D is solvable without binary search.

If $$$t_j<max(⌈pref_i/i⌉)$$$ then there is no solution.

Otherwise the answer is $$$⌈sum(v)/t_j⌉$$$.

There is my solution 161288493

I solved C without prefix or suffix, but the starting idea is the same.

Solution with explanation why it works: just store the difference between 2 consecutive elements in an array, a1[2] shows the difference between the 1st and the 2nd element, (note: a1[1] is equal to the 1st element).

in a1[1] we keep the number to which all the other elements will be equal to at some point(initially it is the 1st element) and after that I can make all equal to 0 by just a1[1] moves, now let's see how do we keep it. Let's have a look on 2 different cases.

1st case: we have something like 3, 4, 5 etc. In this cases the difference is positive. So I can just simply do ans += a1[i], so at 1st ans will be 1 because I need 1 move to make 4 equal to 3 and also 1 move to make 5 equal to 4, so in total 2 moves to 1st make 5 equal to 4 and then both 4 and 4 equal to 3 in just 1 move that's why this solution is optimal.

2nd case: etc we have something like 5, 4, 3 where we have a1[i] < 0, in this case we can make ans = ans — a1[i], so ans is again going to be positive, but the difference is that now we have to make a1[1] += a1[i], because since we found out that there is a smaller number then we have to make them all equal to a1[1] + a1[i] which makes a1[1] smaller since a1[i] is negative so we still get the number a1[1] which is what we make all the elements equal to.

Code link — 161258270

my solution for problem C but I am not able give some formal proof for this

int n ;
cin>>n;
int mn ;
int pre = 0 , suf= 0 ;
for(int i = 0 ;i<n;i++)
{
int x;
cin>>x;
if(i==0)
{
mn =x;
continue;
}

int val = x-suf;//this is current value after applying suffix operation some number of times
if(val>=mn)
{ 
suf+=(val-mn);
}
else 
{ 
pre+=(mn-val);
mn = val;
}
}
cout<<abs(mn)+pre+suf<<endl;

Problem C. I understand why the given method works, but what is the proof/intuition behind the solution that the given method is minimum number of steps??

How is this pref[i]/i coming, I understand that if all i-1 are filled, then each second i unit of water falls into ith one. But the the i-1 vessels would not be filled simultaneously, may be (i-1) gets filled first and some amount falls from that to ith one, I mean it's vaguely clear how it's coming but i dont feel any strong maths , can anyone please explain

I didn't understand why above method works in problem no. C Can anyone please explain me what is the intuition behind the solution why it works?

If you are/were getting a WA/RE verdict on problems from this contest, you can get the smallest possible counter example for your submission on cfstress.com. To do that, click on the relevant problem's link below, add your submission ID, and edit the table (or edit compressed parameters) to increase/decrease the constraints.

• A: Optimal Path
• B: Palindromic Numbers
• C: Helping the Nature
• D: River Locks
• E: Serega the Pirate
• F: Puzzle

If you are not able to find a counter example even after changing the parameters, reply to this thread (only till the next 7 days), with links to your submission and ticket(s).

In problem no. C Here is my understanding

lets take a example a = [1, -2, 3, -4, 5]

1. now we want convert this array into such an array where every element is same whether it is 0 or not;
2. there is one clue here is that while dec or increasing the diff bw elements will remain same.

so lets prepare an array of diff b where b[i] = a[i] — a[i-1]; b = [1, -3, 5, -7, 9]

lets initialize our res = 0;

now for a[0], a[1] we want to make this 2 element equal. how can we do that ? if -> 1st element is less than 2nd element we will perform dec operation on right side of the array else -> we will perform dec operation on left side of the array

so for a[0], a[1] diff is b[1] = -3 (a[0] < a[1]); so we have to dec left side of array by 3 units ( res += 3 units); now a[0] and a[1] will be equal to -2 so we will update b[0] = -2 because in the end we have to make every element to 0;

and for a[1] and a[2] diff = 5 a[2] > a[1] so we have to dec right side of array by 5 units since diff will remain same bw elements so we don't have to bother for actual number (res += 5 units) since right side will dec so no need to update b[0]

now after perfoming all the operation every element will be equal to updated b[0]; and at last we will do (res += abs(b[0]);

This is My Solution — 161315402

Please add this editorial in the Contest materials.

Could someone figure out why my submission to E is getting TLE on test case 48? I suspect that it has something to do with the sets I'm using inside my loop, but I'm not sure. I've tried pruning it for a while, but to no avail. Help would be appreciated! Submission to E

Problem E.

Example #2:

2 3

1 6 4

3 2 5

$$$6>4$$$ and $$$5>4$$$ , so "4" is bad cell. But if we choose 4 , there isn't any valid way to make the puzzle solvable.Then we will get wrong answer "2" .

Did I get it wrong ? Thx.

_1 _2 |3 _ — сумма от 1 до m — 1 {(m — 1) * m / 2} 4 5 |6 | — сумма кратных на m от m до n * m {(n + 1) * n / 2 * m} 7 8 |9

Bit confused by this statement in the editorial for question div2 C:

Let's calculate the final array using prefix and suffix sums for $$$O(n)$$$. Note that it will consist of the same numbers. Add $$$|x|$$$ to the answer, where $$$x$$$ is the resulting number.

How are the prefix and suffix sums used to compute the final array? And what is $$$|x|$$$?

Thank you.

how to solve problem c

Problem C, here's my code that follows the editorial idea.

Was here anybody who didn't think about the solution of Problem C and solved it with segment tree?

B — Palindromic Numbers — My Solution

Alternate solution for problem F:

Note: I will be referring to a series of swaps leading to a "$$$1$$$" initially present at position $$$(x_1, y_1)$$$ to be finally present in $$$(x_2, y_2)$$$ as movement of that "$$$1$$$" from $$$(x_1, y_1)$$$ to $$$(x_2, y_2)$$$.

Firstly, any series of swaps can be viewed as $$$1$$$'s moving from their initial to final position in the matrix. So for each $$$1$$$ located at $$$(x, y)$$$ in the initial matrix, let $$$(x_{final}, y_{final})$$$ be the position it will move to in the final array, when some optimal series of swaps is peformed.

Observation: In atleast one optimal series of swaps, the following holds: If there are two $$$1$$$'s, located at $$$(x_1, y_1)$$$ and $$$(x_2, y_2)$$$, and $$$y_1 < y_2$$$, then $$$y_{1_{final}} < y_{2_{final}}$$$. Simply put, there is no such column boundary that some $$$1$$$ crosses it from left to right, and another from right to left.

Let's think of an assignment strategy using this observation. Clearly, greedily assigning final position based on column number would work if there was only one row, but here there are two, so we have to somehow take that into account.

Firstly, we can ignore all $$$1$$$'s at whose position there is a $$$1$$$ in the finally wanted matrix.

Let's call cells which have a $$$1$$$ in the initial array as type-$$$1$$$ cells, and cells which have a $$$1$$$ in the final wanted array as type-$$$2$$$ cells. Now, we essentially have to pair each type-$$$1$$$ cell with a unique type-$$$2$$$ cell such that sum of distances between two cells in a pair of all pairs is minimised.

Assignment strategy Iterate over columns, and within a column iterate over rows, and do the following if the cell is of type $$$1$$$ or type $$$2$$$ (otherwise ignore it):

Let's say current cell is of type-$$$1$$$ (symmetric strategy for type-$$$2$$$)

1. If there are any unassigned cells of type-$$$2$$$ which we have already visited before current cell, then pair current cell with any of them (give preference to unassigned cells of type-$$$2$$$ with same row if they exist). Mark current cell as assigned.
2. If there are no such already visited cells, then mark current cell as unassigned.

Implementation: link

Can someone please hep me that why my code for problem B is now working

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