satyam343's blog

By satyam343, 2 years ago, In English

Thank you for participation! We apologize for problem D that turned out to be harder than expected. Still, we hope that you liked most of the problems.

In case you found C2 tedious to implement or found many cases to deal with, I would recommend you to have a look at the intended solution. I think it is interesting and easy to implement.

1736A - Make A Equal to B

Solution
Code

1736B - Playing with GCD

Solution
Code

1736C1 - Good Subarrays (Easy Version)

Solution
Code

1736C2 - Good Subarrays (Hard Version)

Solution
Code(Offline queries)
Code(Online queries)

1736D - Equal Binary Subsequences

Solution
Code

1736E - Swap and Take

Solution
Code
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2 years ago, # |
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In C1, $$$dp[i]$$$ can be simply interpreted as the number of good sub-arrays that end at index i. The code is precisely the same.

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2 years ago, # |
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Please update the contest announcement page and the contest page to reflect the editorial, some people could not find this and are waiting for it!

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    2 years ago, # ^ |
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    i have never seen a post without your comments and shares (LOL) keep going and good luck.

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2 years ago, # |
Rev. 2   Vote: I like it -18 Vote: I do not like it

Why do editorialists still do this? Making it a challenge to read problem solutions without spoiling solutions of other problems that one hasn't solved. Do they not know of the existence of the spoiler option?

As I read the editorial for problem C, I now have an extra challenge to avoid any sneak peak/peripheral vision from wandering by chance/accident to solution for B and C2, both of which lie directly above and below resp. to the solution to the problem I'm reading.

It's pathetic. I wish editorialists weren't this lazy.

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2 years ago, # |
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Is there any better explanation for problem B?

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    2 years ago, # ^ |
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    Since a2=gcd(b2,b3), both b2 and b3 has to be divisible by a2 Similarly both b3 and b4 has to be divisible by a3 So the common element which has to be divisible by both a2 and a3 is b3 That's why b3=lcm(a2,a3) Now that you have b3 just check if gcd(b2,b3) is equal to a2 or not. If it is not then answer is "no". Otherwise continue till n

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      2 years ago, # ^ |
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      Why can't b3 = a2*a3? Why will this assumption fail?

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        2 years ago, # ^ |
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        Let's start with an example... We have an array like 4 3 2 , According to your assumption. array a = 4 3 2 array b = 4 12 6 2 Here you can see a1 is gcd of b1&b2 but a2 & a3 case fail cause gcd of b2 & b3 is not equal to a2 and similar to a3. Hope you understood.

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      2 years ago, # ^ |
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      Your explanation is greater than the editorial. thanks:)

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    2 years ago, # ^ |
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    Given array a1 a2 a3

    Let the output array as

    b1 b2 b3 b4

    So gcd(b1, b2) = a1

    gcd(b2, b3) = a2

    gcd(b3, b4) = a3

    i.e b1 = x1 * a1

    b2 = x2 * a1

    Such that gcd(x1, x2) = 1

    b2 = x3 * a2

    b3 = x4 * a2

    Such that gcd(x3, x4) = 1

    b3 = x5 * a3

    b4 = x6 * a3

    Such that gcd(x5, x6) = 1

    Now

    x2 * a1 = x3 * a2

    x2/x3 = a2/a1

    => x2 = a2 / gcd(a1, a2)

    => x3 = a1 / gcd(a1, a2)

    Similarly

    x4 * a2 = x5 * a3

    x4/x5 = a3/a2

    => x4 = a3 / gcd(a2, a3)

    => x5 = a2 / gcd(a2, a3)

    X1 and x6 can be any prime number.

    Now check whether gcd (x1, x2) = gcd(x3, x4) = gcd(x5, x6) = 1 and if we have passed this criteria the answer will be yes otherwise no.

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      23 months ago, # ^ |
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      From x2/x3 = a2/a1 how you were able to conclude that => x2 = a2 / gcd(a1, a2)

      => x3 = a1 / gcd(a1, a2)

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2 years ago, # |
Rev. 6   Vote: I like it 0 Vote: I do not like it

C1 binary search approach:

Suppose we have a range [L,R] which isn't a good subarray.

For this range, the worst element is the element a[j] for which a[j]-j is too low (minimum on the range [L,R]).

So we can have another array b for which b[i]=a[i]-i, then we build a Sparse Table over array b. Now we can iterate over all indices of array a, and binary search the farthest length of good subarray [i,R] for all 1<=i<=n.

And because, whatever the minimum b[j] on range [i,R] is going to be, it still shares the same prefix i-1 with all the elements on the range [i,R], then we can check if query(i,R)+i-1 is bigger than or equal to zero, or not, to edit our searched range.

The binary search is going to be like this:

for(int i=1 ; i<=n ; i++)
{int L=i,R=n,h=i;
while(R >= L)
{int mid = R-(R-L)/2;
if (query(i , mid) + i-1 >= 0) {h = mid; L = mid+1; continue;} R = mid-1;}
ANS += h-i+1
}

Here is my submission: https://codeforces.me/contest/1736/submission/175452764

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2 years ago, # |
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shit

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2 years ago, # |
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I think D and C2 indeed had the correct difficulty for the 4th and 5th problem; the only issue was their positions that confused participants. I'm pretty sure that if participants spent more time on D than C2, there would have been more solves for D.

satyam343, thanks for amazing problems and please fix broken \t{good} in the editorial of C1.

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    2 years ago, # ^ |
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    Explanation of c2 with few examples would be helpful.

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    2 years ago, # ^ |
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    Fixed it.

    I am glad that you found the problems interesting.

    C2 was intended to be harder than D. That is why we kept score of C1+C2 high.

    Surprisingly we had predicted that this round would be easier than usual Div 2 rounds.

    My expected ratings(after testing) were :

    A — 800, B — 1200, C1 — 1400, C2 — 2100, D — 1900 and E — 2200

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2 years ago, # |
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Am I the only one who receives RTE on test 4 for no reason? I tried modifing it with NMAX to 1e3, but it's a long long issue. If I change it in int it receives WA

https://codeforces.me/contest/1736/submission/175448716

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2 years ago, # |
Rev. 2   Vote: I like it -9 Vote: I do not like it

Problem C1 ( Good Subarrays ) Easy O(N) Solution:

Code
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    2 years ago, # ^ |
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    some explanation ??

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      2 years ago, # ^ |
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      While traversing in the array, I sum the minimum of the index and the element i.e. min(i+1, a[i]) {i+1, because of 1-indexing)
      But in this the problem is that if we encounter a number which is smaller than its index, then it gives wrong answer.
      E.g. 4 3 1 2 8
      While counting for the element 8, it gives min(5, 8) =5, however it should give 3 only, because a smaller element 2 is present ahead it.
      So, everytime I encounter such small element, I reduce the effective array by "j" length from starting, which is equal to the difference of index and the element.
      So now, when 8 is encountered, the array is thought of 1 2 8, and hence min(3,8) = 3 is added.

      Please upvote if helpful, though I'm not quite good in explaining :'(

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2 years ago, # |
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E is pretty nice, it uses a very similar dp as this problem. link

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2 years ago, # |
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For problem C1, I have a solution using Segment Tree to maintian the maximum value of a segment. share it. click here

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2 years ago, # |
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My solution for C1:

Claim: If $$$a[l], a[l+1], ..., a[r]$$$ is a good subarray, then , for all $$$x, y$$$, such that $$$l \le x \le y \le r$$$, we have that $$$a[x], a[x+1], ... a[y]$$$ is a good subarray.

With that claim in mind, let's make a two pointers approach. For all $$$l, r$$$ that the interval $$$[l, r]$$$ is good, we have $$${r-l+2}\choose 2$$$ sub-intervals of $$$[l, r]$$$ that is good. So, lets make the following algorithm:

  • Start with $$$l = 1$$$ and $$$r = 1$$$, and $$$antr = 0$$$ ($$$antr$$$ is the last $$$r$$$ that we compute for an answer).
  • If $$$a[r] \ge r-l+1$$$ and $$$a[r+1] < r-l+2$$$, it means that $$$[l, r]$$$ is good and $$$[l, r+1]$$$ is not, then, lets add to our answer $$${r-l+2} \choose{2}$$$ $$$-$$$ $$${antr - l + 2}\choose{2}$$$, then make $$$antr = r$$$ and $$$l++$$$. See that, if $$$[l, r]$$$ is a solution, then $$$[l+1, r]$$$ also is, but we are counting $$$[l+1, r]$$$ two times, so, to avoid that, we are going to add a value for an answer if the $$$r$$$ is different. If it is the same, just do $$$l++$$$ and continue the algorithm (That means, don't update the answer and don't update $$$antr)$$$.
  • Else, just make $$$r++$$$

Code: link

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2 years ago, # |
Rev. 4   Vote: I like it +20 Vote: I do not like it

I think I have a simpler code for C2, in which I don't use any complex data structures except simple arrays.

#include <bits/stdc++.h>
using namespace std;

int n;
vector<int> arr, left_most, right_most, right_second_most;
void get_bound() {
    // left_most[i] represents the leftmost point such that subarray arr[left_most[i]:i] is good
    left_most.resize(n + 1);
    // right_most[i] represents the rightmost point such that subarray arr[i:right_most[i]] is good
    right_most.resize(n + 1);
    // right_second_most[i] is similar with right_most[i]
    // but when you start from i, and go right, you have one chance to skip bad point.
    right_second_most.resize(n + 1);
    for (int l = 1, r = 1, r2 = 1; l <= n; l++) {
        while (r <= n and arr[r] - r >= 1 - l) {
            left_most[r] = l;
            r++;
        }
        right_most[l] = r - 1;
        r2 = max(r2, min(r + 1, n + 1));
        while (r2 <= n and arr[r2] - r2 >= 1 - l) {
            r2++;
        }
        right_second_most[l] = r2 - 1;
    }
}

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cin >> n;
    arr.resize(n + 1);
    for (int i = 1; i <= n; i++) cin >> arr[i];
    get_bound();

    // S1: presum of contributions of all points
    // S2: presum of contributions of all points, if each of them has one skip chance.
    std::vector<long long> S1(n + 1);
    std::vector<long long> S2(n + 1);
    for (int i = 1; i <= n; i++) S1[i] = S1[i - 1] + (right_most[i] - i + 1);
    for (int i = 1; i <= n; i++) S2[i] = S2[i - 1] + (right_second_most[i] - i + 1);
    long long total = S1.back();

    int m;
    cin >> m;
    for (int i = 0; i < m; i++) {
        int p, x;
        cin >> p >> x;
        if (x == arr[p]) {
            cout << total << '\n';
        } else if (x < arr[p]) {
            int now_left_most = max(left_most[p], p - x + 1);
            if (now_left_most == left_most[p]) {
                cout << total << '\n';
                continue;
            }
            // now_left_most > left_most[p]
            // we can cut off contributions from left_most[p] to now_left_most-1.
            // but, they are not cut off completely. they can still reach p-1.
            long long cut_off = S1[now_left_most - 1] - S1[left_most[p] - 1];
            long long remain = (long long)((p - now_left_most) + (p - left_most[p] + 1)) * (now_left_most - left_most[p]) / 2;
            cout << total - cut_off + remain << '\n';
        } else {
            int now_left_most = max(int(lower_bound(right_most.begin() + 1, right_most.end(), p - 1) - right_most.begin() - 1) + 1, p - x + 1);
            if (now_left_most == left_most[p]) {
                cout << total << '\n';
                continue;
            }
            // now_left_most < left_most[p]
            // this is the time to use skip chance!
            // for all those points that stuck at p, they can skip p and reach right_second_most
            long long old_sum = S1[left_most[p] - 1] - S1[now_left_most - 1];
            long long now_sum = S2[left_most[p] - 1] - S2[now_left_most - 1];
            cout << total - old_sum + now_sum << '\n';
        }
    }
}
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2 years ago, # |
Rev. 2   Vote: I like it 0 Vote: I do not like it

Here is an explaination for B I came up with Let a = {a1,a2,a3} b={b1,b2,b3,b4} Since a1 = gcd(b1,b2) => b1 and b2 are multiple of a1 similarly b2 and b3 are multiple of a2

Since b2 is multiple of a1 & a2 => b2 = k1 * lcm(a1,a2) (multiple of lcm) since b3 is multiple of a2 and a3 => b3 = k2 * lcm(a2,a3)

Now a2 = gcd(b2,b3) = gcd(k1*lcm(a1,a2) , k2*lcm(a2,a3)) = GCD GCD will be decided by lcm and multiple factors k1&k2 k1 and k2 can be used to multiply an integer to the gcd(lcm(a1,a2) , lcm(a2,a3))

From the above observation a[2] must be divisible by the GCD (we can set k1 and k2 to reach a[2])

Coulnt do this in contest :(

Link to submission

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    2 years ago, # ^ |
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    "From the above observation a[2] must be divisible by the GCD (we can set k1 and k2 to reach a[2])" i'm sorry i'm not able to understand this line

    i get it it's about finding k1 and k2 ....but i'm confused can you explain ?

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      2 years ago, # ^ |
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      see gcd(k1*lcm(a1,a2) , k2*lcm(a2,a3)) will always have the gcd(lcm(a1,a2),lcm(a2,a3)) rest will be contribute by gcd(k1,k2) the contribution from both the terms must take us to a2

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2 years ago, # |
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I code part of c1 why declaring dp with size n + 5 when n + 1 can also do the work.

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    2 years ago, # ^ |
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    It's a good practice to make your array sizes a bit larger than what you expect their maximum size to be. There isn't much downside to this, since the extra few slots in takes negligible time to declare and it shouldn't affect the correctness of the code if it is based on the actual number of elements (as opposed to the max size).

    There are some problems where you discover later that you might need the extra indices, e.g., due to switching between 0-indexing and 1-indexing, or because the nature of your solution to the problem would benefit from stepping further ahead, etc, and it's more convenient to mindlessly declare it as a larger size every time than to rely on consciously adjusting the declaration when you need to, which you might forget (like if you were really deep into writing some part of the solution at the time) or miss (like if there are multiple such arrays affected and you overlook one).

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2 years ago, # |
Rev. 2   Vote: I like it +40 Vote: I do not like it

Waiting for 1736E — Swap and Take Editorial!

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2 years ago, # |
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Can anyone explain me what this code is doing ?

set<ll> found; found.insert(n+1);  
    sort(track.begin(),track.end()); 
    for(auto it:track){
        if(it.s.s){
            ll r=*found.upper_bound(it.s.f);
            ans[it.s.s]=pref[it.s.f-1]+dp[r]+sum[r-it.s.f]+(it.f+it.s.f-1)*(r-it.s.f);
        }
        else{
            ll r=*found.lower_bound(it.s.f); found.insert(it.s.f);
            dp[it.s.f]=dp[r]+sum[r-it.s.f]+(it.f+it.s.f-1)*(r-it.s.f);
        }
    }
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2 years ago, # |
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In problem C2, what is the offline approach?

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2 years ago, # |
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Regarding problem C2 : how to find such q ?

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    2 years ago, # ^ |
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    Suppose $$$adp[p]=z$$$.

    So $$$q$$$ is the smallest index greater than $$$p$$$ such that $$$a_q \leq z + q - p$$$.

    Now $$$a_q-q \leq z-p$$$. Note that $$$z-p$$$ is fixed. So you can make a new array $$$b$$$ such that $$$b_i=a_i-i$$$. You can make $$$b$$$ before reading queries. You can use segment tree to find $$$q$$$ now. In the intended solution, I am processing the values in sorted order. Thus this way I was able to avoid segment tree.

    Edit: I have included segment tree solution in the editorial.

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2 years ago, # |
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In the C1 Editorial, what do you mean by a[i−dp[i−1],i−1] and a[i−dp[i−1],i]? Array a is 1D array. But you used comma between the calculation of the array index.

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    2 years ago, # ^ |
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    $$$a[l,r]$$$ represents the subarray $$$a_l,a_{l+1}, \dots a_r$$$.

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2 years ago, # |
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In C1, how can we solve it using two pointers, why is this approach failing :

l = 0, r = 0
ans = 0
while(r < n)
   if s[r] >= (r -l + 1):
      r += 1
   else:
      length = r - l
      ans += length * (length-1) / 2
      while (a[r] < (r - l + 1)):
        l += 1
length = r - l
ans += length * (length - 1) / 2
print(ans) 
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2 years ago, # |
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Is the algorithm mentioned in the editorial for C2 always correct?

Edit after reply: Yes, it is; I just misunderstood the track[i] definition :p

Take n = 10 and a = [8, 8, 8, 8, 8, 8, 1, 8, 8, 8].
Consider the query: p = 7, x = 10
The new a would be b = [8, 8, 8, 8, 8, 8, 10, 8, 8, 8].

dp[i] = min(dp[i-1] + 1, a[i])
adp[i] = min(adp[i-1] + 1, b[i])

dp:
dp[1] = 1
dp[2] = 2
dp[3] = 3
dp[4] = 4
dp[5] = 5
dp[6] = 6
dp[7] = 1
dp[8] = 2
dp[9] = 3
dp[10] = 4

adp:
adp[1] = 1
adp[2] = 2
adp[3] = 3
adp[4] = 4
adp[5] = 5
adp[6] = 6
adp[7] = adp[p] = 7
adp[8] = 8 = a[8]
adp[9] = 8 = a[9]
adp[10] = 8 = a[10]

Let q be the smallest index greater than p such that adp[q] = a[q]. So q = 8.

Indeed, adp[1] + adp[2] + ... + adp[p-1] = dp[1] + dp[2] + ... + dp[p-1].

However, adp[q] + adp[q+1] + ... + adp[n] = adp[8] + adp[9] + adp[10] = 24,
while dp[q] + dp[q+1] + ... + dp[n] = dp[8] + dp[9] + dp[10] = 9.
So, $$$\sum_{\ i=q}^{\ n} adp[i] \neq track[q]$$$!

To further increase my confusion, I ran the code for offline queries with this data and it gave me the right answer. (52)

Is there an error in the solution written in words? Is there an error in my understanding of the editorial? And how is the offline queries code different than the solution in words? (e.g. what is "dp" in the offline queries code, if "len[i] = min(len[i-1]+1, a[i])"?)

Any help would be appreciated!

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    2 years ago, # ^ |
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    Here's how you calculate $$$track[8]$$$.

    First assume $$$dp[8]=a[8]$$$.

    Now $$$dp[9]=min(dp[8]+1,a[9])=8$$$ and $$$dp[10]=min(dp[9]+1,a[10])=8$$$.

    So $$$track[8]=dp[8]+dp[9]+dp[10]=24$$$ if $$$dp[8]=a[8]$$$.

    For confusion between $$$len$$$ and $$$dp$$$, I have edited the code.

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      2 years ago, # ^ |
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      Ah, I had misunderstood the assumption part — I thought track[i] is set to 0 if dp[i] != a[i] and track[i] = dp[i] + dp[i+1] + ... + dp[n] if dp[i] == a[i]... Thanks a lot!

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2 years ago, # |
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For C2, Could you please explain how to find q both online and offline.

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    2 years ago, # ^ |
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    I have figured it out now. But another thing is, how to precalculate the track[i]

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      2 years ago, # ^ |
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      Use similar idea which we are using while answering all queries.

      Suppose $$$dp[i]=a[i]$$$. Now let $$$q$$$ be the smallest index greater than $$$i$$$ such that $$$dp[q]=a[q]$$$ if $$$dp[i]=a[i]$$$. So $$$track[i]=\sum_{j=i}^{q-1} dp[j] + track[q]$$$. Please note that I am assuming that $$$dp[i]=a[i]$$$(it is hypothetical, so actual values of $$$dp[i]$$$ are different from what we have when we assume $$$dp[i]=a[i]$$$).

      Now note that $$$dp[j]=a[i]+j-i$$$ for $$$i \leq j < q$$$ if $$$dp[i]=a[i]$$$. So you can easily compute the sum of this arithmetic progression.

      Please refer to "Code(Online queries)" in editorial. I hope it would make sense now.

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2 years ago, # |
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In C2, how can I find q (let q be the smallest index greater than p such that adp[q]=a[q])?

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    2 years ago, # ^ |
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    Refer to this comment

    You need to find the smallest index $$$q$$$ greater than $$$p$$$ such that $$$b_q \leq z-p$$$. You can easiy do it using binary search and range minimum query(which I have used in "Code(Online queries)").

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      2 years ago, # ^ |
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      One more question, why do we need track? Because I thought track[i] = prefix[n]-prefix[i-1] (with prefix[i] calculate i first dp value)

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      2 years ago, # ^ |
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      If we're using binary search and RMQ using segment tree, then the time complexity would be $$$n log^2(n)$$$ instead of $$$n logn$$$ right. I tried something similar, but it was giving TLE, so I had to use sparse table for RMQ to make it $$$nlogn$$$. Is there any other workaround?

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2 years ago, # |
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I find C2 pretty damn good, but would be better if they just get rid of C1 entirely and place C2 as E instead.

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    2 years ago, # ^ |
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    I am glad that you liked it (:

    I thought putting C1 would result in a balanced round, but sadly it didn't help much :(

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19 months ago, # |
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I have a way way simpler solution(online) for C2.

We can basically find out for each index i, if i is the start of a subarray, what are the first two bad points (bad points being points at which the given condition gets violated). We can do this in O(n) using queues (just observe monotonicity of bad points wrt starting points). Why is calculating only till the second point important? Because the only increase in answer for some start point happens when its first bad point gets resolved.

Now for each query, we can do some casework, and use prefix sums to calculate delta (this is the easy part so I wont go into it).

Implementation: link

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17 months ago, # |
Rev. 4   Vote: I like it +8 Vote: I do not like it

Alternate solution for problem E which has same time complexity as the editorial solution, but uses only $$$O(n^2)$$$ memory:

Key observation:

In atleast one of the optimal solutions, the result of all swap operations can be viewed as the following:

  • Some subset of elements $$$S$$$ will move.
  • Each element in $$$S$$$ moves in the following manner- Element with index $$$i$$$ moves to the left till some point $$$L_i$$$ and then moves to the right till some point $$$R_i$$$. Finally all the points with indices between $$$L_i$$$ and $$$R_i$$$ will have $$$a_i$$$ added as their contribution to the score. (There are quite a few tiny observations to be made to prove this, so try it yourself)
  • If $$$i$$$ and $$$j$$$ are indices which belong in subset $$$S$$$ and $$$i < j$$$, then $$$L_i < L_j$$$ (Note that $$$i < L_j$$$ may or may not hold).

So the array can be modelled as something like:

Coming up with DP States and Transitions:

Define the dp state as $$$dp[i][j]$$$, which holds maximum score we can get from the last $$$i$$$ elements, if we need to make atleast $$$j$$$ swaps before the $$$i$$$'th element to be able to transition from this state.

The transitions can be made in the following manner: Iterate over $$$i$$$ from $$$n$$$ to $$$1$$$, and then iterate over $$$L_i$$$ from $$$i$$$ to $$$1$$$, notice that all that really matters about the state which you transition from is its second parameter, so basically maintain an array which stores best possible transition for each value of second parameter, and keep trying to improve it as $$$L_i$$$ decreases.

Implementation: link

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14 months ago, # |
Rev. 5   Vote: I like it 0 Vote: I do not like it
Nooby editorial for B

the number $$$b[i]$$$ ($$$i \neq 1$$$) is constrained to be a multiple of both a(i) and a(i-1). Thus, the smallest value that can be given to $$$b(i)$$$ = $$$LCM(a[i],a[i-1])$$$. Any value given to be $$$b[i]$$$ must be a multiple of this $$$LCM$$$. That is to say $$$b(i) \geq k\cdot LCM(a[i],a[i-1])$$$ for some $$$k \geq 1$$$

wishful thinking: maybe if setting $$$b[i]$$$'s equal to $$$LCM$$$ does not work, then no other answer can work out. Turns out this is true The risk with setting equal to $$$lcm$$$ is that the evaluated gcd might turn out bigger than the required gcd (the $$$a(i)$$$ value), but it is guaranteed to not turn out smaller than required.

proof-ish: since $$$b(i)$$$ is constrained to be a multiple of both $$$a(i)$$$ and $$$a(i-1)$$$, any value I assign to $$$b(i)$$$ must be a multiple of the $$$lcm(a(i),a(i-1))$$$.

What we can say about the evaluated $$$a(i) = gcd(lcm_1,lcm_2)$$$ is that it will be atleast equal to the required/intended $$$a(i)$$$ or more but never lesser, since $$$lcm_1$$$ and $$$lcm_2$$$ are atleast guaranteed to be multiples of required $$$a(i)$$$.

Suppose we set $$$b(i)$$$'s as the lcms. If the gcd of lcms is turning out greater than required. That means $$$gcd(lcm_1,lcm_2) = X > required$$$ then gcd of any other values assigned to b(i) will also turn out greater than required since b(i) must be a multiple of the lcm value. That is $$$gcd(k\cdot lcm_1, l\cdot lcm_2) = gcd(k,l)\cdot gcd(lcm_1,lcm_2) \implies gcd(k,l)\cdot X$$$

Thus, if answer does not exist from setting $$$b(i) = lcm(a(i), a(i-1))$$$ recursively. Then, no answer exists anyways.

Handle case $$$b(0)$$$ and $$$b(n)$$$ separately.

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    10 months ago, # ^ |
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    Very nice explanation, thank you! How do you get the equation: $$$gcd(k* lcm_1, l*lcm_2) = gcd(k,l) * gcd(lcm_1, lcm_2)$$$ I have never seen that before and I don't know how to prove it.

    In fact, for your explanation it will be enough to prove that $$$gcd(k* lcm_1, l*lcm_2) > gcd(lcm_1, lcm_2)$$$

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      10 months ago, # ^ |
      Rev. 4   Vote: I like it 0 Vote: I do not like it

      Oh god, I fucked up. The equality does not hold in general. It should be $$$LHS >= RHS$$$. To see why it holds we can consider the full prime-factorisations of $$$k\cdot lcm_1$$$ and $$$l\cdot lcm_2$$$ VERSUS $$$lcm_1$$$ and $$$lcm_2$$$ respectively.

      So, if we think how $$$gcd(a,b)$$$ is found out by taking the lowest common-frequency of each prime factor in the factorisations of $$$a$$$ and $$$b$$$.

      With this perspective we can see that that taking $$$k\cdot lcm_1$$$ instead of just $$$lcm_1$$$ is only going to include "extra" exponents to the primefactors (or introduce new ones). Thus the new $$$gcd(k\cdot lcm_1, y)$$$ should only increase or stay the same as $$$gcd(lcm_1, y)$$$ for any $$$y$$$. Whatever $$$lcm_1$$$ had in common with $$$y$$$ still remains and there might only be extra stuff that becomes common with the introduction of the $$$k$$$.

      For example imagine: $$$k = 6, l = 8$$$ for $$$k\cdot lcm_1$$$ and $$$l\cdot lcm_2$$$ respectively.
      * The exponent of $$$2$$$ will increase by $$$+1$$$ for $$$lcm_1$$$-term while for $$$lcm_2$$$-term it will increase by $$$+3$$$. Thus the exponent of $$$2$$$ in the new $$$gcd$$$ must also increase by $$$+1$$$ for sure or else the gcd won't really be the greatest common divisor and thats a contradiction.
      * Similarly, if exponent of $$$3$$$ in $$$lcm_1$$$ was say, $$$2$$$ less than that in $$$lcm_2$$$. Then, the new $$$gcd$$$ can also now have an increase of $$$+1$$$ in the exponent of $$$3$$$

      Thanks a lot for mentioning this. Please point out if you find any discrepancy/contradictions again in the arguments above.

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12 months ago, # |
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What is second best approach for Problem D ? If One is unable to get the idea of claim ?

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7 months ago, # |
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Alternative way to solve C2 (actually it is pretty similar to the editorial):

First we need to talk about C1. Let $$$dp_i$$$ is the first position to the left of $$$i$$$ such that $$$a_{dp_i}, a_{dp_i + 1}, ..., a_i$$$ is invalid. Then $$$dp_i = max(dp_{i - 1}, i - a_i)$$$. The answer to C1 is sum of $$$i - dp_i$$$ for all $$$1 \leq i \leq n$$$, or $$$n \cdot (n + 1) / 2 - \sum_{i = 1}^{n} {dp_i}$$$. Submission for C1. If you notice, the $$$dp$$$ array is actually the prefix maximum of $$$i - a_i$$$ for each $$$i$$$. So I use an array $$$b$$$ with $$$b_i = i - a_i$$$ for future usage in C2.

Now turn to C2. Let's call $$$pref_i$$$ is the prefix sum of $$$dp_i$$$, and $$$suf_i$$$ is the sum of $$$\sum_{i = j}^{n}{dp[i]}$$$ supposed we start calculating $$$dp$$$ from $$$j$$$ (this is actually similar to the $$$track$$$ array in editorial, but with a different definition). Calculating $$$suf$$$ can easily be done by monotonic stack with the help of array $$$b$$$. Now, for each query, applying $$$a_p = x$$$ equivalents to changing $$$dp_i = max(dp_{i - 1}, p - x) = d$$$. Then $$$d$$$ will propagate among our $$$dp$$$, and we want to find the first position $$$q$$$ such that $$$d$$$ stops propagating at $$$q - 1$$$. For example, if $$$dp = [0,0,2, 3, 3, 3, 4, 6, 7]$$$ and we want to change $$$dp_5 = d = 5$$$, then $$$q = 8$$$ and our new $$$dp$$$ will be $$$[0,0,2,3,5,5,5,...,...]$$$. Here we can't assure that the last two elements in $$$dp$$$ are $$$6$$$ and $$$7$$$ after the change, but we can calculate the sum of those $$$'...'$$$, and it actually is $$$suf_q$$$. So, the sum of $$$dp$$$ after the change will be: $$$\sum{dp} = pref_{p - 1} + d * (q - p) + suf_{q}$$$, and answer for each query is: $$$n \cdot (n + 1) / 2 - \sum{dp}$$$.

To find the position $$$q$$$, I use binary search and max range query using sparse table. Hence the time complexity will be $$$O(nlogn)$$$. Submission to C2

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6 months ago, # |
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Is it possible to solve D without rotations in polynomial time?