F was Perfect!!

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 $$$Case_3$$$.

ypa

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 ?

Для задачи B есть алгоритм на порядок проще для понимания. Просто обходим массив слева направо и считаем количество положительных и отрицательных интервалов, заканчивающихся в k. Допустим мы прошли k элементов и на них xp положительных и xo отрицательных интервалов.

Если k+1 элемент больше 0, то он дает еще xp положительных и x0 отрицательных (с каждым из предыдущих, заканчивающихся на k), заканчивающихся в k+1. Ну и + сам по себе элемент может быть положительной подстрокой.

Если k+1 элемент меньше 0, то он меняет знак и дает уже x0 положительных и xp отрицательных + 1 отрицательный.

На каждом шаге цикла плюсуем полученное количество к итоговому. Профит.

Т.е. так: код

This is my code for D which simulates the game optimally .My Code

60628371 Here is my code and I wanna know why the runtime error

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.

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.

Sorry, I'm incorrectly understand Um_nik's solution:

60611022

But it's great)

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!

In problem B > editorial No. of subsegments should be n*(n+1)/2

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)*9

Now do this for 2nd array also.

If maxl == maxr and minl == minr answer is bicarp coz bicarp can adjust accordingly else answer is monocarp

My submission https://codeforces.me/contest/1215/submission/60683863

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.me/contest/1215/submission/60690503 Please point out where I am going wrong.

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).

Code : https://codeforces.me/contest/1215/submission/60696313

Can anyone provide intuitive proof for problem D, mohamedeltair's was nice but i was only able to understand the 1st part of it.

Can anyone provide more intuitive proof for problem D, mohamedeltair's proof was nice but i only got the 1st part of it

In C, what would the solution look like if the strings were allowed to have characters from a..z

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.

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

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.

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?

Problem D (again... but with illustrations): Solution: https://codeforces.me/contest/1215/submission/60849161

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.

Can anyone give a proof for C?

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.

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 :)

E is one of the most beautiful solutions, I have ever seen in CP, thanks all

Can anybody tell me where am I wrong in problem D? My submission 174629795

I have a much simpler solution for problem B which is using dp. First we need to for an length n array(assumen count from 1 to n),we can know the subarray of the last element in this subarray is within 1 to n.what is the meaning?For example A={1,2,3,4,5,6},{1,2,3} is a subarray for A and the position of the last element 3 in the original array A is 3 (count from 1 to n),this is a important nature. Why important?Let consider Si,Si is a subarray of A and the position of last element in Si in A is i,so we from the previous nature we know the number of subarray of A is s1+s2+s3+...+sn,so in otherword,we can know that the last element of positive/negative subarray is also within 1 to n(count from 1 to n) and it sum up is the total number of positive/negative subarray,so we can run up a dp to solve this by using below state equation dp[i][0] represent positive subarray end with i th position(count from 1 to n) element in the original array dp[i][1] represent negative subarray end with i th position(count from 1 to n) elemetn in the original array

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