Fun to see no constructive algorithms this time XD

Please make more rounds like this one, they make me feel a bit better about myself

can someone explain the idea of problem E with DP pls, I've seen many solutions with an array like f[N][3] or f[N][N][3], can anyone explain the approach pls.

I really liked problem E, the solution is very interesting for me(also like a problem)

In the problem E's editorial while calculating the answer for prefix, the substring notations used were t[0;m-suf] and s[0;n-pos], then acc to editorial we have to build the z-function on $$$s[0;pos)^{-1}$$$ + # + $$$t^{-1}$$$, shouldn't it be n-pos instead of pos?

Video Solutions of Problem C and Problem D.

Btw F can be solved using SQRT decomposition on queries in $$$O(NsqrtN)$$$

submission

One way of doing F without storing $$$f(i)^2$$$ would be to observe that $$$\frac{(f(i)+1)f(i)}{2} - \frac{f(i)(f(i)-1)}{2} = f(i)$$$. In other words, if a point is added in its range, the change in its contribution to the answer is simply $$$f(i)$$$. Thus, for each point $$$i$$$, we can store only two values: whether it is currently in the set, and what its value of $$$f(i)$$$ is.

To change the answer when a point $$$j$$$ is added, it suffices to query the sum of $$$f(i)$$$ over the interval $$$[j-d,j)$$$ and to calculate the value of $$$f(j)$$$ itself by querying how many points in $$$(j,j+d]$$$ are in the set. Removing a point works similarly.

I have a solution to problem C in linear time.163331694

There's an $$$\mathrm{O}(q\log q)$$$ solution for Problem F with balanced search trees (Splay, Treap, etc).

We may consider new/deleted beautiful triples when point $$$x$$$ is added/deleted from the set as three parts, in which:

• $$$x$$$ is treated as the right endpoint;
• $$$x$$$ is treated as the left endpoint;
• $$$x$$$ is caught in the middle.

It's quite simple to calculate the first two types, since we may find the number of $$$y$$$'s that satisfy $$$y \in (x-d, x)$$$ or $$$y \in(x, x+d)$$$, then add two arithmetic progressions of common difference $$$1$$$. But we find it not very easy to add the number of type 3 triples instantly. So it's necessary to maintain 'if we add/delete point $$$x$$$ from the set, the number of $$$x$$$-as-the-middle triples being added/deleted' in some way.

Let's discuss how many triples will turn invalid when we remove a point from the set. One may notice that if we're adding $$$x$$$ to the set $$$S$$$, then for each $$$y \in S, y \in (x-d, x)$$$, for every point $$$z$$$ satisfying $$$z \in [x-d, y), z \in S$$$, $$$(z, y, x)$$$ will be a new beautiful triple in which $$$y$$$ sandwiching in between $$$x, z$$$. Let $$$c_y$$$ be the number of such $$$z$$$'s for each $$$y$$$. Obviously if we consider all valid $$$y$$$'s in increasing order (denoting $$$x-d<y_1<y_2<\cdots<y_k<x$$$), then $$$c_{y_i}=i-1$$$. It's similar for $$$y \in (x, x+d)$$$. Thus we may maintain set $$$S$$$ with a binary search tree, and add arithmetic progressions on a specific segment — which is a classic trick — when adding an element to $$$S$$$. Once we want to remove a point $$$x$$$ from the set, substract the number of first 2 types along with the count of type 3 triples from the answer, which we've stored on the BST. Also, its deletion will lead to a negative contribution of $$$c_{y_i}$$$ for all valid $$$y$$$'s mentioned above, so don't forget to add the opposite number of a progression on the BST maintaining $$$S$$$.

The next part is considering 'if $$$x$$$ is not in the set, how many $$$x$$$-in-the-middle triples will turn valid after adding it'. Suppose we're adding $$$x$$$ to the set $$$S$$$. We still consider $$$x-d<y_1<y_2<\cdots<x$$$; then for each $$$i \in \lbrace{1,2,\cdots,k-1}\rbrace$$$, for all $$$z \in (y_i,y_{i+1})$$$, $$$(y_j,z,x), j \in \lbrace{1,2,\cdots,i}\rbrace$$$ will be new beautiful $$$z$$$-in-the-middle triples if $$$z$$$ is added after $$$x$$$. It's obvious that for all $$$z$$$ in the same segment whose endpoints are adjacent $$$y$$$'s, the numbers of potential new valid triples are the same. Similarly, for each segment, the sequence of new potential valid triples is an arithmetic progression of common difference $$$1$$$. Thus we may use another BST to maintain all segments $$$(l,r), l\in S, r\in S, \nexists p \in S, p \in (l,r)$$$ and proceed the sum of progressions. If we want to add $$$x$$$ to set $$$S$$$, then add the sum of contributions of type 1 & 2 triples and the segment's to the answer, split the segment it belongs to to two (empty segments should be preserved), and do operations above.

When removing $$$x$$$ from $$$S$$$, we may first add the opposite number of contributions of $$$x$$$ on two BST. What about merging two segments on the second BST? Do they have the same potential new triples after $$$x$$$ is deleted? The answer is YES. Apparently, for $$$x,y$$$ in segment $$$(l, r)$$$, if $$$(p, x, q), p\leq l, q \geq r$$$ is a valid potential triple, so does $$$(p, y, q)$$$ and vice versa. So it's ok to set the contribution of new $$$(l, r)$$$ (in which $$$x$$$ has just been removed) as $$$(l, x)$$$'s.

A careful and efficient implementation is required for this solution. Submission #163416351

Sorry for any grammar/vocabulary mistakes above.

anyone please tell me why i am getting TLE on test case 4 in problem C . 163421348

edit — the problem accepted when i used priorty_queue instead of multiset 163422803

Let's calculate value of $$$\sum f(i)$$$ in every node (it's simple). If we know this sum, we can restore value of $$$\sum f^2(i) - f(i)$$$. Let's say, we already somehow calculate it after some query: $$$F(i) = \sum f^2(i) - f(i)$$$.

If we add $$$k$$$ to all $$$f(i)$$$ we need to update our $$$F(i)$$$ value. Now the right value must be equal to $$$\sum (f(i) + k)^2 - (f(i) + k)=\sum f^2(i) + k^2 + 2 \cdot kf(i) - (f(i) + k)=\sum f^2(i) - f(i) + \sum k^2 + 2 \cdot kf(i) -k$$$.

$$$\sum f^2(i) - f(i)$$$ we already know, it is $$$F(i)$$$. Let's find $$$\sum k^2 + 2 \cdot kf(i) - k$$$. Value of $$$\sum 2 \cdot k f(i)$$$ we know too, because we store value of $$$\sum f(i)$$$ in every node. Value of $$$\sum k^2 - k$$$ we can calculate if we know how many points are now "active" in our node.

So now we get $$$F(i) = F(i) + 2 \cdot k \cdot \sum f(i) + cnt \cdot (k^2 - k)$$$; $$$f(i) = f(i) + cnt \cdot k$$$, where $$$cnt$$$ equals to number of active points.

good round

In Problem 2, whats the difference between these two codes ?? Wrong Code- void solve() { int n;cin>>n: cout<<2<<endl;//d will be always 2

for(int i=1;i<(n+1);i++){
    if(i%2==1){
       cout<<i<<" ";
       while(i*2<=n){
         cout<<i*2<<" ";
       }
    }
    else continue;
}
cout<<endl;

}

Right Code- void solve() { int n;cin>>n; cout<<2<<endl;//d will be always 2

for(int i=1;i<(n+1);i++){
    if(i%2==1){
       int j=i;
       while(j<=n){
         cout<<j<<" ";
         j*=2;
       }
    }
}
cout<<endl;

}

In problem D, sweep line is not needed. Simply sorting the range in increasing order of ending point is enough. AC code: https://codeforces.net/contest/1701/submission/163624301

It would be nice if you added some comments to the solution code for F.

Problem C is quite similar to 781C — Tree Infection, and can similarly be solved by simply simulating the entire process.

Alternative solution for C:

We just simulate the whole process, second by second. We always try to allocate the current task to the most efficient worker. If there are any free workers (they do not have any more good tasks for them), then we have to take a task from the worker with the most tasks. This suggests that we need to use a treeset for fast insert / delete in log(n) time.

The code goes like this: while we have tasks remaining, we let all the busy workers do their task first, then for the free workers, we transfer the task from the worker with the most tasks to it.

We notice that to make the free worker do a task for 2 seconds, we just skip their turn and transfer the task from the busy worker to the free worker. So the next moment that same worker will be busy and will do the task.

Submission: https://codeforces.net/contest/1701/submission/207853000