### awoo's blog

By awoo, history, 17 months ago, translation, 1701A - Grass Field

Idea: BledDest

Tutorial
Solution (vovuh)

1701B - Permutation

Idea: BledDest

Tutorial
Solution (Neon)

1701C - Schedule Management

Idea: BledDest

Tutorial
Solution (awoo)

1701D - Permutation Restoration

Idea: BledDest

Tutorial
Solution (Neon)

1701E - Text Editor

Idea: vovuh

Tutorial
Solution (vovuh)

1701F - Points

Idea: BledDest

Tutorial
Solution (BledDest)  Comments (35)
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 » 17 months ago, # | ← Rev. 2 →   Please make more rounds like this one, they make me feel a bit better about myself •  » » 17 months ago, # ^ | ← Rev. 2 →
•  » » » orz
•  » » Please No more binary search :(. I'm bad at identifying binary search in the wild. If someone suggests me to use binary search then I'm able to solve the problem easily but when I encounter them in the wild they ravish me. Any suggestion on how to handle them?
•  » » » Just think that someone has suggested you to use binary search
•  » » » » correct xd
•  » » ye, me too. i just love the concept behind each problem :3
 » 17 months ago, # | ← Rev. 2 →   can someone explain the idea of problem E with DP pls, I've seen many solutions with an array like f[N] or f[N][N], can anyone explain the approach pls.
•  » » (sorry if this is a bit long/confusing, I'm not the most experienced at explaining)First off, imagine choosing some subset of the letters in $s$ that form $t$: for example if $s$ is zzzabczzz and $t$ is abc, then the abc is useful but all the zs are useless fluff.Then you want to delete all the fluff. Here's one approach: first, you delete fluff from the end of the string until you come to useful letters. Then, you iterate past the useful letters, until you come to more fluff to delete. Repeat until all the fluff is gone.This may not always be optimal, for example if $s$ is zaaaa and $t$ is aaaa: then you want to go to the start, go forward, and delete the singular z.When you're deleting fluff from the back, it takes $1$ operation to iterate past useful letters and $1$ operation to delete the fluff. When you delete fluff from the beginning, it takes $1$ operation to iterate past useful letters, but $2$ operations to delete fluff (you need to iterate forward then delete).Then, the optimal solution will look something like this: there is some prefix (possibly empty) of the string where you've deleted from the beginning, then some contiguous segment of useful letters you haven't iterated over, and finally, some suffix of the string where you've deleted from the end.If the prefix isn't empty, then it needs $2x+y+1$ operations, where $x$ is the number of fluff characters in the prefix, and $y$ is the number of useful characters. Otherwise it takes $0$ operations.Next, the contiguous segment of useful letters, this takes $0$ operations since you don't iterate over it at all.Finally, for the suffix, it takes $x+y$ operations.At least in my solution, the dp[N] was something like this: 0 for the prefix, 1 for the contiguous segment, and 2 for the suffix. The transitions were somewhat tedious to implement but fairly straightforward.
•  » » » Thanks! P/S : It's a very nice explanation.
 » I really liked problem E, the solution is very interesting for me(also like a problem)
 » 17 months ago, # | ← Rev. 2 →   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?
•  » » Yeah, this is a typo. It should be $s[0; pos)$ and $t[0; m - suf)$. Thanks.
•  » » » In the second para there is a typo I guess then, the suffixes rather being $s[n-pos,n)$ and $t[m-suf,suf)$, shouldn't it be $s[pos,n)$ instead ? I am confused about pos, is it assumed as a forward iterator or a backward iterator?
•  » » » » Yeah, it is $s[pos; n)$. Forgot to fix that along with previous typos.
 » 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
 » 17 months ago, # | ← Rev. 3 →   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 $x-d  » 17 months ago, # | ← Rev. 5 → anyone please tell me why i am getting TLE on test case 4 in problem C . 163421348edit — the problem accepted when i used priorty_queue instead of multiset 163422803 •  » » The problem is in the method 'findMax' and 'findMin' you are copying the entire data structure each time the method is calling.int findMin(multiset my_set) this means that the first argument will be copied, so in the worst scenario it will copy a huge data structure each time.The solution is this add '&' before the identifier of the argument: int findMin(multiset &my_set), now the multiset won't be copied.There are two ways to pass an argument, by reference or by coping the entire argument.This is your solution only adding '&' in those two methods: https://codeforces.com/contest/1701/submission/163423017I hope this could help, greeting bro •  » » » thank u  » 17 months ago, # | ← Rev. 2 → Let's calculate value of$\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<>n; cout<<2<
•  » » You are not updating value of iwhile(i*2<=n){ cout<
 » In problem D, sweep line is not needed. Simply sorting the range in increasing order of ending point is enough. AC code: https://codeforces.com/contest/1701/submission/163624301
•  » » Why increasing order of end point and not increasing order of starting point?
•  » » » You should sort in increasing order of end point because let's say if at a point there are multiple points to choose from then choosing the one which ends later may make it possible to make an answer. For example, consider the ranges [1,2],[2,2],[1,3]
•  » » » » Thanks understood.
 » 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.