### BledDest's blog

By BledDest, 3 months 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)

1215F - Radio Stations

Editorial
Solution (BledDest)

• +87

 » 3 months ago, # |   +6 F was Perfect!!
 » 3 months 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 $Case_3$.
 » 3 months ago, # |   0 ypa
 » 3 months 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 months ago, # ^ |   0 That's just for input error or testcase error
•  » » 3 months 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 months ago, # |   0 This is my code for D which simulates the game optimally .My Code
•  » » 3 months ago, # ^ |   -9 Nice solution!
 » 3 months ago, # |   0 60628371 Here is my code and I wanna know why the runtime error
•  » » 3 months ago, # ^ |   0 for problem B
•  » » » 3 months 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 months ago, # ^ |   0 thx
•  » » » » 3 months ago, # ^ |   0 That's stupid. xd
 » 3 months 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 months 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.
 » 3 months 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.
•  » » 2 months ago, # ^ |   0 Why you initialized pos=1 ?
•  » » » 2 months 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.
•  » » 6 weeks 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 weeks 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;
•  » » 3 weeks ago, # ^ |   +3 Wow, beautiful solution brother
 » 3 months ago, # | ← Rev. 3 →   0 Sorry, I'm incorrectly understand Um_nik's solution:60611022But it's great)
 » 3 months 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 months ago, # |   +5 In problem B > editorial No. of subsegments should be n*(n+1)/2
 » 3 months 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 months ago, # ^ |   +1 whoa... just briliiant. Thanks for sharing.
•  » » 3 months ago, # ^ |   0 How did you interpret the equation of maxvalue on each side to be (number of ?/2)*9 + (?%2)*9 ??
•  » » » 3 months 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 months 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 months 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 months 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 months ago, # ^ |   +10
 » 3 months 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
•  » » 7 weeks 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 months ago, # |   0 In C, what would the solution look like if the strings were allowed to have characters from a..z
 » 3 months 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 months ago, # ^ |   0 Can u pls explain ur approach in detail??
•  » » » 3 months 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
 » 3 months ago, # |   0 Can anyone explain problem B?
 » 3 months 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 months 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 months ago, # ^ |   0 Sorry, I had messed up the transitions. They're correct now. Thanks for the feedback :D
 » 3 months 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 months 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 months ago, # |   0 Can someone provide an alternative explanation for problem C?
 » 3 months ago, # |   +3 Problem D (again... but with illustrations): Solution: https://codeforces.com/contest/1215/submission/60849161
 » 2 months 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.
•  » » 7 weeks 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.
 » 6 weeks 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.