Thanks for the fast editorial!

Why the approach for F with euler tour+Lazy prop is giving tle on tc 21? here is the code-https://codeforces.com/contest/1891/submission/230589277

any help will be appreciated.

I came up with the author's solution F, but I didn't have enough time to debug((

Как автор задачи E, мне жаль, что E и F были не в том порядке.

F can be solved without reversing the queries. It is offline though.

What does the "transition" mean in D?

For the B problem what i did was that i just added condition that any element in x should be taken one time but according to the editorial this condition can fail in some cases.So is my code correct https://codeforces.com/contest/1891/submission/230580104

achha contest

nice round for a beginner like me!!

Прощу прощения, можно ли найти где-то авторский код?

In C problem I did as proposed in editorial, but in one test case answers differ. Here is sequence of 14 hits from my implementation:

[1, 1, 2, 5, 6, 6]
=1h small stack=1 0/6
=2h small stack=1 1/6
=4h small stack=2 2/6
=6h big stack=5 4/6
crushing 6/6
=7h -> [3, 6] combo=0
[3, 6]
=10h small stack=3 0/6
crushing 3/6
=11h -> [3] combo=0
1 stack with 3 left. hit by 1
[3]=14h

Can someone provide a sequence to win with just 13 hits?

How to avoid tle for C https://codeforces.com/contest/1891/submission/230581501

Here is an alternative solution for F, using Fenwick Tree.

What was the reasoning for putting problem F at that position?

Anyone have an answer for D in python?

This is probably the easiest F I've ever seen :(

C can be implemented not using two pointers. But during the contest i forgot to check if n == 1 and a[i] == 1 :\

void solve() {
    scanf("%lld", &n);
    int s = 0;
    for (int i = 1; i <= n; i ++) {
        scanf("%lld", &a[i]);
        s += a[i];
    }
    if (n == 1 && a[1] == 1) {
        puts("1");
        return;
    }
    int sp = s / 2;
    sort(a + 1, a + 1 + n);
    int ss = 0, cnt = 0;
    for (int i = n; i >= 1; i --) {
        ss += a[i];
        cnt ++;
        if (ss >= sp){
            break;
        }
    }
    printf("%lld\n", cnt + s - s / 2);
}

I have a problem on C . In 230619057 the code id accepted , but in 230619275 , the code is wrong . The only difference between them is merely in "lower_bound(pres,prew+n+1)" or "lower_bound(pres+1,pres+n+1)" . But the pres is the prefix sum of a which indicates that the sum divided by 2 (floor,except for n = 1) must be positive . So "+1" should not make a difference to my solution . I am quite confused . Please help we .

Can someone please check out why my submission for Problem D is giving TLE?

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define nl '\n'
#define all(v) v.begin(), v.end()
#define IOS ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
ll const mod = 1000000007;
ll inf = 1000000000000000000;

int iter = 0;
ll binexp(ll a, ll b){
    ll ans = 1;
    while (b){
        iter++;
        if(b % 2){
            if(ans>inf/a) return inf+5;
            ans = (ans * a);
        }
        if(a>inf/a) 
            a = inf+5;
        else 
            a = (a*a); 
        b /= 2;
    }
    return ans;
}
ll LOG(ll x, ll p=2){
    if(p == 1) return 0;
    for(ll t=1, i=0;; t*=p, i++){
        iter++;
        if(t > x) return i-1;
        if(t > inf / p) return i;
    }
}
ll func(ll x){
    ll res = 0, L = 4, R, p, val;
    while(L <= x){
        iter++;
        p = LOG(L);
        val = LOG(L, p);
        R = min({(1ll<<(p+1))-1, binexp(p, val+1)-1, x});
        // cout << L << ' ' << R << nl;

        res += (val * ((R - L + 1) % mod)) % mod, res %= mod;
        L = R+1;
    }
    return res;
}
void Heisenberg(){
    ll q; cin >> q;
    while(q--){
        ll l, r; 
        cin >> l >> r;
        cout << (func(r) - func(l - 1) + mod) % mod << '\n';
        // cout << iter << '\n';
    }
}
signed main(){
    auto begin = std::chrono::high_resolution_clock::now();
    IOS
    Heisenberg();
    auto end = std::chrono::high_resolution_clock::now();
    auto elapsed = std::chrono::duration_cast<std::chrono::nanoseconds>(end - begin);
    cerr << "Time measured: " << elapsed.count() * 1e-9 << " seconds.\n"; 
}

Approach: Lets say $$$y = \log_{\log_2(x)}(x)$$$ then the value will remain same till $$$tempR = \min(R, 2^{\log_2(x)+1}-1, \log_2(x)^{y+1}-1)$$$. So I will keep updating answer as $$$ answer += y \cdot (tempR - x + 1)$$$ $$$x = R + 1$$$.

I think swap E and F is a good idea

Can someone please tell me why is my submission for Problem D giving TLE?

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define nl '\n'
#define all(v) v.begin(), v.end()
#define IOS ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
ll const mod = 1000000007;
ll inf = 1000000000000000000;

int iter = 0;
ll binexp(ll a, ll b){
    ll ans = 1;
    while (b){
        if(b % 2){
            if(ans>inf/a) return inf+5;
            ans = (ans * a);
        }
        if(a>inf/a) 
            a = inf+5;
        else 
            a = (a*a); 
        b /= 2;
    }
    return ans;
}
ll LOG(ll x, ll p=2){
    if(p == 1) return 0;
    for(ll t=1, i=0;; t*=p, i++){
        if(t > x) return i-1;
        if(t > inf / p) return i;
    }
}
ll func(ll x){
    ll res = 0, L = 4, R, p, val;
    while(L <= x){
        p = LOG(L);
        val = LOG(L, p);
        R = min({(1ll<<(p+1))-1, binexp(p, val+1)-1, x});
        // cout << L << ' ' << R << nl;

        res += (val * ((R - L + 1) % mod)) % mod, res %= mod;
        L = R+1;
    }
    return res;
}
void Heisenberg(){
    ll q; cin >> q;
    while(q--){
        ll l, r; 
        cin >> l >> r;
        cout << (func(r) - func(l - 1) + mod) % mod << '\n';
    }
}
signed main(){
    auto begin = std::chrono::high_resolution_clock::now();
    IOS
    Heisenberg();
    auto end = std::chrono::high_resolution_clock::now();
    auto elapsed = std::chrono::duration_cast<std::chrono::nanoseconds>(end - begin);
    cerr << "Time measured: " << elapsed.count() * 1e-9 << " seconds.\n"; 
}

This round was too easy, though people still couldn't solve problems, I don't know why. I managed to solve D in time and saw my rank go way up. A handful of 20 minutes was still left for the contest but I convinced myself I wouldn't be able to solve E. I saw E later after the contest ended and it was the easiest Division 2 E I've ever seen.

Mathforces

why is E is much difficult than F. Or why is F much easier than E.

Maybe the simplest implementation for C:

sort(a+1,a+1+n);
for(int i=1;i<=n;i++) sum[i]=sum[i-1]+a[i];
int tmp=(sum[n]+1)/2;
int c=0;
for(int i=1;i<=n;i++) if(sum[i]>tmp) c++;
printf("%lld\n",tmp+c);

can anyone explain why the greedy solution works in C? also for the case when i==j and the last number is an odd number and x=0, then there is no way we can use the 2nd method on this horde

I came up with using offline query in problem F. My idea is first save the query and build the completed tree which is presented in the end. Then, I just need to loop from q to 1 and update normally using Euler's tour when type 2 is meet otherwise if type 1 is the current query, output it's node value. But unfortunately, I got WA immediately at 2nd test case although this idea is kind of nature (or maybe it's wrong in some case), anyone suggests me why am i wrong here?

My submissions: 230648660

can anyone help in f...getting wa on 2..230657380

How the total complexity for problem c is O(nlogn) ,

it should be O(n) .

#

Can someone give a simple idea about F? Also may I know what are the prerequisites to learn to solve problem F?

Is there any online solution for F?

In problem D

and on each segment there are at most O(logn) transitions

shouldnt there be like at most 2 transitions per segment because if we make 3 transitions that would mean jump to next segment?

My logic is similar to one in editorial but I am having issues with MOD. What changes should be done in my code ?

https://codeforces.com/contest/1891/submission/230655160

Also what are some good sources for topics like modular arithmetic and other topics that will help me remain stable expert. Thanks

Hi Codeforces, this was my second ever contest and this time i was able to solve problem A my submission: 230544275 i found this to be interesting because it was difficult for me to comprehend what the question was asking, then i just read the problem and analyzed the sample testcases for about 15-20 mins, I found that if the index that is not sorted (a[i]>a[i+1]) lies in the powers of 2(0,1,3,7,15) then it is possible to sort it otherwise not, I just implemented this and got it through. I was elated after the first submission passed the pretests so my confidence boosted for the second problem, I just implemented a brute force solution 230561923 and it passed the sample cases so I decided to submit it, Obviously it gave a TLE at pretest 3 but overall it was a great experience, looking forward to solving more and growing. Please if possible, share your helpful feedbacks on my solutions. happy coding!!

For problem C I kept on overthinking that the solution would involve DP, however I did notice that the constraints on $$$x$$$ was too big and that it would most probably lead to TLE. However, can anyone think if DP is feasible for problem C if $$$x$$$ was actually some small number?

Out of curiosity following problem F , how will this problem be solved? Initially i have only one vertex which is numbered 1.There are 1e5 qeuries where in one type of query i can add a vertex to the tree with number sz+1 (sz is the no of nodes in the tree currently).Can i answer other type of query where i need to tell the size of subtree rooted at a given vertex in logn time or what will be the most optimized algorithm for this.

For Problem B, I wrote my code on Python, and I am getting correct answers when I run it on my system, but wrong answers when I run on CF. Here is my code — https://codeforces.com/contest/1891/submission/230716694

I understood the editorial algorithm for C, but can someone help me with why the greedy algorithm is correct? I am unable to prove it. Thanks alot!

PROBLEM D

Here is the implementation of sqrt decomposition for F. It is giving TLE as warned by author. Any optimisation in provided code is welcomed.

For D problem can someone provide me simple explanation ? (Even its implementation is complicated even tried) https://codeforces.com/contest/1891/problem/D

is there a way to solve problem c with binary search

Can F be solved online?

https://codeforces.com/contest/1891/submission/230866197

can anyone help me with problem F

i cant find any error

https://codeforces.com/contest/1891/submission/230866197

can anyone help me with problem F? cant find any error

I think it is better with spoilers, and code.

Can someone explain the solution of problem B? I didn't understand the idea of the editorial.

Thank you, 127.0.0.1

127.0.0.1 I am having a little problem in D. Its working on nearly all the test cases and logic is to point. But I think there is some problem with modulus and its giving wrong answer on one test case-v 179 1000000000000000000 Can any body help I cant seem to figure it our. This is my code where maxi is modulus ll l,r; cin>>l>>r;

ll te=l;
 ll ans=0;
 while(te<=r)
 {
    ll p=log2(te);
    ll up;
    // if(p==63)
    // up=r;
    // else
    up=min((ll)(pow(2,p+1)-1LL),r);
    //cout<<up<<endl;
    ll ct=0;
    ll x=1;

   while(x*p<=te)
   {
      x*=p;
      ++ct;
   }
  // cout<<te<<" "<<x<<endl;
   while(true)
   {
    ll nx;
    //cout<<x*p<<e ndl;
    x*=p;nx=min(x,up+1);
    ll nxm=nx%maxi;
    ll  tem=te%maxi;
     ans=(ans+(ct*((nx-te)%maxi))%maxi)%maxi;
     if(nx>up)
     break;
     ct=(ct+1)%maxi;
      te=nx;
   }
   if(up==r)
   break;
   te=up+1;



 }
 cout<<ans<<endl;

https://codeforces.com/contest/1891/submission/231287946[submission:231287946] why i am getting tle in this approach can anyone pls help

can someone explain for me more about how to solve problem D , please ?

in problem b i have brute forces just first 100 element and i got ac why thats true?

can someone help me,i am getting WA on test 2 of F https://codeforces.com/contest/1891/submission/232563023

Did anyone try (probably the dumbest) solution to F by reversing queries and doing segtree with range updates on the euler tour of the final tree? That's the first thing that comes to mind

I have a problem in D. I got a wrong anwer on test 8, and I have trouble finding where I went wrong. My submission is Your text to link here... . Can someone help me figure out what the problem is? Thanks a lot if you can help!

Can anyone please tell me if the error is in the steps of modular operations or in the map in the following solution. It would be a big help. My submission for Problem D

I am quite stuck on the 1891D task. I have checked the solution explanation, and I can't figure out one key element. The explanation says, that on each part with f(i)==f(i+1) we will have at max O(logn) transitions, such that g(i)!=g(i+1).
My question: how can g(i)!=g(i+1) be on the segment where f(i)==f(i+1), when f(x)=log2(x), while g(x) = logf(x)x, which means that it's base is bigger than 2, and that it should change its values significantly more rare that log2(x).

Helpfull

In Problem C, can someone explain why we are taking the attack iteratively rather than the subarray smaller to the largest horde (in this case the current j pointer)? Why not distributing attacks (of type I) on the remaining hordes beneficial since we can keep the largest hordes larger? and also, why are we minimizing the attacks of type II, wouldn't they be more effective if use them when the combo is higher than 1?