solution.

mridul1809

You don't need to find that and nor need to sort it. Since all students are between 1 to N and occur exactly once just create an array to point out their location in the linked list.Whenever you remove a student just clear this pointer.Keep a pointer to last maximum player used and decrement this until a player which exists in the linked list!

dra2421

6 years ago, hide # ^ |

you can see my solution it o(n) and do not use the sorting 90431019

Honestly for the first time I saw the use of linked list making a solution optimal. Very good approach

alis_jain_fake

could you tell me about solution you mentioned in first line or provide a link (in problem G).

anupamshah_

Such a fine editorial !

+13

My approach to problem G :

let G = gcd(A,B)
so we can write
A = GR
B = GS
Now, lcm(A,B) = AB/gcd(A,B) = AB/G = GRS
We need to minimize GRS Here G is common factors between two numbers
R is uncommon factors in number A
S is uncommon factors in number B

So we iterate on G (which will be <= 10^7) and for each G we store smallest two multiples of G which are present in our array. Say those multiples are A,B.
Now G(A/G)(B/G) (lets denote it with Z) is our best possible answer with G as gcd.

We find minimum value of Z for all the possible values of G and print corresponding indices as required.

NOTE
The declaration of A = GR and B = GS requires R and S to be coprime (gcd(R,S) = 1).
While finding Z for G , We may have gcd(A,B) > G but it will always be multiple of G (think why). Thus gcd(R,S) = k
Now gcd(A,B) = kG. Even if answer Z stored for G will be G(A/G)(B/G) = Gk²RS ,
when calculation answer Z for kG (G') as gcd we will get Z as G'(A/G')(B/G') = kGRS which is actual answer.

Link to my solution!

vivek8420

Ca you exaplain your code little bit?? Thanks in advance.

RamiZk

take a look at my code , I believe its the same idea for each x in range of 1e7 get the minimum two numbers ( a , b ) which are divisible by x
the answer will be ( a * b / x )

C++ Solution

My code was a bit messy! Thanks RamiZk for providing an excellent implementation! vivek8420 you can have a look at his implementation it's super clean and easy to understand.

My pleasure

sahal

5 years ago, hide # ^ |

Thanks a lot for this nice implementation.

PhantomAssassin

i m hack your solution with test case: 2 7888889 7888890 easy

You realize that it's not a valid input , don't you ?

not. everything is correct. n >= 2 && a[1] <= 1e7 && a[2] <= 1e7

oh sorry. it turns out that your solution works correctly)

Vishal_verma

What is the time complexity of your solution?

O(Nlog(N))

Thanks for the explanation! It helped :)

Glad I could help! :)

rupav

Really easy to understand :).

Successfully implemented my submission based on your approach. As per your Note: So basically we are ignoring those G whose gcd(A, B) = kG, because, there will be a gauranteed G' which is gcd of same (A,B) but provides a better(minimum) LCM.

Also, you don't need to check for if(m < a[i]) and if(m < b[i]), as this will never happen. (since m is directely proportional to j given a fixed i). i.e. (my soln)

Yup you got it right!

Yes I could have ignored that. Just a messy implementation on my part.

Thanks for sharing your implementation! :)

Eltoney

in problem c: why we take the modules day:=(day+1)%7 ?

UserIsUndefined

That because we want the index of the current day If you just increase it, day will be bigger than the last index of the vector (last index =6) so you want the day to go to the first element after the last one, so you put day%7 Let's say that the day =6 and there is nothing after 6 and you want it to go back to first element, day=(6+1)%7 = 0, and you are again on the index 0 Actually you also can do it like this : If day == 7: day = 0 :))

_bom_bom_

the problem G. As far as I know, $$$\sum_{i \le x}d(i) = O(x \log x)$$$, where $$$d(i)$$$ is the number of divisors of n. So time complexity is equal $$$O(a \log a)$$$, maybe it is equal $$$O(n \log a + a)$$$. Can you point me other $$$O(a \log a)$$$ simple solution?

ak281999

In problem D, in the second example 6 2 1 1 0 0 1 0 1

How is the answer 3? It should be 5 right? Because we can use a battery in first segment so battery decreases by one and accumulator increases by one ==> b=1 , a=2. For the next 2 segments we use 2 accumulators ==> b=1,a=0. The for the fourth segment we use a battery ==> b=0,a=1.For the fifth we use the accumulator!

Can anybody explain how the answer is 3?

It is mentioned in the question that charge of accumulator cannot exceed it's maximum capacity. "If the current segment is exposed to sunlight and the robot goes through it using the battery, the charge of the accumulator increases by one (of course, its charge can't become higher than it's maximum capacity)."
So when we make first move using battery the charge of accumulator wont increment to 2 as it is already at it's maximum capacity (1).

I just figured it out and saw your comment. Thanks :)

Because it was said about optimization, I would like to give some. First, the provided code find all divisors any number $$$a$$$ in not $$$O(d(a))$$$, because to find one divisor required between $$$\Omega(k)$$$ and $$$O(d(n))$$$, where $$$k$$$ is the amount of unique prime divisors. it needs to be changed by itarative algorithm. Second it's necessary to sort the array a. I thinks both tips will reduce the running time.

sodafago

Solution of E NlogN: https://mirror.codeforces.com/contest/1154/submission/52892001 Create segment Tree with = on segment and take maximum on segment Also create segment tree with add on segment and sum on segment And segment tree for calc Ans, with = on segment, and take value in the point

it's very hard way to solve this problem)) We can also solve it using set of pair (value, index) and two array left[n], right[n], which point for unchosen elemnts indexes of their left and right closest unchosen elements.

I know this, but I write them, because this hard, special for this))

kitchent

Thanks for hosting the nice round. I don't quite understand the iterator part of the problem E. If I didn't read it wrong, the iterator --it and ++it are supposed to refer to the original index of the students. If it is so, how do we deal with cases when the next students the couch has to choose is out of reach of k steps (i.e. some of them in between have be chosen in previous rounds)?

flxwu

The students that already been chosen in between are getting removed from the set at the end of each turn. In the editorial, it is written that we iterate the "add" array (that holds all students that were picked in the current turn) and delete all students in the "add" array from the total students set. Thus, in the next turn, the "students in between" that have already been chosen aren't in the students set anymore.

I have changed my opinion about Div3 contests. In this contest the last 3 problem was intresting ans usefull for some experts, imho

Jety

Can Anyone help me finding bug in my Recursive DP approach on problem F : https://mirror.codeforces.com/contest/1154/submission/52944910 Thanks in Advance..

Update: Done !! Recursive DP solution: https://mirror.codeforces.com/contest/1154/submission/52951559

sugar_crush

In problem F editorial can someone please explain this statement

"If shovel A is for free, then we may "swap" shovels A and C, otherwise we may swap shovels A and B, and the answer won't become worse"

Can't understand the same :(

I think two swap A and C would mean include C in previous offer instead of A and buy A in next offer. Similarly for A and B.

norty

F can be solved in O(k^2). First, any deals requiring us buy more than k shovels can be immediately discarded since the problem tells us to buy exactly k shovels. We are left with a subset of deals that have us buy between 1 and k shovels. This means that if we have an array deals[] with each element i corresponding to us having to buy i number of shovels we can simply find the maximum free for every index of this array. therefore, we since this array is size k, instead of running through every single deal, we only have to run through these, giving the solution a time complexity of k^2

Can you please elaborate your solution? I don't think only knowing the maximum amount of free shovels that can be bought will solve the problem.

https://mirror.codeforces.com/blog/entry/66560?#comment-506047

I tried to explain a similar solution to problem F here. Hope it helps!

Thanks a lot! It helped :)

Juuzou-San

6 years ago, hide # |

How we can solve problem E if we have to choose a student which the summation of k closest students to the left and right is maximized?

Sorry for my english

izoomrood

Could anyone explain a "very easy" O(a log a) solution to G?

AFAIK, all the comments here are somewhat related to the official solution in the editorial, but I'd like to know the other one(s). I just can't understand the solutions from the round as is, so thanks in advance!

AjaySabarish

This is a very simple O(MAXlog(MAX) + Nlog(N)) solution, where MAX is 1e7

92523482

XDD_____

Someone please explain my segfault for D https://mirror.codeforces.com/contest/1154/submission/95521763

Focus_2000

5 years ago, hide # |

E can also solve using Doubly linked list and Hashmapmy solution: 127237017

Ashwanth.K

3 years ago, hide # |

My NlogN solution works for the G problem.
Code
Idea: Similar to sieve, assume K = gcd() , iterate over all possible K values.

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