Codeforces Round 918 (Div. 4) Editorial

#	User	Rating
1	jiangly	3976
2	tourist	3815
3	jqdai0815	3682
4	ksun48	3614
5	orzdevinwang	3526
6	ecnerwala	3514
7	Benq	3482
8	hos.lyric	3382
9	gamegame	3374
10	heuristica	3357

#	User	Contrib.
1	cry	169
2	-is-this-fft-	166
3	Um_nik	161
3	atcoder_official	161
5	djm03178	157
6	Dominater069	156
7	adamant	154
8	luogu_official	152
9	awoo	151
10	TheScrasse	147

We hope you enjoyed the contest!

1915A - Выброс

Tutorial

Solution

#include <bits/stdc++.h>

using namespace std;

const int MAX = 200'007;
const int MOD = 1'000'000'007;

void solve() {
	int a, b, c;
	cin >> a >> b >> c;
	cout << (a ^ b ^ c) << '\n';
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int tt; cin >> tt; for (int i = 1; i <= tt; i++) {solve();}
	// solve();
}

1915B - Не совсем латинский квадрат

Idea: flamestorm

Tutorial

Solution

#include <bits/stdc++.h>

using namespace std;

const int MAX = 200'007;
const int MOD = 1'000'000'007;

void solve() {
	int cnt[3] = {};
	for (int i = 0; i < 9; i++) {
		char c;
		cin >> c;
		if (c != '?') {cnt[c - 'A']++;}
	}	
	for (int i = 0; i < 3; i++) {
		if (cnt[i] < 3) {cout << (char)('A' + i) << '\n';}
	}
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int tt; cin >> tt; for (int i = 1; i <= tt; i++) {solve();}
	// solve();
}

1915C - Можно ли построить квадрат?

Idea: SlavicG

Tutorial

Solution

#include "bits/stdc++.h"
using namespace std;
 
#define ll long long
 
#define          all(v)              v.begin(), v.end()
#define         rall(v)              v.rbegin(),v.rend()
 
#define            pb                push_back
#define          sz(a)               (int)a.size()

bool is_square(ll x) {
    ll l = 1, r = 1e9;
    while(l <= r) {
        ll mid = l + (r - l) / 2;
        if(mid * mid == x) return true;
        if(mid * mid > x) r = mid - 1;
        else l = mid + 1;
    }      
    return false;  
}
void solve() {
    ll n; cin >> n;
    ll s = 0;
    for(int i = 0, x; i < n; ++i) {
        cin >> x; s += x;
    }
    if(is_square(s)) cout << "YES\n";
    else cout << "NO\n";
}
 
int32_t main() {
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    int t = 1;
    cin >> t;
    while(t--) {
        solve();
    }
}

1915D - Нестандартная обработка языка

Idea: flamestorm

Tutorial

Solution

#include <bits/stdc++.h>

using namespace std;

const int MAX = 200'007;
const int MOD = 1'000'000'007;

void solve() {
	int n;
	cin >> n;
	string s;
	cin >> s;
	string res = "";
	while (!s.empty()) {
		int x;
		if (s.back() == 'a' || s.back() == 'e') {x = 2;}
		else {x = 3;}
		
		while (x--) {
			res += s.back();
			s.pop_back();
		}
		res += '.';
	}
	res.pop_back();
	reverse(res.begin(), res.end());
	cout << res << '\n';
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int tt; cin >> tt; for (int i = 1; i <= tt; i++) {solve();}
	// solve();
}

1915E - Романтические стаканы

Idea: flamestorm

Tutorial

Solution

#include "bits/stdc++.h"
using namespace std;
 
#define ll long long
 
#define          all(v)              v.begin(), v.end()
#define         rall(v)              v.rbegin(),v.rend()
 
#define            pb                push_back
#define          sz(a)               (int)a.size()

void solve() {
    int n; cin >> n;
    vector<int> a(n);
    for(int i = 0; i < n; ++i) cin >> a[i];
    map<ll, ll> m;
    ll s = 0;
    m[0] = 1;
    for(int i = 0; i < n; ++i) {
        a[i] *= ((i % 2) ? -1 : 1);
        s += a[i];
        if(m[s]) {
            cout << "YES\n";
            return;
        }
        ++m[s];
    }
    cout << "NO\n";
}
 
int32_t main() {
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    int t = 1;
    cin >> t;
    while(t--) {
        solve();
    }
}

1915F - Приветствия

Idea: mesanu

Tutorial

Solution

#include <bits/stdc++.h>
 
using namespace std;
 
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
 
typedef __gnu_pbds::tree<int, __gnu_pbds::null_type, less<int>, __gnu_pbds::rb_tree_tag, __gnu_pbds::tree_order_statistics_node_update> ordered_set;
 
int t, n;
 
vector<pair<int, int>> arr;
 
long long ans;
 
ordered_set st;
 
void solve(){
    cin >> n;
 
    arr.assign(n, {});
 
    for(auto &p : arr) cin >> p.second >> p.first;
 
    sort(arr.begin(), arr.end());
 
    ans = 0;
    st.clear();
 
    for(auto p : arr){
        ans += st.size() - st.order_of_key(p.second);
 
        st.insert(p.second);
    }
 
    cout << ans << "\n";
}
 
int main(){
    ios_base::sync_with_stdio(false);cin.tie(NULL);
 
    cin >> t;
 
    while(t--){
        solve();
    }
}

1915G - Велосипеды

Idea: SlavicG

Tutorial

Solution

#include "bits/stdc++.h"

const int64_t inf = 1e18;

void solve() {
    int n, m; std::cin >> n >> m;
    std::vector<std::pair<int, int>> adj[n];
    for(int i = 0; i < m; ++i) {
        int u, v, w; std::cin >> u >> v >> w; --u, --v;
        adj[u].emplace_back(v, w);
        adj[v].emplace_back(u, w);
    }
    std::vector<int> s(n);
    for(int& i: s) std::cin >> i;

    std::vector<std::vector<int64_t>> dist(n, std::vector<int64_t>(1001, inf));
    std::vector<std::vector<bool>> vis(n, std::vector<bool>(1001, false));
    
    dist[0][s[0]] = 0;
    std::priority_queue<std::array<int64_t, 3>> q;
    q.push({0, 0, s[0]});
    while(!q.empty()) {
        int u = q.top()[1], k = q.top()[2];
        q.pop();
        if(vis[u][k] || dist[u][k] == inf) continue;
        vis[u][k] = true;
        for(auto x: adj[u]) {
            int v = x.first, w = x.second;
            int c = std::min(s[v], k);
            if(dist[v][c] > dist[u][k] + 1LL * w * k) {
                dist[v][c] = dist[u][k] + 1LL * w * k;
                q.push({-dist[v][c], v, c});
            }
        }
    }
    int64_t ans = inf;
    for(int k = 1; k <= 1000; ++k) 
        ans = std::min(ans, dist[n - 1][k]);
    std::cout << ans << "\n";
}   
 
int main() {
    std::ios_base::sync_with_stdio(0);std::cin.tie(0);
    int t = 1; std::cin >> t;
    while(t--) {
        solve();
    }
}

for _ in range(int(input())): n=int(input()) arr=list(map(int,input().split())) newarr=[] sl=0 sd=0 for i in range(n): if i%2==0:sl+=arr[i];newarr.append(sl) else:sd+=arr[i];newarr.append(sd) # print('newarr',newarr) d={0,-arr[0]} f=False for i in range(n-1): if i%2==0: diff=newarr[i+1]-newarr[i] if diff in d:f=True;break else:d.add(diff) else: diff=newarr[i]-newarr[i+1] if diff in d:f=True;break else:d.add(diff) if f:break if f:print("YES") else:print("NO")

int main() { tc { ll n; cin >> n; vector<ll> v(n); for(int i = 0; i < n; i++) cin >> v[i]; if(n == 1) { cout << "NO" << endl; continue; } bool flag = false; for(int i = 0; i < n - 1; i++) { if(v[i] == v[i + 1]) { flag = true; break; } } if(flag) { cout << "YES" << endl; continue; } flag = true; unordered_map<ll, int> mp; ll pre = 0; for(int i = 0; i < n; i++) { if(i & 1) pre += v[i]; else pre -= v[i]; if(pre == 0) { cout << "YES" << endl; flag = false; break; } mp[pre]++; } if(flag) { for(auto it : mp) { if(it.second > 1) { cout << "YES" << endl; flag = false; break; } } if(flag) cout << "NO" << endl; } } return 0; }

void solve() { std::cin >> n; std::vector<int> a(n); for (int i = 0; i < n; i++) std::cin >> a[i]; long long s = 0; std::unordered_set<ll> st; st.insert(0); for (int i = 0; i < n; i++) { a[i] *= ((i % 2) ? -1 : 1); s += a[i]; if (st.find(s) != st.end()) { std::cout << "YES\n"; return; } st.insert(s); } std::cout<<"NO\n"; }

bool comp(pair<int64, int64> a, pair<int64, int64> b) { return a.first < b.first; } void solve() { inp(n); multiset<int64> mst; vector<pair<int64, int64>> ab(n); lp(i, n) { inpp(ab[i].first, ab[i].second); } sort(all(ab), comp); uint64 ans = 0; lp(i, n) { ans += (uint64)distance(mst.upper_bound(ab[i].second), mst.end()); dbg2(i, ans); mst.insert(ab[i].second); } out(ans); return; }

from sys import stdin; T = int(stdin.readline().strip()); for _ in range(T): N = int(stdin.readline().strip()); AB = dict() for i in range(N): a,b = map(int,stdin.readline().split()); AB[a]=b; AB=dict(sorted(AB.items(), key=lambda e: e[1])) G=0; for i,K in enumerate(list(AB.keys())): G+=len([x for x in list(AB.keys())[:i] if (x>K)]) print(G)

juice[i] = sc.nextInt(); } HashMap<Integer, Integer> mppp = new HashMap<Integer, Integer>(); int quantity = 0; mppp.put(0, 1); int bit = 0; for(int i =0; i < glasses; i += 1) { if(i % 2 == 1) { juice[i] *= -1; } quantity += juice[i]; if(mppp.containsKey(quantity)) { System.out.println("YES"); bit = 1; break; } mppp.put(quantity, mppp.getOrDefault(quantity, 0)+1); } if(bit == 0) System.out.println("NO"); } }

Comments (122)

Show archived | Write comment?

ismailfateen

13 months ago, # |

This was a standard div 4, I believe previous div 4s were harder than this. Good contest regardless!

→ Reply

Pretest2

For anyone who wants to know more about similar problems like G.

Here is a great blog on shortest path modelling:

https://codeforces.me/blog/entry/45897

jonnnnnsinaga

13 months ago, # ^ |

Never thought there is something like that. Thank you !!!

humblebose

why does this give TLE in E?anyone

mohmmdali

I am not sure but in python checking if f in d: takes enough time to make it TLE, I have encountered this while doing a problem in CSES. Everything else looks fine to me.

d is a set, its O(1) to my knowledge

_Kee

Lookup and insertion of set can be O(n) if it's filled with values that cause hash collisions. Apparently it's possible to craft a very bad input that makes Python's (or Pypy's) set run very slowly.

Garavitey

int the third question if i put r = sum(the sum of all number) instead of 1e9 then it fails in testcase 3. why is that happening?

T1.OTTO

The maximum of sum is 2E14, and 4E28 after squared it.

i can't understand? does it mean it cause overflow error?

long long maximum is about 9E18

Otherworldly

I kept my promise, I can die with a clear conscience.

MallaSailesh

Can someone please tell where i am going wrong in this solution wirtten by me for G. 239395920

Can someone tell why is this approach is wrong

robostac

Also you are marking nodes as visited and not clearing it when the value is in your hashtable blocking that path until the next test case.

Understood the test case

Can you also give a better explanation why is it failing and why if we do like this it fails for this question

DFS is awkward for shortest paths. In the test case the first time you visit node 3 you've already visited 1 and 2. This doesn't leave any options for continuing from 3 so you cache it's value as INF. Then when you start to follow the shortest path (1->3->2->4) you reach 3 and return the cached value (which is incorrect).

You'll notice that my example test case all the bicycles have the same value making the problem into a very simple shortest path on weighted graph problem.

D_coder22

← Rev. 2 →

why my soln of F fails for this test case:

from bisect import * I=lambda:map(int,input().split()) rr=range for _ in rr(int(input())): n,=I() an=0 l=[list(I()) for _ in rr(n)] l.sort() k=[] for i in rr(n): ind=bisect(k,l[i][1]) an+=len(k)-ind k.insert(ind,l[i][1]) print(an)

test case: 1 200000 -999999999 999999999 -999999998 999999998 -999999997 999999997 -999999996 999999996 -999999995 999999995 -999999994 999999994 -999999993 999999993 -999999992 999999992 -999999991 999999991 -999999990 999999990 -999999989 999999989 -999999988 999999988 -999999987 999999987 -999999986 999999986 -999999985 999999985 -999999984 999999984 -999999983 999999983 -999999982 999999982 -999999981 999999981 -999999980 999999980 -999999979 999999979 -999999978 999999978

Valentino_01

trying using set

but where i m using set in the code

Terror_404

here is my code for f using binary index tree. https://codeforces.me/contest/1915/submission/239413594

Anurag_51

Why my E's solution got hacked for this testcase? Both the editorial's and my solution is giving "NO" as output.

int main(){
    cout<<"1\n200000\n";
    for (int i = 0; i < 100000; ++i)
    {
        if (i != 0) cout << ' ';
        cout << "1 107898";
    }
    cout <<endl;

    return 0;
}

My code:

moraprex

I am not sure, but maybe because of unordered map? Here: https://codeforces.me/blog/entry/62393

aman098

Yes, when I used unordered_map it was giving TLE but when I used simple map it got accepted. But why?? Also unordered_map complexity is better as compared to map.

gladiator2win

Because average TC of unordered_map is O(1) but in worst case it can be O(n) in case of collisions, test case is specifically created to have collisions, that's why read the above blog and learn from it.

stefanvil3

Insert in unordered_map is usually O(N), so your overall complexity goes from O(N) or O(NlogN) to O(N^2). Unordered_map uses hash tables, so if you come up with tests anti-hash tables you’re gonna get hacked because of TLE

coding_parrot

Can anyone give the reasoning, on why a normal dijkstra won't work for problem G ? We are asked to find the path that takes minimum time from node 1 to node n.

can't we just update the distance to a node based on this? d[v] > d[u] + min(k, s[u]) * w[u][v]

why are we storing d[v][s] ? why do we need 2d array to calculate shortest distance to node v using slowness s

-is-this-hk-

you might go to a node just to achieve it's slowness factor and then come back from where you came.

captain_india1

Can anyone help in question 6. How is this standard problem

Basically we just want to count how many pair (i, j) that a[i] > a[j] and b[i] < b[j]. Because this is the necessary and sufficient condition when people i and people j meet at the same point.

animesh2403

e can also be solved easily by creating prefix sum array for even and odd indices we need to find even[j]-even[i]=odd[j]-odd[i] which is equivalent to even[j]-odd[j]=even[i]-odd[i] so just store the difference array of even and odd prefix sum if any value repeats answer is "yes" else "no"

Adityavb05

Can anyone please explain why the below solution fails for problem G:

Approach: A slight modification of the Dijkstra's Algorithm. We calculate DIS [ i ] [ j ] to be the shortest path to reach from node 1 to node ' i ' with bike of slowness factor ' j ' . Update the DIS array when a new path having shorter distance is encountered with bike of same slowness factor.

Submission link 239402842

Ishaan_23

You are updating mins. But when going a -> b where mins at a was 10 and b would become 5, you are calculating time using 5 * edge weight. Note that to reach b you should use 10 * edge weight, use updated mins after that.

Edit: code here's mine with the same idea

AJ__20

I can't understand the sentence in the E problem, " Be careful about using hash tables, as they can be hacked.". Could anyone explain this?

boxlesscat

Blowing up unordered_map, and how to stop getting hacked on it

Zenitsiuuu

Can someone show me how to solve F without ordered_set? Maybe with Fenwick tree or something else?

granadierfc

you can sort by leftmost point (i.e. a[i]) and then iterate from back to front. You will first need to do coordinate compression of b[i] values (for that you can use a map). After that when you are at i , all j > i satisfy a[j] > a[i] and you will query fenwick(map[b[i]]) , to get number the elements which are less than b[i].

OMARB

https://codeforces.me/contest/1915/submission/239302718 this is my solution

blaze

Can someone hack this solution for problem G? 239328319

AmanMittal_21

If I get a input for which like everyone gets hacked due to TLE, i can only apply in a contest to a limited people but whosoever i put they get hacked,

Shouldn't there be system testing for all the successful hacks afterwards with those inputs on all users???

Ain't it unfair?

ne_justlm

to my knowledge solutions do get retested after the hacking phase with the use of new hack-inspired tests

No, it doesn't atleast what i know for div 3 and 4, you can do as much hack as you can with the same generator code, but even if successful it won't be tested on ohters, me and my friend had same logic my got hacked his not.

last div3 got retested tho. maybe it will be retested a bit later?

Maybe, let's wait and watch

← Rev. 3 →

oh, did your solution use unordered_set or unordered_map? it might be the reason you got hacked, average comlexity of it is linear, but if there are too much collisions it can be o(n), people use this to make others solutions get tle, so try not to use it, there was a post about unordered map and etc. so read it if you have free time

upd.: here is a link https://codeforces.net/blog/entry/62393 upd.2: average complexity is constant, not linear, sorry

Ahmed_Ashraf_f

good contest but some solutions for g should not be accepted due to overflow

look at this one

https://codeforces.me/contest/1915/submission/239403382

curr is int and it should be long long upd: didn't know system testing hadn't been finished

Can anyone explain why me solution got AC? it has a time complexity of n^3 https://codeforces.me/contest/1915/submission/239328701

audi_007

in this Submission,the verdict says wrong answer and the reason is wrong output on the testcase

ABC ?CA CAB (4th case of test 2)

However , my program works exactly well with this testcase as well .

Is there any error from codeforces side?

Daft_Dimbo

← Rev. 4 →

Code where error exists

while (t--)
    {
 
        cin >> s;
        f(i, 0, 3)
        {
            if (s[i] == '?')
            {
                flag = true;
                break;
            }
        }
        if (flag == true)
            break;
    }

Hint

The error

Shavkat_Rakhmonov

Damn, my solution for the E got hacked because I was using the unordered_map :( I know how it can be fixed just so funny that I was getting caught on this again

Atul-1706

Why does unordered_map get hacked?

You can take a look more in depth in this post https://codeforces.me/blog/entry/62393

phoenix_vasu

[1st submission][2nd submission]

these are my two submissions to problem F , the only difference between these two submissions is that one has (#define int long long) and in other submission I commented that line , the one with (#define int long long ) got TLE whereas the submission without this got AC, why so and how can we avoid this TLE?

porker2008

Your solution is O(N^2) due to erasing elements from vector.

Your 1st solution is lucky to barely passed the time limits.

Thanks but why that solution is getting AC everytime when using int and TLE when using long long instead of int , does using long long increase execution time of a programme?

indresh

11 months ago, # ^ |

Why my seg tree sol gives TLE it's nlogn https://codeforces.me/contest/1915/submission/244464158

Prince2618

Can we use DP in any way to find the solution for problem G?

I know we can go with some modification of dijkstra's algorithm, but was wondering if any one has done it differently.

can anyone explain how can i find number of pairs of segment such that one completely lies inside anohter , inorder to solve F

alpha1215

sort the input based on the starting position then for segment i all the segments j such that i < j < n and ending position of j is less than i will completely lies inside segment i

NoahAT

Can someone explain to me why do I get TLE when I use unordered_set in PROBLEM G? When I changed it to set, I got AC.

link to blog. unordered map set work on hashing so it can tle if inputs are intentionally generate to cause large no of collisions it's better to use so it's better to use set and map in contest

WaveBreaker

We can solve problem F by using merge sort algorithm.

You can count the number that smaller the current number and before it in the second element of pair after sort it by nondecreasing sort.

This is the code: https://codeforces.me/contest/1915/submission/239384276

TheEgoist

Problem-F can also be solved using Fenwick tree? If yes, then can someone please explain the solution using Fenwick tree.

i assume that you know that in problem can be simplified to counting the inversion in an array to do this you can assign each element A[i] to it's index in sorted order like A = {5, 7 3, 2, 9} sorted A = {2, 3, 5, 7, 9} now you assign each element to it's relative index like (5 => 3), (7 => 4) and so on now A becomes A = {3, 4, 2, 1, 5} notice that all the elements in A are now between 1 and n. now assume you have visited array vis and a an array B which stores the prefix sum of vis array now iterate on A in reverse order and add B[A[i]] to you'r ans. why? it's because if you see carefully B[i] stores no. of elements < A[i] that are already visited and update the vis array vis[A[i]] = 1 and recalculate the prefix sum array B. the complexity of this O(n^2) now you can do the same thing in fenwick tree using point updates so it will be O(n*long(n)). you can learn fenwick tree from here hope it helps and sorry for my bad english.

Ylandolsi

10 months ago, # ^ |

what u did is cordinate compression ?

yes

can u explain please how the problem is transformed into counting the inversion in an array ( and what is inversion in an array exactly )

as everybody is moving with same speed 1 unit / sec relative velocity between any two person is 0. ~~~~~ let 2 person be x (ax, bx), y (ay, by) greeting between x and y happen only when

case 1 :- ax < ay < by < bx

case 2 :- ay < ax < bx < by

no. of greetings will be no. of ordered pairs(x, y) satisfying one of the above condition

now to do this you sort the people based on their starting point

now person i and j (i < j) greet only when b[i] > b[j]

so you just have to calculate no. of pairs (i, j) (i < j) such that b[i] > b[j]

these pairs are called inversion and count of all such pairs called no. of inversions in array

now you can think of solving this simplified problem

hope it help's let me know if you have any other query. ~~~~~

we need an array sorted to count the no. of pairs (i, j) (i < j) such that b[i] < b[j] ?

but we don t have that array that s why we used that technique ?

yes we have make an array whose i'th element denotes starting and ending points.

sorry it's b[i] > b[j]

ok thanks bro

Harsh_Mohta_007

please dont freeze the contest need to upsolve it not able to submit code now

after system testing it will be open for practice

Typed_SIGTERM

Problem C, why sqrt doesn't get TLE?

bautroimoi

sqrt is logn

ramakrishnalattuipally

IN problem E. Romantic Glasses TEST case 6: 2 5 10 4 4 9 6 7 8

why answer "YES"

there is no possiblity of equal subarray from (l,r)???????????????????

cosenza

subarray [4, 4]

ibrosh

Did u mean [4,5]?

ah, yes

Bluepanda_x

In problem E, can someone explain why using a hashtable leads to a TLE/hack?

has anyone solved F without using segement or fenwick tree in python? using bisect insort to keep a sorted list gives TLE on 11

bisect.insort is O(n), because inserting a value into the middle of a list is O(n).

WarriorOfLiberation

Yeah use merge sort for inversion coutning

what s is inversion counting bro

Inversion in a array is number of positions I<j such that a[I]>a[j], merge sort is widely used to find that in nlogn, search on internet for it

01_01_1998

Need help with C. Took inputs as long long, added them. Take l=0, r=total, k=0. While(l<=r){ m = l+(r-l)/2; if(m*m <= total) k=m, l=m+1; else r=m-1; } At the end if(k*k != total) print No else print Yes

I am getting Wrong Answer at case 3. Please, point out what I did wrong. Submission

I think the upperbound of your binary search is too big. you should change the upperbound to somewhere at 2e7 (or sqrt root of 2e14)

ACforever

Why did my F TLE？ https://codeforces.me/contest/1915/submission/239528521

Insertion for vector is O(N) on average.

Your solution's time complexity is O(N^2) imo.

theRealChainman

A-G solutions https://therealchainman.blogspot.com/2023/12/codeforces-round-918-div-4-problem-use.html

asnecemnnit

Can someone please look at the following code snippet (for problem 1915-F), and tell me why it gives TLE on test case #3: (And please bear with my excessive usage of macros, i hope you understand what the code is doing.)

Ianlsam

maybe give full code, impossible to understand anything when everything is #defined

b_a_d_m_a_n

Your solution has worst case $$$O(N^2)$$$ because you use std::distance which is linear for containers that aren't random access. As the example solution in this editorial shows, you can instead use an order statistic tree to solve this problem without TLE.

AhmetKaan

+33

First of all thanks for the nice problemsetting. SlavicG, mesanu, flamestorm

I have a slightly different solution for problem G, here is how it goes:

not_wiz

In the tutorial of F, the explanation says "for each of them add the number of segments that we have already passed that have an a value larger than the one of the current segment." Shouldn't it be smaller instead of larger as we have B sorted in increasing order ? Edit: I got it

TARIQ_123

In Problem E can anybody tell why solution gives TLE when I use unordered map and gets accepted when I use map

void solution(){ ll n; cin>>n; vector a(n); unordered_map<ll,ll> mpe; ll sume = 0,sumo = 0; for(auto &it:a) cin>>it; for(ll i = 0; i < n; i++){ if(i % 2 == 0){ sume += a[i]; }else{ sumo += a[i]; }

    ll diff = sumo - sume;
    if(diff == 0 or (mpe.find(diff) != mpe.end())){
       cout<<"YES"<<endl;
       return;
    }

    mpe[diff] = i;
}

cout<<"NO"<<endl;

}

int main(){ ll t; cin>>t; while(t--)solution(); return 0; }

SNX03

Bad hashing in unordered map can lead to collisions which increases the complexity of map's functions from $$$O(1) -> O(N)$$$ hence the TLE.

Find more here:

Спойлер

Tirth11

A more efficient (and prolly simpler) code for G as it does not require a 2D array: https://codeforces.me/contest/1915/submission/239576580

It's just Dijkstra with time as the parameter, but when I push cities into the priority queue, I also store the minimum slowness with which I will leave the city, once I visit it (the minimum between the slowness with which I entered the city and the slowness of that city's bike). Instead of a "visited" array, I made an array which stores the minimum slowness with which I have left the city, if I have visited it. So when I visit a city again, I carry on with that iteration only if the slowness with which I am entering the city is strictly lesser than the minimum slowness with which I have exited the city before, otherwise I continue with the next iteration.

Can someoone teach me G in a little more detail I don't know much about creating a new graph with state data.

carolinux

My G did not withstand system test.

https://codeforces.me/contest/1915/submission/239379244

Subtle bug in Djikstra implementation, try to find it if you like. (Note: I know what it is and I think it was a great way to learn something through the contest, won't make the same mistake again now.)

himanshu_swn

why is unordered_map not working but normal map working?

nasir_bashirov

+11

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

Blue_BEE

Can anyone help me to find what's wrong with my solution please? (My code for 1915E — Romantic Glasses). It's getting wrong answer on test case 3. My code is in JAVA. I've tried my best but can't find anything :(

naitikvarshney

In your last for loop, compare in the following way because on comparing def.get(i)==def.get(i-1), it is comparing the references not the value.

for(int i=1;i<=n;i++){
   long a=def.get(i);
   long b=def.get(i-1);
   if(a==b){
   ans = "YES";
   break;
 }
}

D347H

12 months ago, # |

Can someone explain why Dijkstra in G problem is not getting TLE verdict? We have n^2 vertices, for each vertex we traverse m edges, which should be at least O(n^2 * m) yet author's solution's passing with no probs. Where my logic gone wrong? Or I guess don't understand Dijkstra at all LOL

Not-Afraid

12 months ago, # ^ |

Dijsktra's time complexity, when using priority queue is O(E + VlogV), you can refer to the : proof here

nika-skybytska

This is a good question, and flamestorm or SlavicG should've presented a more elaborate complexity analysis in the editorial.

We have $$$V = 1000n$$$ vertices, but we do not traverse all $$$m$$$ edges for each vertex. Instead, each edge $$$(u, v, w)$$$ is traversed at most $$$2000$$$ times: once from each of the vertices $$$(u, 1), (u, 2), \ldots, (u, 1000)$$$ and $$$(v, 1), (v, 2), \ldots, (v, 1000)$$$. Hence, the total number of edges is $$$E \le 2000m$$$.

The time complexity of Dijkstra's algorithm as implemented in the editorial is $$$O((E + V) \log V)$$$, which for $$$n = m = 1000$$$ has an upper bound of roughly $$$60 \cdot 10^6$$$.

Wow, that was insightful, thanks!

abhimaurya-dev

In tutorial of 1915E what does the author mean by hash map can be hacked?

mahir1

Check this

sipsyong

Can someone solve the question about 1915G Problem's time complexity? I knew the time complexity of the dijkstra algorithm as O((V+E)logV). My question is that assuming n*1000 nodes, the number of edges will be (n*1000)^2? I need help with how the number of E is calculated.

yuchuan

How did the first example of G come out, and why did I calculate and simulate it as 23

mtpedro

11 months ago, # |

Why does this give TLE in F? Thanks.

Krishna_Vamsi_29

import java.util.*;

public class Main { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int cases = sc.nextInt(); while(cases-- > 0) { int glasses = sc.nextInt(); int[] juice = new int[glasses]; for(int i =0;i< glasses; i=i + 1) {

Why is this code giving wrong answer for problem E?

_Haki_Soul_Master

Someone help me out with Problem G of this contest. This is my respective solution 245688650. It is giving the wrong answer in test case 3.

variety-jones

Take a look at Ticket 17335 from CF Stress for a counter example.

Your current strategy is 1 --> 2 --> 1 --> 6 --> 7.

However, the optimal strategy is 1 --> 4 --> 1 --> 2 --> 1 --> 6 --> 7.

vinhhocptit

8 months ago, # |

In problem F how can we actually solve it without using order_of_key, with some implementations I did try to use lower bound it is still TLE. Here is my code https://codeforces.me/contest/1915/submission/261686634

arjunKunwar

2 weeks ago, # ^ |

Without using order_of_key

I solved this way , it might be helpful

https://codeforces.me/contest/1915/submission/239333098

Though with the help of order_of_key it is better to understand and easy to implement

minhvu0401

7 months ago, # |

I've used binary search in "Can I Square?" and it was overflow. What should I do?

supercookie

Is B wrongly tagged as bitmasks ? I think it shouldn't be. If you have an explanation for why it's bitmasks please tell me. I am very new to this.

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