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By awoo, history, 21 month(s) ago, translation,

1251A - Broken Keyboard

Idea: BledDest

Tutorial
Solution (Ne0n25)

1251B - Binary Palindromes

Tutorial

1251C - Minimize The Integer

Idea: Roms

Tutorial
Solution (Roms)

1251D - Salary Changing

Idea: Roms

Tutorial
Solution (Roms)

1251E1 - Voting (Easy Version)

1251E2 - Voting (Hard Version)

Idea: Roms

Tutorial
Solution (Roms)

1251F - Red-White Fence

Idea: Neon and BledDest

Tutorial
Solution (BledDest)

• +73

 » 21 month(s) ago, # |   +9 in c tutorial what does thirst means ?
•  » » 21 month(s) ago, # ^ |   +6 I think it's first rather than thirst.Maybe just another typo...
 » 21 month(s) ago, # |   +6 Can someone please explain the DP solution of E1?
•  » » 21 month(s) ago, # ^ | ← Rev. 3 →   +11 Sort all the voters in the order of non-decreasing values of $m$Let $dp[ind][team]$ store the minimum amount of money you need to spend to make first $ind+1$ voters vote for you, when already $team$ number of voters are voting for you. So our final answer will be stored in $dp[n-1][0]$.Now let's see the transitions from $dp[ind][team]$ if $team+ind>=voter[ind].m$ then it is possible that in the future, we'll be able to take him for free, move to $dp[ind-1][team]$ just pay him $voter[ind].p$ and then move to $dp[ind-1][team+1]$ value of $dp[ind][team]$ will be the minimum of all these three. For our base condition, when we have only one voter to consider $(ind=0)$, then we'll have to either pay him or not, depending on the number of voters we currently have with us.The 1st transition works because, that state will be filled only when all of first $ind$ voters have agreed to vote us.Implementation: 63362330
•  » » » 21 month(s) ago, # ^ |   +6 Actually, you don't need the first transition at all. Taking a voter for free is covered by the second transition, and cases when team value for some states is larger than needed are handled by the nature of DP automatically. You only decrease the number of extra votes by paying the voters (as in the third transition).The successful submission of your code without the first transition http://codeforces.com/contest/1251/submission/63500118
•  » » » » 21 month(s) ago, # ^ |   0 Can u suggest me some similar but beginer level dp problems?
•  » » » » 21 month(s) ago, # ^ |   0 Thanks! edited.
•  » » » 21 month(s) ago, # ^ |   0 could you explain why team + ind > voter[ind].m means its possible to take said voter in the future ? I don't understand that part.
 » 21 month(s) ago, # | ← Rev. 3 →   0 Can someone please help me finding bug in my solution for the problem E2. It's giving WA on test case 3, the test case is really large so I'm not able to find out which input is giving WA.Link to my code
•  » » 21 month(s) ago, # ^ | ← Rev. 2 →   +3 may be you need lower_bound instead of find, and totalcost+=cheapest.f for adding the cost
 » 21 month(s) ago, # | ← Rev. 2 →   0 My approach for F is very dirty. (I gave up to code this)Let's make some definitions: $R$ is the height of red board. $Q$ is the query number. Let's define unique whiteboard which whiteboard has unique height. Let's define non-unique whiteboard which is whiteboard but not unique whiteboard. Suppose there are $x$ unique whiteboards and $y$ non-unique whiteboards, and you picked $x_{1}$ unique whiteboards, $y_{1}$ non-unique whiteboards once and $y_{2}$ non-unique whiteboards twice. ($0 \le x_{1} \le x$, $0 \le y_{1} + 2 y_{2} \le y$) Then you have to satisfy this formula: $target = \frac{Q}{2} - R - 1 = x_{1} + y_{1} + 2 y_{2}$Since $y_{2}$ non-unique whiteboards should be at both side, the side of $x1$ and $y1$ boards only matters. So there are $2^{x_{1} + y_{1}}$ cases in state $(x_{1}, y_{1}, y_{2})$.Now for $y_{2} = [1, 2, 3, \ldots, \frac{target}{2}]$, calculate number of possible $(x_{1}, y_{1})$ pairs, and add it. And you have to apply dirty casework to calculate this fast.
 » 21 month(s) ago, # | ← Rev. 2 →   +1 How to prove Monotonicity in D?
•  » » 21 month(s) ago, # ^ |   0
 » 21 month(s) ago, # | ← Rev. 3 →   0 what if we get a mid such that only mid < l is there for all salary ranges, since that will be an invalid median how do we check that we should go higher or lower?
•  » » 21 month(s) ago, # ^ |   0 the ans must large or equal than the median of all salary's liso you could calculate the value of median of all salary's li ，then start binary search
 » 21 month(s) ago, # |   0 What's the dp sol of E1?
•  » » 21 month(s) ago, # ^ |   0
•  » » » 21 month(s) ago, # ^ |   0 Thank you.
 » 21 month(s) ago, # |   +3 I cannot find a way to prove the correctness of gready algorithm in E2. If we are considering voters with m = i, and we need more voters with m >= i. Suppose that we pick a voters with m = x, so the need of additional picking of this voter before may be not neccessary. I don't know how it can effect to the process of greedy?
•  » » 21 month(s) ago, # ^ |   0 let's notice to the inverse iteration of i. If we are considering voters with m=i, we can still use voters with m=x>i because those voters (with m=x>i) were inserted into the multiset before (in the order of iteration) and notice that if they are not chosen to pay money, they will be still in the multiset. HUST student!
 » 21 month(s) ago, # |   +5 In E2, when pref[x] + cnt < x we can buy additional x−pref[x]−cnt votes or buy all the members of the group x. But in the solution buying all the members of group x is not considered. I don't understand why. Can someone explain please?
•  » » 21 month(s) ago, # ^ |   +8 because the limit mi is 0<=mi
•  » » 21 month(s) ago, # ^ |   0 If we are considering $m_i=x$, the number of $m_i=x$ is $s$, the number of $m_ix$, we have already considered it, denote $y=m$, obviously $y>x$ and $p+s+cnt\ge y$Consider $x$, $need=x-p-cnt$$s-need=s-x+p+cnt=y-x>0$ so $s>need$Proved.
 » 21 month(s) ago, # |   0 In E why is this strategy wrong? we choose the minimum m value each time having maximum price. Then we try to buy people such that cnt(no of people that have been voted for us till now) becomes greater than current minimum m value choosen. We'll buy diff = min(m)-cnt cheapest people i.e. we choose diff people from remaining set of people having minimum price. I'll maintain two multiset for it. One with comparison on m value first and other with price. Moreover, I could understand what editorial is trying to say but the accompanying code doesn't seem to do the exact same thing that has been explained.
 » 21 month(s) ago, # |   0 Problem D has a greedy tag on it. Can anyone please provide a greedy solution?
•  » » 21 month(s) ago, # ^ |   0 The greedy are Embedded in binary search https://codeforces.com/contest/1251/submission/63414228 The function named 'ok' is greedy solution
 » 21 month(s) ago, # |   0 in problem D, could someone help me find out what will happen if $mid$ is not in any salarie $[l_i, r_i]$? Thanks! :)
•  » » 21 month(s) ago, # ^ | ← Rev. 3 →   0 mid > ri ---> The salary of people must less than medianmid < li ---> The salary of people must large than median
 » 21 month(s) ago, # |   0 C no is a very nice problem ...
 » 21 month(s) ago, # |   0 Can someone explain me the solution to c problem
•  » » 21 month(s) ago, # ^ | ← Rev. 2 →   0 Since you cannot swap two adjacent odds or even digits, make two separate lists one for odd and other for even digits in the order in which they appear in the original string. Now when we merge these two, compare the elements from odd and even lists since they can be swapped. We maintain the order in which the digits appear in their respective lists.Python solution: 63358004
 » 21 month(s) ago, # |   +1 Could someone please explain problem D solution? I am having a hard time getting the editorial.
•  » » 21 month(s) ago, # ^ |   0 I can explain my solution which is similar. I'm assuming you know about binary search and sorting.Let $ok(mid)$ be a function that returns:-1 if $mid$ is impossible to make as a median because it is too low.0 if $mid$ is possible to make as a median in some way.1 if $mid$ is too big to make as a median (there isn't enough money).Then we can binary search $mid$ to find the maximum such that $ok(mid)$ returns $0$.How will the function $ok$ work? Assume that we will assign the number $mid$ to some interval. Also, let initially $k = s$ from the statement (the total money). Subtract $mid$ from $k$, since that's the cost to assign the number $mid$ to the interval that will be the median. We can put the $i$-th interval $[l_i, r_i]$ into 1 of 3 groups: $r \lt mid$: the number we assign must be smaller than $mid$ (no other choice). Let the count of these numbers be $cntsmall$. Also, since we want to use as little money as possible (greedy), might as well assign $l_i$ to this number. Don't forget to subtract that from $k$. Those with $l \gt mid$: the number we assign must be bigger than $mid$ (no other choice). Let the count of these numbers be $cntbig$. Do the same stuff as with the previous group. Those with $l \le mid \le r$: for these numbers, we will choose whether we assign a number less than, equal to, or more than than $mid$ (whether they will be "before" or "after" $mid$ in the sorted list of numbers). Now, if $cntsmall \gt \lfloor n/2 \rfloor$ or $cntbig \gt \lfloor n/2 \rfloor$ we can't use $mid$ as a median because we must always have exactly $\lfloor n/2 \rfloor$ numbers left and right of the middle element (the median). So return $-1$ if there are too many bigger numbers ($mid$ is too small), or $1$ is there are too many smaller numbers ($mid$ is too big).If everything is alright, then we need to assign numbers to the intervals from the 3rd group. From the first group (smaller numbers), we need to assign $\lfloor n/2 \rfloor - cntsmall$ numbers (The first $cntsmall$ numbers were forcefully assigned, now we care about the remaining numbers). Since we want to minimize the cost, we should sort all intervals so that their $l$ is increasing, then assign $l_i$ for the first $\lfloor n/2 \rfloor - cntsmall$ of these numbers (again, by assigning we mean subtracting that number from $k$, the total money remaining).From the second group (bigger numbers), we need to assign $\lfloor n/2 \rfloor - cntbig$ numbers. Since they will come after $mid$, they need to be bigger or equal to $mid$. Obviously, it's best to use as little money as possible, so we assign $mid$ to all of them, so the total cost for this part is just $mid * (\lfloor n/2 \rfloor - cntbig)$.Now, all intervals have been assigned a number. If $k \geq 0$, then we had enough money to do it, so return $0$. If $k \leq 0$, then we didn't have enough money, or in other words $mid$ was too big, so return $1$.
•  » » » 13 months ago, # ^ |   0 Too good. Thank you so much for a nice explanation!
 » 21 month(s) ago, # |   0 Can someone explain E2 solution better? I don't understand the two cases, and also why we iterate from x=n-1 to x=0.
 » 21 month(s) ago, # |   0 Excuse me for interrupting you. I think the Tutorial of Problem D may have something wrong. The ok function has two return values.if(need > v.size()) return false;It is because var mid too small.return sum <= s;It is because var mid too large.So how to keep monotonicity?May someone help me? Thanks for your kindness.
•  » » 21 month(s) ago, # ^ |   0 oh
 » 20 months ago, # |   0 My thought to problem F:unique whiteboard means whiteboard has unique height.non-unique whiteboard means whiteboard has non-unique height.Let's use the "generating function".For each unqiue whiteboard, its generating function is $(1+2x)$For each non-unique whiteboard, its generating function is $(1+2x+x^2)=(1+x)^2$So, the generating function of the answer is $(1+2x)^\text{how many unique whiteboards} \cdot (1+x)^{2\times \text{how many non-unique whiteboards}}$We will get a $O(n \log^2 n)$ solution if we use FFT/NTT to calculate it directly.To optimize it, we can pre-calc the coefficients of $(1+2x)^\text{how many unique whiteboards}$ and $(1+x)^{2\times \text{how many non-unique whiteboards}}$, by binomial theorem, then use FFT/NTT to merge them. The complexity is $O(n \log n)$.Sorry for my poor English...
•  » » 12 months ago, # ^ |   0 Could you elaborate more about why the generating function for each unique whiteboard and non unique whiteboard are (1 + 2*x) and (1 + x)^2 respectively? I would really appreciate that.
 » 20 months ago, # | ← Rev. 2 →   0 Can someone please tell me what is wrong with this solution to the first problem?string solve(string s) { vector count(26, 0); for (char c: s) { count[c-'a']+=1; } set ans; for (int i = 0; i < s.size(); i++) { if (i == 0) { if (i+1 == s.size() || s[i] != s[i+1]) { ans.insert(s[i]); } } else if (i == s.size()-1) { if (i-1 == 0 || s[i] != s[i-1]) { ans.insert(s[i]); } } else { if (s[i-1] != s[i] && s[i] != s[i+1]) { ans.insert(s[i]); } } } for (int i = 0; i< count.size(); i++) { if (count[i]%2 == 1) { ans.insert(i+'a'); } } vector toRet; set::iterator it; for (it = ans.begin(); it != ans.end(); it++) { toRet.push_back(*it); } sort(toRet.begin(), toRet.end()); return string(toRet.begin(), toRet.end());}
 » 17 months ago, # |   0 how to solve B ?
 » 16 months ago, # | ← Rev. 2 →   0 Is it just me or are the sidebar links to the announcement of tutorial to this round e.g on the page https://codeforces.com/contest/1251/problem/E1 not appearing?EDIt: yeah seems to work now
•  » » 16 months ago, # ^ |   0 Seem to work for me
 » 14 months ago, # |   0 In problem F how do we get $\binom{2m}{k}$ when no white board has unique length? We can chose say x1 boards once and x2 boards twice(such that x1 + 2*x2 = k) and for each of x1 boards we can place it either to the left or right of red board, right?
•  » » 12 months ago, # ^ |   0 Yes. We are counting the ways to put k non unique whiteboards using 2*m in total. One way to prove this is thinking that you're using x non unique whiteboards on the left side of the red board and k — x of the others on the right side. As you can see the answer is $\sum\limits_{i = 0}^{i = k}{{m} \choose {i}} * { {m} \choose {k - i}} = {{2*m} \choose k}$. There is an obvious bijective proof of this. However you can proof it by using the Snake Oil method too!.
 » 14 months ago, # | ← Rev. 2 →   0 Super simple idea for E2:Suppose we choose exactly $k$ voters to pay. Then the sequence of sorted $m$ values of the remaining voters must be non-strictly bounded from above by the sequence $k,k+1,...,n-1$. We can also think of this as being bounded from above by the sequence $0,1,...,n-1$, with allowed "empty slots" in the sequence.So, the solution is simply to choose a subset of voters satisfying the bounded property, such that the sum of their $p$ values is maximum. We can do this by greedily filling slot $0$ (which only accepts voters with $m\leq 0$), then filling slot $1$ (which only accepts voters with $m\leq 1$), and so on, always choosing the voter with maximum $p$ value if it exists, or just leaving an empty slot if no voters satisfy the requirement for that slot.This can be done easily with a priority queue in $O(n\log n)$.This leads to the same algorithm as the editorial by the way, just a little clearer to think about in my opinion.
 » 13 months ago, # |   0 In A what if a key working correctly was pressed twice consecutively by polycarp ? Wont the question break then ?
 » 12 months ago, # |   0 Why in Problem F it is better to use NTT than to use FFT?
 » 9 months ago, # |   0 the solution of A was confusing to me. in tutorial it said if the length is odd than that character is correct key. but it checks (j-i)%2==0 instead of (j-i)%2==1. that is because for loop starts form 0; even though it is correct I got confused because of this. it should be (j-i+1)%2==1. while(j+1
 » 3 months ago, # |   0 I really liked E! Thank you.