BledDest's blog

By BledDest, 3 years ago, translation,

1215A - Yellow Cards

Editorial
Solution (fcspartakm)

1215B - The Number of Products

Editorial
Solution (fcspartakm)

1215C - Swap Letters

Editorial
Solution (fcspartakm)

1215D - Ticket Game

Editorial
Solution (fcspartakm)

1215E - Marbles

Editorial
Solution (fcspartakm)

Editorial
Solution (BledDest)

• +87

 » 3 years ago, # |   +6 F was Perfect!!
 » 3 years ago, # |   +77 Another proof for $D$:Consider first the cases where there are question marks only in one side, assume their count is $C$ ($C$ is $even$). Let the side without question marks be $S_1$ and the other side be $S_2$, and let $Sum(S_1)-Sum(S_2)$ be $d$: $Case_1$ ($d=\dfrac{C}{2}*9$): When Monocarp puts $x$, Bicarp puts $9-x$, and Bicarp wins. $Case_2$ ($d>\dfrac{C}{2}*9$): Monocarp always puts $0$, so $d$ will always be $>0$, and Monocarp wins. $Case_3$ ($d<\dfrac{C}{2}*9$): Monocarp always puts $9$, so $d$ will be $<0$ at the end, and Monocarp wins. Now consider the cases where there are question marks in both sides. Let the side with less question marks be $S_1$ and the other side be $S_2$, and let $Sum(S_1)-Sum(S_2)$ be $d$. Assume $S_1$ has $C_1$ question marks and $S_2$ has $C_2$ question marks ($C_1+C_2$ is $even$), and let $C$ be $C_2-C_1$: If $d=\dfrac{C}{2}*9$: When Monocarp puts $x$ in one side, Bicarp puts $x$ in the other side. Thus $d$ never changes, and we reach $Case_1$. If $d>\dfrac{C}{2}*9$: Monocarp first puts $9$ in $S_1$. And when Bicarp puts $x$ in one side, Monocarp puts $x$ in the other side, until $odd$ number of question marks remain in $S_2$. At this point, $d>\dfrac{C}{2}*9+9$. So after Bicarps's turn, $d$ must be $>\dfrac{C}{2}*9$ and we reach $Case_2$. If $d<\dfrac{C}{2}*9$: Monocarp first puts $0$ in $S_1$. And when Bicarp puts $x$ in one side, Monocarp puts $x$ in the other side, until $odd$ number of question marks remain in $S_2$. At this point, $d<\dfrac{C}{2}*9$. So after Bicarps's turn, $d$ still must be $<\dfrac{C}{2}*9$ and we reach $Case_3$.
•  » » 2 years ago, # ^ |   0 God level explanation!
•  » » 9 months ago, # ^ |   0 I think we only have to look at cases where Monocarp adds digits 0 or 9, since, if he adds any other digit d, his chances of winning can improve by either changing d to 0 or changing d to 9 (for intuition, we can think that Monocarp will want to take "extreme" steps because he wants to make the two sides unequal). Also, we can just look at "surplus" question marks, since, if ? exist on both sides, Bicarp will always add the same digit that Monocarp added, but to the opposite side. This can be proved as well (I think it has something to do with Nash Equilibrium, but for intuition, we know that Bicarp will not like whatever Monocarp does, and will try to nullify it). The above statements (if properly proved) will greatly simplify your proof.
 » 3 years ago, # |   0 ypa
 » 3 years ago, # |   +5 In Problem B the input is The second line contains n integers $a_1,a_2,…,a_n (−10^9≤a_i≤10_9;a_i≠0)$ — the elements of the sequence. but the Tutorial with many if(a[i]==0) why ?
•  » » 3 years ago, # ^ |   0 That's just for input error or testcase error
•  » » 3 years ago, # ^ |   +9 I think he just considered a[i] == 0 case too. But as given in question there are no element with 0 value so we can just skip this condition , rest of the solution is same .
 » 3 years ago, # |   0 This is my code for D which simulates the game optimally .My Code
•  » » 3 years ago, # ^ |   -9 Nice solution!
 » 3 years ago, # |   0 60628371 Here is my code and I wanna know why the runtime error
•  » » 3 years ago, # ^ |   0 for problem B
•  » » » 3 years ago, # ^ |   +6 The write to arr[(n << 1)] on line 18 caused array index overflow.Increasing the size of the array seems to work.  long ans[2]; cin >> n; //vector arr(n << 1); - long *arr = new long[(n << 1)]; + long *arr = new long[(n << 1) + 1]; cin >> next; if (next > 0) { ans[0] = arr[0] = 1; 
•  » » » » 3 years ago, # ^ |   0 thx
•  » » » » 3 years ago, # ^ |   0 That's stupid. xd
 » 3 years ago, # | ← Rev. 2 →   +21 I have some questions for E: When we place all marbles of type i in front of all marbles of type j, the relative positions between type i and other types can change. So when we do transition in the dp state, how can we ensure that summing up all cnt[j][i] is the number of extra moves that we need? Say we need to add marble i to dp[mask], after we move all the marbles in mask to the front of type i, the positions of all the marbles in mask is potentially scrumbled. In that case, how is dp[mask] still the number of moves needed to reorder the marbles in mask after moving them to the front of type i marbles? For example, say the order is {2, 3, 1, 2}. When we move all type 1 and 2 marbles to the front of type 3, the number of moves needed to reordred type 1 and 2 is from state {2, 1, 2, 3}, and is different from reordering them from state {2, 3, 1, 2}, which is the initial state. Hope my questions make sense.
•  » » 3 years ago, # ^ |   +17 For those of you who are as confused as I was, I asked my friend who ACed this problem and the explaination is as follow: When we transit from one dp state to another, we only consider types that are in the mask. So when we try to place type i as the last marbles in this dp state, each swap of type i marble with the other marbles is counted in cnt[j][i], for some j in the mask. So the summation of cnt[j][i] is indeed the number of extra steps that we need to take. The cue to understand the dp state properly is to know that dp[mask] means we only consider marbles that are in the mask and ignore other types. Take the same example {2, 3, 1, 2} here. Say we wanna transit from dp[011] to dp[111] (adding type 3 marbles), and want the last marbles to be of type 3. dp[011] means the number of steps needed to reorder {2, 1, 2}, as we ignore type 3 marbles. Thus dp[111] equals to summation of cnt[j][3] + dp[011]. I hope this explanation helps.
•  » » » 22 months ago, # ^ |   0 can you make me understand why does cnt[j][i] is indeed the number of swaps required. because as far as i know in the mask the positions of the 1s' in the mask have already changed. And the precalculated cnt[j][i] will then remain invalid. Where is the flaw in my reasoning??
•  » » » » 22 months ago, # ^ |   0 cnt[j][i] would only factor in the relative positions between type i and j. i.e. only the swaps between type i and j are counted, swaps between type i and another type k would be taken care of in cnt[k][i], and so is the case between type j and type k.When you try to move type j marbles in front of all the marbles in mask, each swap would be counted only once in cnt[j][i], for some i in the mask, because we only consider type j and types that are in the mask.Hope the explanation helps.
•  » » » » » 22 months ago, # ^ |   0 Correct me if i am wrong this then means, that when bit i is moved ahead of all bits marked 1 in mask then later on when the bit k is added is already taken care of despite the change in positions or in other words cnt[i][k] when added already takes care of bits that are marked 1.
•  » » » » » » 22 months ago, # ^ |   0 It's not proper for you to think that way because $dp[mask]$ only means types in mask is "sorted" in some order, not necessarily means type i is at the very beginning. We don't really have to care about the order of how it is "sorted".Maybe it's better to think of the dp transition this way: when you try to "sort" all marbles of type k and types in mask, one way you can do is to put all types in mask in front of type k, this is what $\sum_j^{j\in mask} cnt[j][k]$ accounts for. Then you "sort" all the marbles that are in mask, which is what $dp[mask]$ accounts for. You sum these two terms up to get $dp[mask|1«k]$ under this configuration.
•  » » » » » » » 22 months ago, # ^ | ← Rev. 2 →   0 Sorry if I am being very stupid,if the mask elements are already sorted, then isn't it possible that cnt[j][i], doesn't give the right value to add to the dp transition.I mean to say if you are swapping then don't you have to swap types of colour not in the mask . How are they accounted for?
•  » » » » » » » » 22 months ago, # ^ |   0 Swaps involving types not in mask will be accounted in a later stage where that type is being considered in the state transition
 » 3 years ago, # | ← Rev. 2 →   0 solution for B. easier way is to build prefix function with 1 and -1 and calculate how many pair give positive multiplication,i.e, count[1]C2+count[-1]C2. let the number of pairs be X,then nC2 — X will be ans for negative.
•  » » 3 years ago, # ^ |   0 Why you initialized pos=1 ?
•  » » » 3 years ago, # ^ |   0 i am assuming 1 beforehand and considering n+1 elements,u could see that for prefix conventions if p[r]*p[l-1] denotes the sign of segment [l,r].so for 0 u need something before it.
•  » » 3 years ago, # ^ |   0 Can you explain me why you can caculate the number of pos subsegment by (pos*(pos-1))/2+(neg*(neg-1))/2;
•  » » » 3 years ago, # ^ |   +3 consider we have a sequence like this +-+-+ Then he is increasing pos whenever curr is +; initially, curr = 1, pos = 1; (*) + ->curr = 1, pos = 2; (*) - ->curr = -1, pos = 2; + ->curr = -1, pos = 2; - ->curr = 1, pos = 3; (*) + ->curr = 1, pos = 4; (*) Therefore, neg = 5+1 - pos = 2; Notice the (*) denotes all the indexes where the curr == 1, Now see that on choosing any indexes from all the (*) and forming a segment of it will form a positive segmentExample here (*) = {0,1, 4,5} choosing (l,r] as subsegment of these indices will give you positive segment. Number of ways = (pos)C2;Now, similarly you can see why he is adding negative count as (neg)C2;
•  » » » 23 months ago, # ^ |   0 He is just multiplying the prefix values of same signs by taking nc2=n(n-1)/2. This, is in turn ,directly giving him all the positive values.
•  » » 3 years ago, # ^ |   +3 Wow, beautiful solution brother
 » 3 years ago, # | ← Rev. 3 →   0 Sorry, I'm incorrectly understand Um_nik's solution:60611022But it's great)
•  » » 2 years ago, # ^ |   0 Can u explain um_nik's solution please? i am not able to understand how multiplying the positive and negative gives the answer for all the subsegments
 » 3 years ago, # | ← Rev. 2 →   +3 Hello, Can anyone explain me solution of problem C I'm not able to see why it should be proceeded this way! Thank you so much!Edit: Sorry for this comment — I understood what was happening, Could anyone tell if I can delete this comment.Thank you!
 » 3 years ago, # |   +5 In problem B > editorial No. of subsegments should be n*(n+1)/2
 » 3 years ago, # |   +6 Whoa editorial for D is so hard to interpret. I did in a easy way. See in left side what's the max value u can get it will be (number of ?/2)*9 + (?%2)*9 if monocarp wants to make it max and for min we calculate (number of ?/2)*9Now do this for 2nd array also.If maxl == maxr and minl == minr answer is bicarp coz bicarp can adjust accordingly else answer is monocarp
•  » » 3 years ago, # ^ |   +1 whoa... just briliiant. Thanks for sharing.
•  » » 3 years ago, # ^ |   0 How did you interpret the equation of maxvalue on each side to be (number of ?/2)*9 + (?%2)*9 ??
•  » » » 3 years ago, # ^ |   +6 Greedily, so suppose monocarp try to make the left side max he will get one more chance than bicarp to do so. so if it's odd he will get the last chance to increase it. Same goes for bicarp. If suppose monocarp tries to increase left bicarp will get one more chance to increase right side. And then after the ? is finish on either side bicarp will start minimizing the other side. It just works lol. Also upvote if u find it helpful :p
•  » » » » 3 years ago, # ^ |   0 So you are saying that Monocarp will try to increase the sum of one side by changing '?' into 9, and bicarp will change '?' on the same side into 0. So Monocarp has at most (number of '?' on that side)/2 chances and then +1 if the number of '?' is odd. But though your solution is fairly simple but it is hard to see why this actually "works"...
 » 3 years ago, # | ← Rev. 2 →   0 In problem D, my thought was to first divide the array in two halves (namely left and right). Then calculate the cumulative sum in both sides (suml and sumr) considering any '?' = 0. Then Monocarp will pick the side where the cumulative sum is larger and put 9 in place of '?' and try to make this side even larger. And then (if he has more moves) he would try to put 0 on the other side to prevent bicarp from catching up. So after that if the side with "less sum to start with" becomes greater or equal than the other side that means bicarp can successfully catch up. But I am getting wrong answer at test case 23: https://codeforces.com/contest/1215/submission/60690503 Please point out where I am going wrong.
 » 3 years ago, # | ← Rev. 3 →   +11 Solution of D in simple terms: Because the size of the array(lets call it array for simplicity) in input is EVEN, we can always divide this array into two equal parts. Then for bicarp to win he must make sure the two parts have equal sum. So first of all we will find the sum of the left side of the array() "lsum" and we will find the sum of the right side of the array "rsum" (consider '?' as 0). Now there are 3 cases — 1) lsum == rsum , 2) lsum > rsum 3) lsum < rsum. Now we will use the '?' marks in both sides and try to make both sides equal so that bicarp wins. But if Monocarp has some strategy such that he always prevents bicarp from doing so then Monocarp wins. 1) Lets consider lsum == rsum:: If there are equal number of '?' on both sides Bicarp can nullify any moves of Monocarp and keep the two sides equal. If monocarp changes one '?' into X on any side, bicarp can change one '?' on the other side into X, so both sides remain the same. As there are equal number of '?' in both sides, bicarp can always do the same. Now if there are unequal number of '?' on both side, then one side will have more '?' than the other side but the initial sum is equal on both sides(as we assumed for now). In that case Monocarp can choose the side with more '?' and keep changing them into 9 so that the sum of this side gets bigger and bigger. To keep things equal bicarp must keep changing '?' marks on the other side by 9. But as Monocarp's side has more '?' (at least two more than bicarp's side, because the number of '?' is even), bicarp will eventually run out of '?' on his side, but there will AT LEAST be 2 '?' on Monocarp's side. So Monocarp can change one of these into 9 , now even if bicarp chooses 0(lowest) for the other '?' , Monocarp wins. (If instead of keeping things balanced, bicarp chose to change '?' into 0 on Monocarp's side to lessen the difference between two sides, then when Monocarp's side has no '?' left, Monocarp will start making bicarp's '?' into 0 so that bicarp cannot catch up with Monocarp's side, and as there is already less '?' in bicarp's side, bicarp stands no chance.) Now that we realize there is a connection of number of '?' on each side with our solution let's denote lcnt = (number of '?' on left side) and rcnt = (number of '?' in right side). 2) Now consider that lsum > rsum && lcnt > rcnt , we have one side with more sum (initially) and more '?' marks. So Monocarp will choose this side and keep changing '?' into 9. And bicarp will try to keep things balanced but will eventually run out of '?' and loose (by the same logic applied in 1). Same goes for (rsum > lsum && rcnt > lcnt). 3) Now consider lsum > rsum but lcnt < rcnt :: Now Monocarp will change the '?' on left side(I arbitrarilly picked the left side to have more sum but less '?', vice versa is ofc possible) into 9 to make it even larger and bicarp will do the same on the right side and not let the difference to increase (So lsum-rsum remains constant). Eventually Monocarp will run out of '?'. So bicarp has some '?' left. If he can reach lsum using these '?' then bicarp wins. So basically we do this, sum = lsum-rsum; cnt = rcnt-lcnt; Now before the tricky part let's make sure you understand two things: i) the very first move of the game was made by Monocarp, so after some moves if "moves by Monocarp" = "moves by Polycarp", clearly the next move will be made by Monocarp. So Monocarp will change any available '?' and bicarp has the chance to observe and move carefully. ii) Right now (after Monocarp uses all of his sides '?') Monocarp and bicarp will have equal amount of moves and equal amount of '?' to change (think about cases where Monocarp has i) EVEN or ii) ODD number of '?' at the very begining). Now the tricky part, Monocarp has two options to win, if he can overflow this side, means that if he changes all the '?' available for him (if there are 4 '?' available Monocarp can touch only 2) into 9, then even if bicarp chooses 0 for his '?', rsum becomes greater than lsum. In that case Monocarp's best choice is to use 9 for all '?'. Or if he can underflow this side, that is choose 0 for all '?' (he can touch) and thus even if bicarp chooses 9 for his '?', rsum remains less than lsum. The only way bicarp can win is if rsum + (cnt/2 * 9) = lsum, that is he needs to make all '?' on his control 9 in order to win. Why this works? If Monocarp tries to overflow and chooses 9 for one '?' bicarp can choose 0 for one of his '?'. If Monocarp chooses 0 , bicarp can choose 9. If Monocarp chooses 6, bicarp chooses 3. (notice as Monocarp moves first, bicarp can observe and choose optimally).
•  » » 3 years ago, # ^ |   +10
•  » » 23 months ago, # ^ |   0 Your approach is failing at s="441???" Monocarp has more chances here he wins in every condition. But your output is Bicarp
•  » » » 23 months ago, # ^ |   +3 Unfortunately you missed the last sentence of the input section which states that "The number of "?" characters is even".
 » 3 years ago, # |   0 Can anyone provide more intuitive proof for problem D, mohamedeltair's proof was nice but i only got the 1st part of it
•  » » 3 years ago, # ^ |   0 For the second part, let us suppose C1>C2. Let C1 = x+C2. So for both sides upto the first C2 positions on either side -> i.e. C2 positions in first half and C2 positions in second half, if Monocarp gives value 'val' on side, Bicarp gives value 'val' on the other side. So they even out each other. This happens till C2 positions on both sides are filled.So finally we have x positions remaining in the first half after this process is done. This reduces the second part of the solution to the first part of the solution. Similar logic if C2>C1 for the second part also.
 » 3 years ago, # |   0 In C, what would the solution look like if the strings were allowed to have characters from a..z
•  » » 2 years ago, # ^ | ← Rev. 2 →   0 In C, what would the solution look like if the strings were allowed to have characters from a..z Enchom BledDest spookywookyThanks!
•  » » » 16 months ago, # ^ |   0 its strange , Its been 16 months still no comment on @rishi's commentIn C, what would the solution look like if the strings were allowed to have characters from a..z __
 » 3 years ago, # |   0 Problem B can be done using a much simpler logic and code by iterating through the array and maintaining count of pair indices having positive product and negative product having current index as R.
•  » » 3 years ago, # ^ |   0 Can u pls explain ur approach in detail??
•  » » » 3 years ago, # ^ | ← Rev. 8 →   0 I used a DP approach for this problem; keeping track of two values: $pos_i$ = number of subsegments with positive product with last index $= i$ $neg_i$ = number of subsegments with negative product with last index $= i$ Transitions should be quite elementary if you think about what adding a positive/negative element would do to the product of a subsegment.Final answer is $N = \sum neg$ and $P = \sum pos$Code: 60664552
•  » » » » 2 years ago, # ^ |   0 very good & easy process. thanks Spheniscine
 » 3 years ago, # |   0 Can anyone explain problem B?
 » 3 years ago, # | ← Rev. 3 →   0 Another way to think about the solution for B is:Let $pos[i]$ be the count of sub arrays that end in i ($[k..i]$ for $k \geq 0$) whose product is positive, and $neg[i]$ the count of subarrays that end in i ($[k..i]$ for $k >= 0$) whose product is negative.The base case for them is simple: pos[0] = 1 if v[0] > 0, otherwise 0 neg[0] = 1 if v[0] < 0, otherwise 0 Given that $pos[x-1]$ and $neg[x-1]$ are calculated, we can easily calculate $pos[x]$ and $neg[x]$ as follows: pos[x] = 1+pos[x-1] if v[x] > 0, pos[x] = neg[x-1] if v[x] < 0  neg[x] = neg[x-1] if v[x] > 0, neg[x] = 1+pos[x-1] if v[x] < 0 After these arrays are fully calculated, the amount of sub arrays that are positive is the sum of all $pos[i]$, and the amount of arrays that are negative is the sum of $neg[i]$My submission 60721761
•  » » 3 years ago, # ^ |   0 What you are saying is not always true. For example, let's have an array 5, -3 Then pos[0] = 1, neg[0] = 0. neg[1] = 2, pos[1] = neg[0] + 1 = 1. But pos[1] cannot be 1 because there is no subsegment that ends at index 1 with product being positive.
•  » » » 3 years ago, # ^ |   0 Sorry, I had messed up the transitions. They're correct now. Thanks for the feedback :D
•  » » 2 years ago, # ^ |   0 Can you please explain your approach to this problem B
•  » » » 2 years ago, # ^ |   0 Consider this example:a = {1, 2, -2, -5, 4}Now you have got two arrays pos[n] and neg[n]. Each of them counts the number of subarrays that end at position i and have the product of their elements as positive and negative respectively.For the above example, pos[0] = 1 and neg[0] = 0Now for any position i, if a[i] is negative then you will have: neg[i] = 1 + pos[i - 1] and pos[i] = neg[i - 1].Otherwise, if a[i] is positive then: neg[i] = neg[i - 1] and pos[i] = pos[i - 1] + 1So, in the above example you will have pos[n] = {1, 2, 0, 3, 4} and neg[i] = {0, 0, 3, 1, 1}
•  » » » » 5 days ago, # ^ |   0 This is the great solution for B ..Thank you so much ..
 » 3 years ago, # |   0 Can anyone explain the editorial solution of E where it is calculating the number of swaps required for each pair of marbles? I didn't understand how the two pointer approach is working here.
 » 3 years ago, # | ← Rev. 2 →   0 BledDest, Just reading and building the 2-Sat graph for F in Java gives MLE.... 60748712. It doesn't even make it to the Solve2Sat() call where there might be DFS related MLEs. Could you boost the Memory Limit?
 » 3 years ago, # |   0 Can someone provide an alternative explanation for problem C?
•  » » 2 months ago, # ^ |   0 Case:(i) s1="aa" & s2="bb" then s1==s2 can be done by just a single swap(0,1) => s1="ba" & s2="ba".(ii) s1="ab" & s2="ba" then s1==s2 can be done by 2 swaps (1,1) => s1="aa" & s2="bb" now repeat the above process that is swap(0,1) s1="ba" and s2="ba".Cases like s1="a" s2="b" , s1="aba" s2="bab" s1="aabbb" s2="babaa" & so on can not be make s1 equals s2Now task is to first not consider all such indices i at which s1[i]=s2[i] , then divide the string s1 and s2 in above two cases and store indexes for a in vector v1 and b in vector v2 for s1 or s2 .if(v1.size() is even and v2.size() is also even this imply the case(i) so we have answer v1.size()/2 + v2.size()/2 as we can swap each pair of a & b separetly.if(any size of both vector are odd ) this implies case(ii) so we have answer =v1.size()/2 + v2.size()/2 +2+2 because when we are done with all the swaps we will only left with the case like "ab" and "ba" which can be done in 2 swaps .here is my solution: https://codeforces.com/contest/1215/submission/148853299ps: if you didn't get what i want to convey then you can ask me again . Happy to help:)
 » 3 years ago, # |   +3 Problem D (again... but with illustrations): Solution: https://codeforces.com/contest/1215/submission/60849161
 » 3 years ago, # |   0 It is interesting to hear from the authors how they came up with the idea of problem D so that its solution is based on (L+R)/2 = 0. What was the order? Did you made up a problem statement after having a solution for similar problems or you built a problem and then solved it?Why it is interesting is that I can't see any way how a person can find the property of (L+R)/2 = 0 during a competition.
•  » » 3 years ago, # ^ |   0 Well, initially fcspartakm proposed this problem on lower constraints, so it could be solved with dynamic programming. Then I thought about how this problem can be solved using the special structure of allowed turns (they are symmetric). I can't remember how exactly I came to the equality $\frac{L + R}{2} = 0$, but it was something like that: if the first player wants to win, they should either make the balance very low, or make it very high. So I tried to analyze these two cases (the first player wants the balance to be as low as possible or as high as possible), and then somehow arrived at this answer. I think that this formula is easier to come up with if you have mathematical background, but if you don't have it, it might be much easier to find an approach more related to programming than to maths.I actually wanted to make this problem interactive, where the contestant should play as the second player, but, unfortunately, this was unacceptable for The Quals.
 » 3 years ago, # |   0 Whish me good luck, guys. It has already been 5 weeks that I am trying to understand the solution for problem E. Don't know what I am doing wrong but my brain literally denies to understand it.
 » 2 years ago, # |   0 Can anyone give a proof for C?
 » 2 years ago, # |   0 I got another way to prove D solution.You can let each $?$ have an initial value of $4.5$, thus, on each turn, the player adds $x$ on the value of a question mark, s.t. $x \in [-4.5, 4.5]$.That way, the second player will always be able to mirror the first player's actions.If the first player adds $x$ to some side, the second player can add $-x$ to the same side, or add $x$ to the other side.So if $sum_{left} + 4.5 * q_{left} == sum_{right} + 4.5 * q_{right}$, the second player will win by mirroring.Else, the first player will choose the larger side, and keeps increasing it, or decreasing the other one. The second player can do no better than mirroring, and will lose.
 » 23 months ago, # |   0 I think in Problem A, the checking condition for min value should be: if(n-cnt<=0)__ then min=0. In the editorial its given, cnt<=0__. Forgive me if I am wrong :)
 » 23 months ago, # |   0 If someone wants DP solution of Div 2 B. The Number of Products.Here it is 85855036
•  » » 21 month(s) ago, # ^ |   0 could you explain your states a bit more and also the transitions , because maybe iam not pretty lucid about them thank you!
 » 20 months ago, # |   0 E is one of the most beautiful solutions, I have ever seen in CP, thanks all
 » 20 months ago, # |   0 For 1215A- Yellow Cards explanation above, It may be the condition for the minimum is "if cnt >= n, then the minimum is 0.In Other cases, the minimum is (n — cnt). "
 » 18 months ago, # | ← Rev. 2 →   0 I am seeing an editorial of problem B after failing to implement it. I am very curious how this kind of implementation comes into your mind.
 » 16 months ago, # |   0 I really didn't get the idea of div2 B problem like what was the thought process while thinking of this solution... can someone help me please